NCERT Solutions Class 8 Maths Chapter 05 Number Play

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Question. Can every natural number be written as a sum of consecutive numbers?
Answer: Not all numbers can be expressed this way. Many numbers, like 7 = 3 + 4, can be shown as a sum of consecutive numbers. However, some numbers, especially powers of 2 such as 1, 2, 4, 8, and 16, cannot be written in this form. Testing with various examples demonstrates this pattern.
In simple words: Some numbers work (like 7 = 3 + 4), but powers of 2 do not work this way.

Exam Tip: Try adding two or more consecutive numbers to see if they create your starting number. Powers of 2 never work.

Question. Which numbers can be written in more than one way?
Answer: Numbers with more factors usually have more representations. For example, 15 can be expressed as 7 + 8, or 4 + 5 + 6, or 1 + 2 + 3 + 4 + 5. The number of different ways to write a number as a sum of consecutive terms grows with the number of divisors it has.
In simple words: Numbers with many divisors can be shown as consecutive sums in many ways. Numbers with few divisors have fewer ways.

Exam Tip: Count the divisors of a number to predict how many consecutive-sum representations it might have.

Question. All odd numbers can be written as a sum of two consecutive numbers - is this always true?
Answer: Yes, this is always true. Any odd number of the form 2n + 1 can always be written as n + (n + 1). For instance, 9 = 4 + 5 and 21 = 10 + 11. This works for all odd numbers without exception.
In simple words: Take any odd number. Cut it in half and use the two whole numbers around the middle. They always add up to that odd number.

Exam Tip: Remember the pattern: any odd number 2n + 1 equals n + (n + 1). This is a reliable shortcut.

Question. Can we write even numbers as a sum of consecutive numbers?
Answer: Some even numbers work, but others do not. For example, 10 = 1 + 2 + 3 + 4 and 12 = 3 + 4 + 5 are valid. However, 8 cannot be expressed using positive consecutive numbers. The ability depends on which even number you choose and its factors.
In simple words: Some even numbers can be broken into consecutive numbers (like 10), but others cannot (like 8).

Exam Tip: Test even numbers by trying different starting points and lengths for consecutive sequences.

Question. Can 0 be written as a sum of consecutive numbers?
Answer: If you use only positive numbers, the answer is no, because any sum of positive numbers is positive. However, if you permit negative numbers, then yes - for example, -1 + 0 + 1 = 0. This is a fascinating extension when negative integers are allowed.
In simple words: Zero cannot come from adding positive numbers. But if you mix positive and negative numbers, like -1 + 0 + 1, you get zero.

Exam Tip: Always clarify whether the problem allows negative numbers - it changes what is possible.

Question. Take any 4 consecutive numbers. For example, 3, 4, 5, and 6. Place '+' and '-' signs in between the numbers. How many different possibilities exist? Write all of them.
Answer: There are 8 different possibilities. Since there are 3 spaces between 4 numbers where you can place either + or -, that gives 2^3 = 8 combinations:
(i) 3 + 4 + 5 + 6
(ii) 3 + 4 + 5 - 6
(iii) 3 + 4 - 5 + 6
(iv) 3 + 4 - 5 - 6
(v) 3 - 4 + 5 + 6
(vi) 3 - 4 + 5 - 6
(vii) 3 - 4 - 5 + 6
(viii) 3 - 4 - 5 - 6
In simple words: With 4 numbers, you have 3 spaces to fill with + or -. Each space has 2 choices, so 2 × 2 × 2 = 8 total combinations.

Exam Tip: Use powers of 2: with n numbers, there are n-1 sign positions, giving 2^(n-1) combinations.

Question. Evaluate each expression and write the result next to it. Do you notice anything interesting?
Answer:
(i) 3 + 4 + 5 + 6 = 18
(ii) 3 + 4 + 5 - 6 = 6
(iii) 3 + 4 - 5 + 6 = 8
(iv) 3 + 4 - 5 - 6 = -4
(v) 3 - 4 + 5 + 6 = 10
(vi) 3 - 4 + 5 - 6 = -2
(vii) 3 - 4 - 5 + 6 = 0
(viii) 3 - 4 - 5 - 6 = -12
In simple words: All the answers are even numbers. No odd result appears among the 8 combinations.

Exam Tip: Look for patterns like "all even" or "all divisible by something" - these often reveal deeper mathematical rules.

Question. Now, take four other consecutive numbers. Place the '+' and '-' signs as you have done before. Find out the results of each expression. What do you observe?
Answer: Using 7, 8, 9, 10, the eight expressions give:
(i) 7 + 8 + 9 + 10 = 34
(ii) 7 + 8 + 9 - 10 = 14
(iii) 7 + 8 - 9 + 10 = 16
(iv) 7 + 8 - 9 - 10 = -4
(v) 7 - 8 + 9 + 10 = 18
(vi) 7 - 8 + 9 - 10 = -2
(vii) 7 - 8 - 9 + 10 = 0
(viii) 7 - 8 - 9 - 10 = -20
Observation: All results are even numbers. Just as with the previous set, no matter how you arrange the + and - signs, the answer is always even.
In simple words: Again, every answer is even. This is exactly the same pattern as with 3, 4, 5, 6.

Exam Tip: When a pattern repeats across multiple examples, it suggests a general rule that always holds.

Question. Repeat this for one more set of 4 consecutive numbers. Share your findings.
Answer: Using 11, 12, 13, 14, the eight expressions give:
(i) 11 + 12 + 13 + 14 = 50
(ii) 11 + 12 + 13 - 14 = 22
(iii) 11 + 12 - 13 + 14 = 24
(iv) 11 + 12 - 13 - 14 = -4
(v) 11 - 12 + 13 + 14 = 26
(vi) 11 - 12 + 13 - 14 = -2
(vii) 11 - 12 - 13 + 14 = 0
(viii) 11 - 12 - 13 - 14 = -28
Findings: All results are even numbers, matching the earlier sets. Some results repeat across different sets (like -4, -2, and 0). We get a mix of positive, zero, and negative even numbers. This confirms the pattern holds for any four consecutive numbers.
In simple words: Once again, all answers are even. Some answers appear again (like -4 and -2). The pattern keeps repeating.

Exam Tip: Repeated patterns across many examples strongly suggest a universal rule - look for the underlying reason why.

Question. Do these patterns occur no matter which 4 consecutive numbers are chosen? Is there a way to find out through reasoning?
Answer: Yes, these patterns occur for any 4 consecutive numbers. Using algebra, take numbers n, n + 1, n + 2, n + 3. Adding them without regard to signs: n + (n + 1) + (n + 2) + (n + 3) = 4n + 6 = 2(2n + 3), which is always even. Since plus and minus signs only change which parts add versus subtract, but do not alter whether a number is odd or even, every possible expression made from these four numbers must yield an even result. Therefore, the pattern holds universally.
In simple words: Take any four consecutive numbers, add them up (ignoring signs), and you get an even number. Changing + to - does not make odd numbers, so all answers stay even.

Exam Tip: Use algebra to prove patterns you notice - it shows why the pattern always works, not just that it works for your examples.

Question. Is there a way to explain why this happens?
Answer: Yes. Four consecutive numbers always follow the form n, n + 1, n + 2, n + 3. Among these, n and n + 2 have the same parity (both even or both odd), and n + 1 and n + 3 also have the same parity (both even or both odd). When you add or subtract numbers with the same parity, the result is always even: even ± even = even, and odd ± odd = even. Since you get two even results from pairing the numbers this way, and then you combine those two even results (even + even = even), the final answer is always even. This reasoning works regardless of the signs you use.
In simple words: Pair up the four numbers: the first with the third (same parity), and the second with the fourth (same parity). Each pair gives an even answer. Two even answers combine to make an even answer.

Exam Tip: Breaking a problem into pairs or groups with shared properties often reveals why a pattern holds.

Question. Replace any negative sign in the expression a + b - c - d with a positive sign and find the difference between the two numbers.
Answer: Original expression: a + b - c - d. If we change the sign of c from negative to positive, the new expression becomes a + b + c - d. The difference is (a + b + c - d) - (a + b - c - d) = 2c. This shows that changing a minus sign to a plus sign increases the whole expression by twice the value of that number.
In simple words: When you flip a number from minus to plus, the answer jumps up by twice that number. This is because you add the number twice: once you stop subtracting it, and once you start adding it.

Exam Tip: Remember that changing a sign from - to + adds twice the number to your result, not just the number itself.

Question. What do you conclude from this observation?
Answer: When you change a negative sign to a positive sign in an expression like a + b - c - d, the value of the entire expression grows by twice the number whose sign you switched. For example, if you change the sign of c from -c to +c, the new expression is larger by 2c. This is because moving from subtraction to addition requires two steps: stop taking away c, and start adding c. Since changes of signs always shift the value by an even amount (twice something), the parity of the result never changes. Therefore, all expressions formed from the same set of numbers must share the same parity (all even or all odd).
In simple words: Flipping a sign changes the answer by an even amount. Even changes keep the parity the same. So all combinations of signs give answers of the same type (all even or all odd).

Exam Tip: When a quantity changes by an even number, its parity stays the same - this is the key to understanding why patterns repeat.

Question. Is the phenomenon of all the expressions having the same parity limited to taking 4 numbers? What do you think?
Answer: No, this phenomenon applies to any number of terms. The key reason is that changing any sign flips a number from +x to -x (or vice versa), which always changes the total by 2x - an even number. This reasoning holds regardless of how many terms you have. For 3 numbers, 5 numbers, 10 numbers, or 100 numbers, the same logic applies: since changing a sign always shifts the result by an even amount, the parity can never change. All expressions made from the same set of numbers must have the same parity. This is a universal principle in number theory.
In simple words: Whether you have 3, 4, 5, or 100 numbers, flipping a sign always changes your answer by an even amount. So the parity always stays the same.

Exam Tip: Look for principles that work in general, not just for specific cases - they often reveal deeper mathematical truths.

Question. Using our understanding of how parity behaves under different operations, identify which of the following algebraic expressions give an even number for any integer values for the letter-numbers.
Answer:
(i) 2a + 2b: Since 2a is always even and 2b is always even, their sum is even. Always even.
(ii) 3g + 5h: The value depends on whether g and h are even or odd. If both are odd, the sum is even; if both are even, the sum is even; but if one is odd and the other is even, the sum is odd. Not always even.
(iii) 4m + 2n: Since 4m is even and 2n is even, the sum is even. Always even.
(iv) 2u - 4v: Both 2u and 4v are even, so their difference is even. Always even.
(v) 13k - 5k: This simplifies to 8k, which is always even. Always even.
(vi) 6m - 3n: Since 6m is always even but 3n may be odd or even, the difference can be odd. Not always even.
(vii) x² + 2: If x is even, x² is even, making the sum even. If x is odd, x² is odd, making the sum odd. Not always even.
(viii) b² + 1: If b is even, b² is even, making the sum odd. If b is odd, b² is odd, making the sum even. Not always even.
(ix) 4k × 3j: Since 4k is always even, any product involving it is even. Always even.
In simple words: Expressions that always multiply by 2 or 4, or add/subtract only even numbers, always give even answers. Expressions mixing odd and even terms do not always work.

Exam Tip: To check if an expression is always even, trace through the parity step-by-step using the rules: even + even = even, odd + odd = even, but odd + even = odd.

Question. Similarly, determine and explain which of the other expressions always give even numbers. Write a couple of examples and non-examples, as appropriate, for each expression. Write a few algebraic expressions which always give an even number.
Answer:
(i) 6p + 4q: Both 6p and 4q are always even. Since even + even = even, the sum is always even.
Examples: (p = 1, q = 2) gives 6 + 8 = 14; (p = -3, q = 5) gives -18 + 20 = 2. Both are even.
(ii) 10x - 2y: Both 10x and 2y are always even. Since even - even = even, the difference is always even.
Examples: (x = 3, y = 4) gives 30 - 8 = 22; (x = -2, y = 7) gives -20 - 14 = -34. Both are even.
(iii) 8a × b: Since 8a is always even, multiplying by anything yields an even result.
Examples: (a = 2, b = 5) gives 16 × 5 = 80; (a = -1, b = -3) gives -8 × -3 = 24. Both are even.
(iv) 2(m² + n²): Any integer squared is an integer. Multiplying any integer by 2 gives an even result.
Examples: (m = 3, n = 4) gives 2(9 + 16) = 50; (m = -2, n = -1) gives 2(4 + 1) = 10. Both are even.
(v) 4(p + q + r): Any multiple of 4 is even.
Examples: (p = 1, q = 1, r = 1) gives 4(3) = 12; (p = -4, q = 2, r = 0) gives 4(-2) = -8. Both are even.
(vi) 12k - 8m: Both are multiples of 4, hence even. Their difference is even.
Examples: (k = 1, m = 1) gives 12 - 8 = 4; (k = -3, m = 2) gives -36 - 16 = -52. Both are even.
(vii) 2(a + b + c + d): Multiplying any sum by 2 gives an even result.
Examples: (a = 2, b = 5, c = -3, d = 1) gives 2(5) = 10; (a = 0, b = 0, c = 0, d = 7) gives 2(7) = 14. Both are even.
(viii) (x - y)² - (x + y)²: Expanding gives (x² - 2xy + y²) - (x² + 2xy + y²) = -4xy, which is always even.
Examples: (x = 3, y = 1) gives -4(3)(1) = -12; (x = -2, y = 5) gives -4(-2)(5) = 40. Both are even.
In simple words: Multiply by 2 or 4, and you always get even. Add or subtract even numbers, and you stay even. Any expression with a factor of 2 is always even.

Exam Tip: Look for factors of 2 or 4 in the expression - if they appear, the result is guaranteed to be even.

Question. Take a pair of even numbers. Add them. Is the sum divisible by 4?
Answer: No, the sum of two even numbers is not always divisible by 4. Whether the sum works depends on which even numbers you pick. Every even number falls into one of two types: multiples of 4 (like 4, 8, 12, 16), which leave remainder 0 when divided by 4, or even numbers that are not multiples of 4 (like 2, 6, 10, 14), which leave remainder 2 when divided by 4. When both numbers are multiples of 4, the sum is divisible by 4 (example: 8 + 12 = 20, divisible by 4). When both are even but not multiples of 4, the sum is still divisible by 4 (example: 6 + 10 = 16, divisible by 4). However, when one is a multiple of 4 and the other is not, the sum is not divisible by 4 (example: 4 + 6 = 10, not divisible by 4).
In simple words: Mixing the two types of even numbers breaks divisibility by 4. Match the types (both multiples of 4, or both non-multiples), and divisibility by 4 works.

Exam Tip: Even numbers come in two types: multiples of 4, and non-multiples of 4. Understanding this distinction is key to predicting divisibility.

Question. When will two even numbers add up to give a multiple of 4?
Answer: Two even numbers add up to a multiple of 4 only when they belong to the same type. Both must be multiples of 4 (such as 4, 8, 12, 16), and then their sum is divisible by 4 (example: 8 + 12 = 20). Alternatively, both must be even but not multiples of 4 (these have the form 4k + 2, such as 2, 6, 10, 14), and their sum is also divisible by 4 (example: 6 + 10 = 16). If you mix one from each type, divisibility by 4 fails. This is because two numbers of form 4k + 2 add to 4k + 2 + 4m + 2 = 4(k + m + 1), which is divisible by 4.
In simple words: Pick two numbers from the same even family - either both from multiples of 4, or both from numbers like 2, 6, 10. Then their sum is always divisible by 4.

Exam Tip: Classify even numbers by their type (form 4k vs. form 4k + 2) to predict divisibility results.

Question. Look at the following expressions and the visualisation. Write the corresponding explanation and examples.
Answer: [This question appears in the source with a visual table that demonstrates how 4p + (4q + 2) combines to form 4(p + q) + 2, with multiple visual representations. The text shows this algebraically and with concrete examples.]
Expression: 4p + (4q + 2) = 4p + 4q + 2 = 4(p + q) + 2
This shows that adding a multiple of 4 to a number of form 4q + 2 always yields a number of the form 4r + 2 (remainder 2 when divided by 4).
Examples: 4 + 6 = 10 (forms 4(1) and 4(1) + 2, giving 4(2) + 2 = 10); 8 + 10 = 18 (forms 4(2) and 4(2) + 2, giving 4(4) + 2 = 18); 8 + 6 = 14 (forms 4(2) and 4(1) + 2, giving 4(3) + 2 = 14).
In simple words: When you add a multiple of 4 to a number like 4q + 2, the answer is always of the form 4r + 2 - it leaves remainder 2.

Exam Tip: Express numbers in the form 4k or 4k + 2 to predict sums and divisibility patterns.

Question. We examine different statements about factors and multiples and determine whether a statement is 'Always True', 'Sometimes True', or 'Never True'. 1. "The product of a multiple of 6 and a multiple of 3 is a multiple of 9."
Answer: This statement is always true. A multiple of 6 has the form 6a = 2 × 3 × a. A multiple of 3 has the form 3b. Their product is 6a × 3b = 18ab = 9 × 2ab, which is always a multiple of 9. Example: 6 × 3 = 18, and 18 is a multiple of 9. No counterexample exists because the product always contains 9 as a factor.
In simple words: A multiple of 6 already has a 3 in it. Multiply by another multiple of 3, and you have at least 3 × 3 = 9.

Exam Tip: Write multiples in factored form to see what factors they contain and predict products.

Question. "The sum of three consecutive even numbers is divisible by 6."
Answer: This statement is always true. Let the three consecutive even numbers be 2n, 2n + 2, 2n + 4. Their sum is 2n + (2n + 2) + (2n + 4) = 6n + 6 = 6(n + 1), which is clearly divisible by 6. Example: 4 + 6 + 8 = 18, and 18 is divisible by 6. The sum always contains 6 as a factor, so divisibility never fails.
In simple words: Three consecutive even numbers always add to a multiple of 6 because their sum follows the pattern 6(n + 1).

Exam Tip: Use algebraic form to find the common factor: if you find a factor in the formula, it divides all instances.

Question. "8(7b - 3) - 4(11b + 1) is a multiple of 12."
Answer: This statement is sometimes true. Simplifying: 8(7b - 3) - 4(11b + 1) = 56b - 24 - 44b - 4 = 12b - 28 = 4(3b - 7). For this to be divisible by 12, the quantity (3b - 7) must be divisible by 3. This happens only for certain values of b, not all. For example, when b = 3: 4(9 - 7) = 4(2) = 8, which is not divisible by 12. When b = 4: 4(12 - 7) = 4(5) = 20, also not divisible by 12. When b = 10: 4(30 - 7) = 4(23), still not divisible by 12. The condition depends on b, so the statement is sometimes true.
In simple words: The expression simplifies to 4(3b - 7). For this to be divisible by 12, you need 3b - 7 to be divisible by 3, which is not always the case.

Exam Tip: When a simplified expression has a parameter, check whether the remaining factor always satisfies the divisibility condition.

Question. 1. If 8 exactly divides two numbers separately, it must exactly divide their sum.
Answer: This statement is always true. If 8 divides both numbers, then each is a multiple of 8. Let them be 8a and 8b. Their sum is 8a + 8b = 8(a + b), which is clearly a multiple of 8. Examples: 16 + 24 = 40, which is divisible by 8; 8 + 56 = 64, divisible by 8; 32 + (-8) = 24, divisible by 8. Since the sum is factored as 8 times an integer, divisibility never fails. Therefore, if two numbers are each divisible by 8, their sum is always divisible by 8.
In simple words: If both numbers are multiples of 8, their sum is also a multiple of 8 because you can factor out the 8.

Exam Tip: Use the factored form to show divisibility: if both numbers contain the factor, their sum does too.

Question. 2. If a number is divisible by 8, then 8 also divides any two numbers (separately) that add up to the number.
Answer: This statement is sometimes true. Suppose the number N is divisible by 8, so N = 8k. The claim is that if a + b = 8k, then both a and b must be divisible by 8. This is false. Counterexample: N = 24 (divisible by 8). Choose a = 10 and b = 14. Then 10 + 14 = 24, but neither 10 nor 14 is divisible by 8. Another counterexample: N = 32 (divisible by 8). Choose a = 20 and b = 12. Then 20 + 12 = 32, but neither 20 nor 12 is divisible by 8. A case where it is true: a = 16 and b = 16. Then 16 + 16 = 32 (divisible by 8), and both 16 and 16 are divisible by 8. Since the statement holds only for special pairs and not for all pairs, it is sometimes true.
In simple words: A number divisible by 8 can be split into two parts in many ways. Only some of those splits have both parts divisible by 8.

Exam Tip: Divisibility of a sum does not force divisibility of the individual terms - they can have different remainders that cancel.

Question. 3. If a number is divisible by 7, then all multiples of that number will be divisible by 7.
Answer: This statement is always true. If a number n is divisible by 7, then n = 7k for some integer k. Now take any multiple of n: m = n × t = (7k) × t = 7(kt). Since m is expressed as 7 times an integer, it is always divisible by 7. Examples: If n = 14 (divisible by 7), then its multiples 14, 28, 42, 56 are all divisible by 7. If n = 21, then 21, 42, 63, 84 are all divisible by 7. If n = -7, then -7, -14, -21 are all divisible by 7. There is no way to construct a multiple of a number divisible by 7 that is not divisible by 7. Therefore, if a number is divisible by 7, every multiple of that number is also divisible by 7.
In simple words: If you multiply a multiple of 7 by something, you still have a multiple of 7, because the 7 is still there.

Exam Tip: Multiples of multiples stay multiples - the factor is preserved through multiplication.

Question. 4. If a number is divisible by 12, then the number is also divisible by all the factors of 12.
Answer: This statement is always true. The factors of 12 are 1, 2, 3, 4, 6, and 12. If a number N is divisible by 12, then N = 12k for some integer k. Since 12 = 2² × 3, the number 12k contains all the prime factors needed. It can be divided by 2, by 3, by 4 (= 2²), by 6 (= 2 × 3), and by 12. Examples: 36 is divisible by 12, and it is also divisible by 1, 2, 3, 4, 6, and 12. Similarly, 60 is divisible by 12 and by all its factors. The statement always holds because divisibility by a composite number guarantees divisibility by its factors. Therefore, this statement is always true.
In simple words: A multiple of 12 contains all the building blocks (factors) of 12, so it is divisible by each of them.

Exam Tip: A number divisible by a composite number is automatically divisible by all its factors.

Question. 5. If a number is divisible by 7, then it is also divisible by any multiple of 7.
Answer: This statement is sometimes true. Let N = 7k be the number divisible by 7, and let M = 7m be a multiple of 7. The claim is that N must be divisible by M. This is not always true. Counterexample: N = 21 (divisible by 7) and M = 14 (a multiple of 7). Then 21 ÷ 14 is not an integer, so 21 is not divisible by 14. Another counterexample: N = 35 (divisible by 7) and M = 21 (a multiple of 7). Then 35 ÷ 21 is not an integer. However, there are cases where it works: N = 42 and M = 14 both work. Then 42 ÷ 14 = 3 (an integer). Since the statement holds only for some pairs and not all, the statement is sometimes true, not always.
In simple words: A multiple of 7 may or may not be divisible by other multiples of 7. It depends on the specific numbers.

Exam Tip: Being divisible by the same number does not mean one must divide the other - you need additional relationships.

Question. 6. If a number is divisible by both 9 and 4, it must be divisible by 36.
Answer: This statement is always true. If a number N is divisible by 9, then N = 9a. If N is also divisible by 4, then N = 4b. For a number to be divisible by 36, it must contain the prime factors 9 = 3² and 4 = 2². The least common multiple (LCM) of 9 and 4 is LCM(9, 4) = 36, since 9 and 4 share no common prime factors. Any number divisible by both 9 and 4 must be a multiple of 36. Therefore, if a number is divisible by both 9 and 4, it is always divisible by 36. This statement is always true.
In simple words: A number divisible by both 9 and 4 must be divisible by their LCM, which is 36.

Exam Tip: When a number is divisible by multiple coprime numbers, it is divisible by their product (or LCM if not coprime).

Question. 7. If a number is divisible by both 6 and 4, it must be divisible by 24.
Answer: This statement is sometimes true. If a number is divisible by 6, it contains factors 2 × 3. If divisible by 4, it contains factors 2 × 2. Together, the number must contain 2 × 2 × 3 = 12. Therefore, any number divisible by both 6 and 4 is divisible by 12, not necessarily 24. Counterexample: N = 12. Then 12 ÷ 6 = 2 (divisible) and 12 ÷ 4 = 3 (divisible), but 12 ÷ 24 is not an integer. Another counterexample: N = 36. Then 36 ÷ 6 = 6 and 36 ÷ 4 = 9, but 36 ÷ 24 is not an integer. Case where it works: N = 48. Then 48 ÷ 6 = 8 and 48 ÷ 4 = 12, and 48 ÷ 24 = 2. Since the statement is true for some numbers and false for others, it is sometimes true.
In simple words: Numbers divisible by both 6 and 4 are always divisible by 12, but not always by 24. Some are, some are not.

Exam Tip: Use prime factorization to find the guaranteed common divisor - in this case, 6 and 4 guarantee divisibility by 12, but not 24.

Question. 8. When you add an odd number to an even number we get a multiple of 6.
Answer: This statement is never true. Adding an odd number to an even number always yields an odd result: odd + even = odd. However, all multiples of 6 (6, 12, 18, 24, 30, ...) are even. Since an odd number cannot be even, the sum of an odd and an even number can never be a multiple of 6. Example: 5 (odd) + 8 (even) = 13 (odd, not divisible by 6). Another example: 7 (odd) + 4 (even) = 11 (odd, not divisible by 6). No matter which odd and even numbers you choose, the result is always odd and therefore never a multiple of the even number 6. This statement is never true.
In simple words: An odd number plus an even number is always odd. But all multiples of 6 are even. Odd can never equal even.

Exam Tip: Check the parity of the result first - if it does not match the parity of multiples of the target number, the statement is never true.

Question. Find a number that has a remainder of 3 when divided by 5. Write more such numbers. Which algebraic expression(s) capture all such numbers?
Answer: Numbers leaving remainder 3 when divided by 5 include 3, 8, 13, 18, 23, 28, 33, and so on. These increase by 5 each time, so all are of the form 5k + 3. Similarly, 5k - 2 also works because 5k - 2 = 5(k - 1) + 3, which gives remainder 3. Checking the options: (i) 3k + 5 does not always give remainder 3; (ii) 3k - 5 does not; (iii) 3k/5 is not an integer; (iv) 5k + 3 always gives remainder 3; (v) 5k - 2 always gives remainder 3; (vi) 5k - 3 gives remainder 2, not 3. The correct expressions are 5k + 3 and 5k - 2.
In simple words: Add 3 to any multiple of 5, and you get remainder 3. Subtract 2 from a multiple of 5, and you also get remainder 3 (because -2 and +3 are the same gap from the next multiple).

Exam Tip: Remainders can be expressed as 5k + 3 or equivalently 5(k ± 1) ± adjusted constant - verify both forms.

Question. Let us consider another expression, 5k - 2, and see the values it takes for different values of k. For k: 1, 2, 3, 4, 5, the expression 5k - 2 gives: 3, 8, 13, 18, 23. Are there other expressions that generate numbers that are 3 more than a multiple of 5?
Answer: Yes, several expressions generate numbers of the form (multiple of 5) + 3. The expression 5k + 3 works: for k = 1, 2, 3, 4, 5, it gives 8, 13, 18, 23, 28 - all leave remainder 3. The expression 5(k + 1) - 2 simplifies to 5k + 3, which also works. The expression 5(k - 1) + 3 simplifies to 5k - 2, which also works. Any linear shift of k combined with 5 times a number will produce the same pattern. As long as the constant term leaves remainder 3 when divided by 5 (like 3, 8, 13, -2, -7, etc.), the expression will capture all such numbers. Examples of other valid expressions include 5k + 8, 5k - 7, 5(k + 2) - 7, etc.
In simple words: Any expression of the form 5k + r, where r leaves remainder 3 when divided by 5 (like r = 3, 8, -2, -7), will generate numbers with remainder 3.

Exam Tip: Equivalent expressions exist when the constant term and variable term are adjusted - verify by checking remainders modulo 5.

Question 1. The sum of four consecutive numbers is 34. What are these numbers?
Answer: Let the four consecutive numbers be n, n + 1, n + 2, n + 3. Their sum is 34, so n + (n + 1) + (n + 2) + (n + 3) = 34. Simplifying: 4n + 6 = 34, which gives 4n = 28, so n = 7. The four consecutive numbers are 7, 8, 9, 10.
In simple words: Use n for the first number. Then add n, (n + 1), (n + 2), and (n + 3) to get 34. Solve to find n = 7, and the numbers are 7 through 10.

Exam Tip: For consecutive numbers, always use the form n, n+1, n+2, ... to make the algebra straightforward.

Question 2. Suppose p is the greatest of five consecutive numbers. Describe the other four numbers in terms of p.
Answer: Since p is the greatest, the other four numbers are immediately below it. Reading backward from p: the next lower is p - 1, then p - 2, then p - 3, then p - 4. So the four numbers are p - 1, p - 2, p - 3, p - 4.
In simple words: If p is the largest, count backward: p - 1, p - 2, p - 3, p - 4 are the other four.

Exam Tip: Work backward from the largest to express all terms in relation to it - this avoids introducing extra variables.

Question 3(i). For each statement below, determine whether it is always true, sometimes true, or never true. Explain your answer. Mention examples and non-examples as appropriate. Justify your claim using algebra. The sum of two even numbers is a multiple of 3.
Answer: This statement is sometimes true. Let the even numbers be 2a and 2b. Their sum is 2a + 2b = 2(a + b). The sum is always even, but whether it is divisible by 3 depends on the values of a and b. Example where it is true: 6 + 12 = 18, which is divisible by 3. Both 6 and 12 are even, and their sum is a multiple of 3. Example where it is false: 2 + 4 = 6... wait, 6 is divisible by 3. Try 2 + 8 = 10, which is not divisible by 3. Both are even, but the sum is not divisible by 3. Since the statement is true for some pairs of even numbers and false for others, it is sometimes true.
In simple words: Some even numbers add to multiples of 3 (like 6 + 12 = 18), but others do not (like 2 + 8 = 10). Divisibility by 3 depends on the numbers, not on being even.

Exam Tip: Divisibility by 3 and being even are independent properties - check both before concluding.

Question 1. If a number is divisible by both 4 and 6, is it always divisible by 24? Justify your answer with examples.
Answer: No, a number divisible by both 4 and 6 is not always divisible by 24. To see why, we need to find the least common multiple of 4 and 6. The LCM of 4 and 6 is 12, not 24. So any number divisible by both 4 and 6 must be a multiple of 12, but it does not have to be a multiple of 24. For example, 12 is divisible by both 4 and 6, but when we divide 12 by 24, we get a remainder. Similarly, 36 is divisible by both 4 and 6, but 36 is not divisible by 24 (since 36 ÷ 24 gives 1 with remainder 12). However, some numbers like 48 and 72 that are divisible by both 4 and 6 are also divisible by 24. So the answer depends on the specific number.
In simple words: Just because a number divides evenly by 4 and 6 does not mean it will divide evenly by 24. You need a common multiple of both 4 and 6, which is 12, not 24.

Exam Tip: Remember that divisibility by two numbers only guarantees divisibility by their LCM, not by their product. Always check LCM values carefully in such problems.

Question 2. (i) If a number is divisible by 4, then it is also divisible by 2 - is this always true, sometimes true, or never true?
Answer: This statement is always true. When a number is divisible by 4, it means the number can be written as 4k for some whole number k. We can rewrite this as 4k = 2(2k). Since 2k is also a whole number, the number is divisible by 2. In other words, every multiple of 4 is also a multiple of 2. For example, 4, 8, 12, 16, 20 are all divisible by 4, and each one divides evenly by 2 as well.
In simple words: Any number that 4 goes into will also be divisible by 2, because 4 itself contains 2 as a factor.

Exam Tip: When checking divisibility by a composite number, remember that divisibility follows the factor chain - if a number is divisible by a larger number, it is automatically divisible by all of that number's factors.

Question 2. (ii) If a number is not divisible by 18, then it is also not divisible by 9 - is this always true, sometimes true, or never true?
Answer: This statement is never true. A number can fail to be divisible by 18 but still be divisible by 9. For a number to be divisible by 18, it must be divisible by both 2 and 9. Therefore, a number that is divisible by 9 but not by 2 (that is, an odd number) will not be divisible by 18. For instance, 27, 45, 63, and 81 are all divisible by 9 but not by 18 because they are odd numbers. So the statement is never true because many numbers exist that are divisible by 9 but not by 18.
In simple words: A number can be divisible by 9 without being divisible by 18. This happens when the number is odd, because 18 requires the number to be even.

Exam Tip: Always break composite numbers into their prime factors (18 = 2 × 9). A number might satisfy one factor requirement but not the other.

Question 2. (iii) If two numbers are not divisible by 6, then their sum is not divisible by 6 - is this always true, sometimes true, or never true?
Answer: This statement is sometimes true and sometimes false. A number divisible by 6 must be divisible by both 2 and 3. Even if two numbers are not divisible by 6, they can still add up to a multiple of 6. Consider the example where 4 is not divisible by 6 and 2 is also not divisible by 6, but their sum 4 + 2 = 6 is divisible by 6. On the other hand, when we add 5 + 7 = 12, neither 5 nor 7 is divisible by 6, and their sum 12 is not divisible by 6. So it is sometimes true that their sum is not divisible by 6, and sometimes false.
In simple words: Two numbers that do not divide by 6 might add up to make a number that does divide by 6, or they might not. It depends on what the two numbers are.

Exam Tip: When evaluating statements about sums, always test with concrete examples to see if the claim holds in all cases or only some.

Question 2. (iv) The sum of a multiple of 6 and a multiple of 9 is a multiple of 3 - is this always true, sometimes true, or never true?
Answer: This statement is always true. Let 6a represent a multiple of 6 and 9b represent a multiple of 9. We can rewrite these as 6a = 3(2a) and 9b = 3(3b). When we add them together: 6a + 9b = 3(2a) + 3(3b) = 3(2a + 3b). This expression is clearly a multiple of 3 because it has 3 as a factor. We can verify this with examples: 12 + 18 = 30, which is divisible by 3. Also, 6 + 45 = 51, which is divisible by 3. So the statement is always true.
In simple words: Both 6 and 9 are multiples of 3, so when you add any multiple of 6 to any multiple of 9, the result will always be a multiple of 3.

Exam Tip: Factor out common divisors algebraically to prove divisibility - this is more reliable than testing individual examples alone.

Question 2. (v) The sum of a multiple of 6 and a multiple of 3 is a multiple of 9 - is this always true, sometimes true, or never true?
Answer: This statement is sometimes true. Let 6a represent a multiple of 6 and 3b represent a multiple of 3. Their sum is 6a + 3b = 3(2a + b). This sum is always divisible by 3, but it is divisible by 9 only when (2a + b) is divisible by 3. That does not always happen. When we test 6 + 3 = 9, which is a multiple of 9, the statement holds true. But when we test 6 + 6 = 12, which is not a multiple of 9, the statement fails. So the statement is sometimes true and not always true.
In simple words: Adding a multiple of 6 to a multiple of 3 always gives you a multiple of 3, but only sometimes gives you a multiple of 9. It depends on the exact values you choose.

Exam Tip: Distinguish between "always," "sometimes," and "never" by looking for both examples that work and counterexamples that fail.

Question 3. Find a few numbers that leave a remainder of 2 when divided by 3 and a remainder of 2 when divided by 4. Write an algebraic expression to describe all such numbers.
Answer: We need to find numbers satisfying both conditions. Numbers that leave remainder 2 when divided by 3 have the form 3k + 2. Numbers that leave remainder 2 when divided by 4 have the form 4m + 2. Setting these equal: 3k + 2 = 4m + 2, which simplifies to 3k = 4m. Since 3k must be a multiple of 4 and 4m must be a multiple of 3, we need the LCM of 3 and 4, which is 12. So 3k = 12t, giving k = 4t, and 4m = 12t, giving m = 3t. Now the number becomes 3k + 2 = 3(4t) + 2 = 12t + 2. Testing values: when t = 0, we get 2; when t = 1, we get 14; when t = 2, we get 26; when t = 3, we get 38; when t = 4, we get 50; when t = 5, we get 62. So the numbers are 2, 14, 26, 38, 50, 62, and all such numbers can be described by the algebraic expression 12k + 2.
In simple words: Numbers that work are 2, 14, 26, 38, 50, 62, and so on. They follow a pattern where each number is 2 more than a multiple of 12.

Exam Tip: When a number must satisfy multiple remainder conditions, find the LCM of the divisors to create a unified formula covering all solutions.

Question 4. "I hold some pebbles, not too many, When I group them in 3's, one stays with me. Try pairing them up - it simply won't do, A stubborn odd pebble remains in my view. Group them by 5, yet one's still around, But grouping by seven, perfection is found. More than one hundred would be far too bold, Can you tell me the number of pebbles I hold?"
Answer: Let the number of pebbles be N. From the poem, we know: when divided by 2, remainder is 1; when divided by 3, remainder is 1; when divided by 5, remainder is 1. The LCM of 2, 3, and 5 is 30. So any number meeting all three conditions can be written as N = 30k + 1 for some whole number k. We also know that when divided by 7, the remainder is 0 (grouping by seven gives perfection, meaning it divides evenly). So N is a multiple of 7. The largest multiple of 7 less than 100 is 98. Testing 98: 98 ÷ 3 leaves remainder 2, 98 ÷ 2 leaves remainder 0, so 98 does not meet our conditions. Testing 91: 91 ÷ 3 = 30 remainder 1; 91 ÷ 2 = 45 remainder 1; 91 ÷ 5 = 18 remainder 1; 91 ÷ 7 = 13 remainder 0; and 91 < 100. Therefore, there are 91 pebbles.
In simple words: We need a number that divides by 2, 3, and 5 with remainder 1 each time, and divides by 7 with no remainder. The answer is 91.

Exam Tip: When multiple conditions involve remainders, use the LCM method to consolidate them, then apply additional constraints (like divisibility by another number) to narrow down the final answer.

Question 5. Tathagat has written several numbers that leave a remainder of 2 when divided by 6. He claims, "If you add any three such numbers, the sum will always be a multiple of 6." Is Tathagat's claim true?
Answer: Yes, Tathagat's claim is true. Numbers that leave remainder 2 when divided by 6 take the form 6k + 2. Examples include 2, 8, 14, 20, 26, 32, 38, and so on. If we pick any three such numbers, say 6a + 2, 6b + 2, and 6c + 2, their sum is (6a + 2) + (6b + 2) + (6c + 2) = 6(a + b + c) + 6 = 6(a + b + c + 1). Since this sum has 6 as a factor, it is always divisible by 6 with no remainder. Therefore, whenever you add three numbers of the form 6k + 2, the result will always be a multiple of 6.
In simple words: If three numbers each leave remainder 2 when divided by 6, adding them gives three remainder 2's, which makes 6 extra, so the total is a perfect multiple of 6.

Exam Tip: Use algebraic expressions to prove general statements about divisibility rather than relying on specific examples - this demonstrates complete understanding.

Question 6. When divided by 7, the number 661 leaves a remainder of 3, and 4779 leaves a remainder of 5. Without calculating, can you say what remainders the following expressions will leave when divided by 7? Show the solution both algebraically and visually. (i) 4779 + 661 (ii) 4779 - 661
Answer: (i) When 4779 is divided by 7, remainder is 5. When 661 is divided by 7, remainder is 3. When we add the two numbers, we add the remainders: 5 + 3 = 8. But 8 is greater than 7, so we divide 8 by 7 to find the final remainder: 8 ÷ 7 gives remainder 1. Therefore, 4779 + 661 leaves remainder 1 when divided by 7.
(ii) When 4779 is divided by 7, remainder is 5. When 661 is divided by 7, remainder is 3. When we subtract one number from the other, we subtract the remainders: 5 - 3 = 2. Therefore, 4779 - 661 leaves remainder 2 when divided by 7.
In simple words: When adding numbers, add their remainders and then simplify if needed. When subtracting, subtract the remainders. This works because the remainders behave the same way as the original numbers.

Exam Tip: The remainder properties for addition and subtraction mirror the operations themselves - if you know the individual remainders, you can find the remainder of the sum or difference without doing the full calculation.

Question 7. Find a number that leaves a remainder of 2 when divided by 3, a remainder of 3 when divided by 4, and a remainder of 4 when divided by 5. What is the smallest such number? Can you give a simple explanation of why it is the smallest?
Answer: We are given that the number leaves remainder 2 when divided by 3, remainder 3 when divided by 4, and remainder 4 when divided by 5. Notice that each remainder is exactly one less than its divisor: 2 = 3 - 1, 3 = 4 - 1, and 4 = 5 - 1. This means N + 1 must be divisible by 3, 4, and 5. The LCM of 3, 4, and 5 is 60. So N + 1 must be a multiple of 60. The candidates are 60, 120, 180, and so on. Subtracting 1 to find N, we get: N = 59 (from 60 - 1), N = 119 (from 120 - 1), N = 179 (from 180 - 1). The smallest number is 59. We can verify: 59 ÷ 3 = 19 remainder 2; 59 ÷ 4 = 14 remainder 3; 59 ÷ 5 = 11 remainder 4. All conditions are satisfied. It is the smallest because 60 is the smallest positive multiple of 3, 4, and 5, and subtracting 1 gives the smallest N satisfying all remainder conditions.
In simple words: Each remainder is one less than the divisor, so the number is one less than a common multiple of 3, 4, and 5. The smallest such common multiple is 60, so the answer is 59.

Exam Tip: When remainders follow the pattern "divisor minus 1," recognize this as N = (LCM) - 1. This insight immediately simplifies the problem.

Divisibility Shortcuts for 5, 2, 4, and 8

Question 8. Similarly, explain using algebra why the divisibility shortcuts for 5, 2, 4, and 8 work.
Answer: Let the number be expressed in place value form: 1000d + 100c + 10b + a, where a is the units digit, b is the tens digit, c is the hundreds digit, and d is the thousands digit. All place-value terms except a contain a factor of 10.

Divisibility by 5: A number is divisible by 5 if its units digit is 0 or 5. Using algebra: N = 10b + a (ignoring higher place values). Rewriting, N = 5(2b) + a. This is divisible by 5 only if a is also divisible by 5, which means a = 0 or a = 5. So a number is divisible by 5 if and only if its units digit is 0 or 5.

Divisibility by 2: N = 10b + a. Since 10b is always even, the even/odd nature of N depends entirely on a. If a is even, N is even and divisible by 2. If a is odd, N is odd and not divisible by 2. So a number is divisible by 2 if and only if its units digit is even.

Divisibility by 4: A number is divisible by 4 if its last two digits form a number divisible by 4. Using algebra: N = 100c + 1000d + ... + (10b + a). The terms 100c, 1000d, and so on are all multiples of 100, hence multiples of 4. Only the part 10b + a matters. If 10b + a is divisible by 4, then the entire number is divisible by 4. So a number is divisible by 4 if the number formed by its last two digits is divisible by 4.

Divisibility by 8: A number is divisible by 8 if its last three digits form a number divisible by 8. Using algebra: N = 1000d + 100c + 10b + a. The term 1000d is divisible by 8 regardless of d. The term 100c might not be divisible by 8 (since 100 ÷ 8 is 12.5). We must rewrite N as N = 1000k + (100c + 10b + a), where 1000k is divisible by 8 for any integer k. The divisibility by 8 depends only on 100c + 10b + a, which is precisely the number formed by the last three digits. So a number is divisible by 8 if the number formed by its last three digits is divisible by 8.
In simple words: All the place-value parts except the ones digit are multiples of 10, so only the last digit (or last few digits) matters for divisibility checks.

Exam Tip: Always express numbers in place-value form (1000d + 100c + 10b + a) when proving divisibility shortcuts - this reveals which digits are actually relevant for each divisor.

Divisibility by 9

Question 9. Can you say, without actually calculating, which of these numbers are divisible by 9: 999, 909, 900, 90, 990?
Answer: A number is divisible by 9 if the sum of its digits is divisible by 9. Let me check each number:

999: Digit sum = 9 + 9 + 9 = 27. Since 27 is divisible by 9, the number 999 is divisible by 9.

909: Digit sum = 9 + 0 + 9 = 18. Since 18 is divisible by 9, the number 909 is divisible by 9.

900: Digit sum = 9 + 0 + 0 = 9. Since 9 is divisible by 9, the number 900 is divisible by 9.

90: Digit sum = 9 + 0 = 9. Since 9 is divisible by 9, the number 90 is divisible by 9.

990: Digit sum = 9 + 9 + 0 = 18. Since 18 is divisible by 9, the number 990 is divisible by 9.

All five numbers are divisible by 9.
In simple words: Add up all the digits in each number. If the total can be divided by 9 with no remainder, the whole number can too.

Exam Tip: The divisibility rule for 9 is faster than doing the actual division - always use the digit sum method to save time on exams.

Question 10. Can we say that any number made up of only the digits '0' and '9', in any order, will always be divisible by 9?
Answer: Yes, this is always true. A number is divisible by 9 if the sum of its digits is divisible by 9. Now consider any number made only of 0's and 9's. Each 9 contributes 9 to the digit sum, and each 0 contributes 0. The total digit sum is 9 + 9 + 9 + ... + 0 + 0 + 0, which equals 9 times the count of 9's in the number. This is always a multiple of 9. Therefore, the number itself must be a multiple of 9. We can verify with examples: 9 has digit sum 9 (divisible by 9); 90 has digit sum 9 (divisible by 9); 909 has digit sum 18 (divisible by 9); 9900 has digit sum 18 (divisible by 9); 999 has digit sum 27 (divisible by 9). All are divisible by 9. Hence, any number made using only the digits '0' and '9' will always be divisible by 9 because its digit sum is always a multiple of 9.
In simple words: Any combination of 0's and 9's adds up to a number of 9's, which is always divisible by 9.

Exam Tip: Recognize patterns - if you see only 0's and 9's in the digits, immediately conclude divisibility by 9 without calculating the digit sum.

Question 11. Is 10 divisible by 9? If not, what is the remainder?
Answer: No, 10 is not divisible by 9. When we divide 10 by 9, we get 10 ÷ 9 = 1 remainder 1. Therefore, if we divide 10 by 9, we get remainder 1.
In simple words: 9 goes into 10 once, with 1 left over, so the remainder is 1.

Exam Tip: The remainder rule for powers of 10 is useful - every power of 10 leaves remainder 1 when divided by 9, and this pattern repeats.

Question 12. Similarly, look at the remainder when the multiples of 100 (100, 200, 300, ...) are divided by 9. What do you notice?
Answer: When we check multiples of 100 divided by 9, we observe the following pattern:

100 ÷ 9: Digit sum = 1, Remainder = 1
200 ÷ 9: Digit sum = 2, Remainder = 2
300 ÷ 9: Digit sum = 3, Remainder = 3
400 ÷ 9: Digit sum = 4, Remainder = 4
500 ÷ 9: Digit sum = 5, Remainder = 5
600 ÷ 9: Digit sum = 6, Remainder = 6
700 ÷ 9: Digit sum = 7, Remainder = 7
800 ÷ 9: Digit sum = 8, Remainder = 8
900 ÷ 9: Digit sum = 9, Remainder = 0 (because 900 is divisible by 9)
1000 ÷ 9: Digit sum = 1, Remainder = 1
1100 ÷ 9: Digit sum = 2, Remainder = 2

The pattern of remainders is 1, 2, 3, 4, 5, 6, 7, 8, 0, 1, 2, 3, ... It repeats every 9 steps. This happens because 100 ÷ 9 = 11 remainder 1, so each multiple of 100 leaves a remainder equal to its coefficient (100 times 1 gives remainder 1, 100 times 2 gives remainder 2, and so on), and the pattern repeats after reaching 9.
In simple words: The remainders follow a repeating pattern: 1, 2, 3, 4, 5, 6, 7, 8, 0, 1, 2, 3, ... They repeat every 9 steps.

Exam Tip: This pattern works for all larger numbers too - it is a fundamental property of how remainders behave with powers of 10.

Question 13. Using this observation, find the remainder when 427 is divided by 9.
Answer: To find the remainder when 427 is divided by 9, we first find the digit sum: 4 + 2 + 7 = 13. Since 13 is greater than 9, we subtract 9 from it: 13 - 9 = 4. Therefore, the remainder when 427 is divided by 9 is 4. We can verify: 427 ÷ 9 = 47 remainder 4.
In simple words: Add the digits to get 13. Since 13 is too large, subtract 9 to get 4. The remainder is 4.

Exam Tip: When the digit sum exceeds 9, you can subtract 9 repeatedly (or just once) to find the final remainder - this works because the digit sum and the original number leave the same remainder when divided by 9.

Question 14. Will this work with bigger numbers?
Answer: Yes, the same pattern works for all larger numbers, and it will always work. We are looking at multiples of 100: 100, 200, 300, 400, 500, ..., 1000, 1100, 1200, and so on. The key observation is that 100 ÷ 9 = 11 remainder 1. This is true regardless of how large the number becomes. This means 100 × 1 leaves remainder 1, 100 × 2 leaves remainder 2, 100 × 3 leaves remainder 3, and so on, up to 100 × 8 which leaves remainder 8, and 100 × 9 which leaves remainder 0. After 100 × 9, the pattern repeats: 100 × 10 leaves remainder 1 again. The pattern always repeats every 9 steps, no matter how big the numbers get. Therefore, this works for all larger numbers and the pattern will always be 1, 2, 3, 4, 5, 6, 7, 8, 0, repeating forever.
In simple words: Because 100 divided by 9 always leaves remainder 1, the same remainder pattern keeps repeating no matter how large the multiples of 100 become.

Exam Tip: This principle extends to any power of 10 - understanding why 10^n always leaves remainder 1 when divided by 9 unlocks the entire divisibility rule for 9.

Question 15. Look at each of the following statements. Which are correct and why? (i) If a number is divisible by 9, then the sum of its digits is divisible by 9. (ii) If the sum of the digits of a number is divisible by 9, then the number is divisible by 9. (iii) If a number is not divisible by 9, then the sum of its digits is not divisible by 9. (iv) If the sum of the digits of a number is not divisible by 9, then the number is not divisible by 9.
Answer: (i) Correct. The divisibility rule for 9 works in both directions. If a number is divisible by 9, the sum of its digits will also be divisible by 9. For example, 729 is divisible by 9, and its digit sum is 7 + 2 + 9 = 18, which is divisible by 9.

(ii) Correct. This is exactly the divisibility rule for 9. If the digit sum is divisible by 9, the number itself is divisible by 9. For example, if the sum is 18 (divisible by 9), then the number 999 is divisible by 9.

(iii) Correct. If a number is not divisible by 9, then by contrapositive logic, its digit sum cannot be divisible by 9 either. If the digit sum were divisible by 9, the number itself would be divisible by 9, which contradicts our assumption.

(iv) Correct. This is the reverse direction of statement (ii). If the digit sum is not divisible by 9, then the number definitely cannot be divisible by 9. For example, 245 has a digit sum of 2 + 4 + 5 = 11, which is not divisible by 9. Therefore, 245 is not divisible by 9.
In simple words: The divisibility rule for 9 works in all four directions - the number and its digit sum always agree on whether they are divisible by 9.

Exam Tip: Understand that divisibility statements have both forward and reverse directions. If one direction is true, the reverse (contrapositive) is automatically true as well.

Question 16. Find, without dividing, whether the following numbers are divisible by 9. (i) 123 (ii) 405 (iii) 8888 (iv) 93547 (v) 358095
Answer: A number is divisible by 9 if the sum of its digits is divisible by 9. Let me check each one without dividing:

(i) 123: Digits are 1 + 2 + 3 = 6. Since 6 is not divisible by 9, the number 123 is not divisible by 9.

(ii) 405: Digits are 4 + 0 + 5 = 9. Since 9 is divisible by 9, the number 405 is divisible by 9.

(iii) 8888: Digits are 8 + 8 + 8 + 8 = 32. Since 32 is not divisible by 9, the number 8888 is not divisible by 9.

(iv) 93547: Digits are 9 + 3 + 5 + 4 + 7 = 28. Since 28 is not divisible by 9, the number 93547 is not divisible by 9.

(v) 358095: Digits are 3 + 5 + 8 + 0 + 9 + 5 = 30. Since 30 is not divisible by 9, the number 358095 is not divisible by 9.
In simple words: For each number, just add all its digits. If the sum divides evenly by 9, so does the original number.

Exam Tip: This method is much faster than actual division and works for numbers of any size - always use it on exams.

Question 17. Find the smallest multiple of 9 with no odd digits.
Answer: We need a multiple of 9 using only even digits (0, 2, 4, 6, 8). The smallest such number would start with the smallest even digit. Let's try 2. The smallest even digits after 2 are 0's and 2's. Testing 2, 20, 22: 2 has digit sum 2 (not divisible by 9); 20 has digit sum 2 (not divisible by 9); 22 has digit sum 4 (not divisible by 9). Continuing with 3 digits starting with 2, we try 200, 202, 204, 206, 208, 220, 222, 224, 226, 228, 240, 242, 244, 246, 248, 260, 262, 264, 266, 268, 280, 282, 284, 286, 288. The digit sum 288 is 2 + 8 + 8 = 18, which is divisible by 9. Therefore, the smallest multiple of 9 with no odd digits is 288. We can verify: 288 ÷ 9 = 32, confirming it is indeed a multiple of 9.
In simple words: Use only even digits and find the smallest number with digit sum 9, 18, 27, etc. The number 288 has digit sum 18, which works.

Exam Tip: When restricted to certain digits, systematically test combinations in increasing order rather than guessing randomly.

Question 18. Find the multiple of 9 that is closest to the number 6000.
Answer: To find the multiple of 9 closest to 6000, we perform some test multiplications. Working around the quotient 6000 ÷ 9 ≈ 667: 9 × 600 = 5400; 9 × 650 = 5850; 9 × 660 = 5940; 9 × 667 = 6003; 9 × 666 = 5994. Now, 6000 lies between 5994 (= 9 × 666) and 6003 (= 9 × 667). Measuring distances from 6000: distance to 5994 is |6000 - 5994| = 6; distance to 6003 is |6003 - 6000| = 3. Since 3 < 6, the closest multiple of 9 to 6000 is 6003.
In simple words: Find the two multiples of 9 that 6000 falls between, then see which one is closer. The answer is 6003.

Exam Tip: When finding the closest multiple, always check distances from both surrounding multiples rather than assuming the lower or upper one is nearer.

Question 19. How many multiples of 9 are there between the numbers 4300 and 4400?
Answer: The first multiple of 9 after 4300 is 4302 (since 4300 ÷ 9 ≈ 477.8, so the next whole quotient is 478, and 9 × 478 = 4302). The last multiple of 9 before 4400 is 4392 (since 4400 ÷ 9 ≈ 488.9, so the quotient is 488, and 9 × 488 = 4392). To count the multiples between 4302 and 4392, we note that 4302 = 9 × 478 and 4392 = 9 × 488. The multiples of 9 in this range correspond to quotients 478, 479, 480, ..., 488. The count is 488 - 478 + 1 = 11. Therefore, there are 11 multiples of 9 between 4300 and 4400.
In simple words: Find the first and last multiples of 9 in the range, then count how many multiples fit between them. The answer is 11.

Exam Tip: Always remember the "+1" when counting inclusive ranges (e.g., from 478 to 488 inclusive is 488 - 478 + 1 = 11 numbers).

Divisibility by 3

Question 20. The shortcut to find the divisibility by 3 is similar to the method for 9. A number is divisible by 3 if the sum of its digits is divisible by 3. Explore the remainders when powers of 10 are divided by 3. Explain why this method works.
Answer: Let us examine what remainder each power of 10 leaves when divided by 3:

10 ÷ 3 gives remainder 1
100 ÷ 3 gives remainder 1
1000 ÷ 3 gives remainder 1
10000 ÷ 3 gives remainder 1, and so on.

Key observation: Every power of 10 leaves remainder 1 when divided by 3. Now, a number written in place value form is abcd = 1000a + 100b + 10c + d. Replacing powers of 10 with their remainders when divided by 3: 1000 becomes 1, 100 becomes 1, 10 becomes 1, and 1 is already 1. So the number becomes equivalent (in terms of remainder) to a + b + c + d. Therefore, a number and the sum of its digits leave the same remainder when divided by 3. Since both the number and the sum of its digits have the same remainder when divided by 3, this means: if the digit sum is divisible by 3, the whole number is divisible by 3; if the digit sum is not divisible by 3, the whole number is not divisible by 3. This is exactly the divisibility rule for 3.
In simple words: Just like with 9, the digits in higher place values don't matter for divisibility by 3 - only the digit sum matters.

Exam Tip: The mathematical reason divisibility shortcuts work is that powers of 10 have predictable remainders - understanding this pattern helps you remember and apply the rules confidently.

Divisibility by 11

Question 21. Using these observations, can you tell whether the number 462 is divisible by 11?
Answer: To check if 462 is divisible by 11, we break it into place values and examine each part's remainder when divided by 11. The number 462 consists of 400 + 60 + 2. Let us find each part's relationship to multiples of 11: 400 is 36 × 11 + 4, so remainder = +4; 60 is 5 × 11 + 5, but actually 60 = 6 × 11 - 6, so remainder = -6; 2 = 0 × 11 + 2, so remainder = +2. Adding the remainders: (+4) + (-6) + (+2) = 0. A remainder of 0 means the number is exactly divisible by 11. Therefore, 462 is divisible by 11. We can verify: 462 ÷ 11 = 42 with no remainder.
In simple words: Break the number into parts based on place value. Add the remainders, treating some as positive and some as negative based on their position. If the total is 0 or a multiple of 11, the number divides by 11.

Exam Tip: The divisibility rule for 11 is more complex because powers of 10 alternate between +1 and -1 remainders, creating the alternating sum pattern.

Question 22. What could be a general method or shortcut to check divisibility by 11?
Answer: The divisibility rule for 11 states: Take the difference of the sum of digits in odd places and the sum of digits in even places (where places are counted from the right, starting at position 1). If this difference is 0 or a multiple of 11, the number is divisible by 11. This rule works for any number, whether it has 2 digits or 20 digits. For example, to check if 462 is divisible by 11: the digits are 4, 6, 2 (from left to right). Odd places from the right are positions 1 and 3, holding digits 2 and 4. Even places from the right are position 2, holding digit 6. Calculate (2 + 4) - (6) = 6 - 6 = 0. Since the difference is 0, the number 462 is divisible by 11.
In simple words: Add the digits in odd positions (from the right) and subtract the digits in even positions (from the right). If the result is 0 or divisible by 11, so is the number.

Exam Tip: Remember to count positions from the RIGHT, not the left - this is where students often make mistakes on the divisibility rule for 11.

Question 23. If this difference is 11 or a multiple of 11, what does that say about the remainder obtained when the number is divisible by 11?
Answer: When we use the divisibility rule for 11, the difference we calculate (sum of digits in odd places minus sum of digits in even places) and the original number leave the same remainder when divided by 11. So if the difference is 11, -11, 22, -22, 33, or any other multiple of 11, then all of these are divisible by 11, which means the remainder obtained when dividing the original number by 11 is 0. Therefore, the number is exactly divisible by 11.
In simple words: If the alternating digit sum gives a multiple of 11 (like 11, 22, 33, or even -11), the original number is divisible by 11 with remainder 0.

Exam Tip: The remainder relationship holds both ways - the difference of the alternating digit sums tells you the remainder of the original number when divided by 11.

Question 24. Using this shortcut, find out whether the following numbers are divisible by 11. Further, find the remainder if the number is not divisible by 11. (i) 158 (ii) 841 (iii) 481 (iv) 5529 (v) 90904 (vi) 857076
Answer: (i) 158: Odd places (from right): 8, 1; sum = 9. Even places: 5; sum = 5. Difference = 9 - 5 = 4. Not divisible by 11; remainder = 4.

(ii) 841: Odd places: 1, 8; sum = 9. Even places: 4; sum = 4. Difference = 9 - 4 = 5. Not divisible by 11; remainder = 5.

(iii) 481: Odd places: 1, 4; sum = 5. Even places: 8; sum = 8. Difference = 8 - 5 = 3. Not divisible by 11; remainder = 3.

(iv) 5529: Odd places: 9, 5; sum = 14. Even places: 2, 5; sum = 7. Difference = 14 - 7 = 7. Not divisible by 11; remainder = 7.

(v) 90904: Odd places: 4, 9, 9; sum = 22. Even places: 0, 0; sum = 0. Difference = 22 - 0 = 22. Since 22 is a multiple of 11, the number is divisible by 11.

(vi) 857076: Odd places: 6, 0, 5; sum = 11. Even places: 7, 7, 8; sum = 22. Difference = 22 - 11 = 11. Since 11 is a multiple of 11, the number is divisible by 11.
In simple words: Calculate the alternating digit sum (odd positions minus even positions from the right). If it is 0 or a multiple of 11, the number divides by 11. Otherwise, that value IS the remainder.

Exam Tip: Double-check your identification of odd and even positions - they are counted from the RIGHT, and the rightmost digit is position 1 (odd).

Question 25. Is this method similar to or different from the method we saw just before?
Answer: The method is actually the same, just written in a different way. Both approaches are based on the same rule for divisibility by 11. Whether you break down the number by place values and add/subtract remainders, or you calculate the difference between the sum of odd-positioned digits and the sum of even-positioned digits, you arrive at the same result because the underlying mathematical principle is identical. The alternating sum of digits naturally captures how the place-value remainders combine, so these two presentations are equivalent.
In simple words: These are two ways of describing the exact same divisibility rule - one breaks down place values, the other uses digit positions. Both give the same answer.

Exam Tip: Understanding that different presentations of a rule are fundamentally the same helps you choose whichever method feels clearer to you on the exam.

Divisibility Table

Question 26. Fill in the following table. Find a quick way to do this?

Answer: The quick way to fill this table is to use the divisibility shortcuts for each divisor rather than doing long division:

- Divisible by 2: Check if the units digit is even (0, 2, 4, 6, 8).
- Divisible by 3: Check if the sum of digits is divisible by 3.
- Divisible by 4: Check if the last two digits form a number divisible by 4.
- Divisible by 5: Check if the units digit is 0 or 5.
- Divisible by 6: Check if the number is divisible by both 2 and 3.
- Divisible by 8: Check if the last three digits form a number divisible by 8.
- Divisible by 9: Check if the sum of digits is divisible by 9.
- Divisible by 10: Check if the units digit is 0.
- Divisible by 11: Check if the difference between the sum of odd-positioned digits and even-positioned digits is 0 or a multiple of 11.

By applying these shortcuts, you can fill the entire table without performing any actual division, making the process much faster and less error-prone.

Exam Tip: Memorize all nine divisibility shortcuts - they save enormous time on any exam involving divisibility questions and are much more reliable than trial division for large numbers.

Question 27. How can we find out if a number is divisible by 6?
Answer: A number is divisible by 6 if and only if it is divisible by both 2 and 3. Both conditions must be satisfied. A number is divisible by 2 if its units digit is even: 0, 2, 4, 6, or 8. A number is divisible by 3 if the sum of its digits is divisible by 3. For example, consider 132: the units digit is 2, so it is divisible by 2. The digit sum is 1 + 3 + 2 = 6, which is divisible by 3. Therefore, 132 is divisible by 6. Consider another number, 124: the units digit is 4, so it is divisible by 2. The digit sum is 1 + 2 + 4 = 7, which is not divisible by 3. Therefore, 124 is not divisible by 6. Hence, a number is divisible by 6 if it is divisible by both 2 and 3.
In simple words: For divisibility by 6, check two things: the number must be even, and its digits must add up to a multiple of 3. Only then is it divisible by 6.

Exam Tip: Always break composite numbers (like 6 = 2 × 3) into their prime factors and check each factor separately - this is far easier than trying to remember rules for composite divisors.

Question 28. Will checking its divisibility by its factors 2 and 3 work? Use the shortcuts for 2 and 3 on these numbers and divide each number by 6 to verify - 38, 225, 186, 64.
Answer: Yes, checking divisibility by the factors 2 and 3 works perfectly to determine divisibility by 6. Let us verify with each number:

38: Units digit is 8 (even), so divisible by 2. Digit sum = 3 + 8 = 11 (not divisible by 3). So 38 is not divisible by 6. Verification: 38 ÷ 6 = 6 remainder 2. Confirmed not divisible.

225: Units digit is 5 (odd), so not divisible by 2. So 225 is not divisible by 6 (no need to check divisibility by 3). Verification: 225 ÷ 6 = 37 remainder 3. Confirmed not divisible.

186: Units digit is 6 (even), so divisible by 2. Digit sum = 1 + 8 + 6 = 15 (divisible by 3). So 186 is divisible by both 2 and 3, which means it is divisible by 6. Verification: 186 ÷ 6 = 31. Confirmed divisible.

64: Units digit is 4 (even), so divisible by 2. Digit sum = 6 + 4 = 10 (not divisible by 3). So 64 is not divisible by 6. Verification: 64 ÷ 6 = 10 remainder 4. Confirmed not divisible.

The shortcuts for 2 and 3 work reliably to determine divisibility by 6 - whenever a number is divisible by both 2 and 3, it is automatically divisible by 6, and the division confirms this.
In simple words: The shortcut method always gives the correct answer - use it instead of doing actual division.

Exam Tip: This principle applies to any composite number - if you need to check divisibility by a composite number, always factor it and check each prime factor using the appropriate shortcut.

Question. Divisibility by 6 — Apply the shortcut to numbers 38, 225, 186, 64
Answer: A number is divisible by 6 only when two conditions hold at the same time - the last digit must be even (so the number is divisible by 2), AND the sum of all digits must be divisible by 3. Testing each number:

38: Last digit is 8 (even, so divisible by 2). Digit sum = 3 + 8 = 11 (not divisible by 3). Result - not divisible by 6.

225: Last digit is 5 (odd, so not divisible by 2). Digit sum = 2 + 2 + 5 = 9 (divisible by 3). Result - not divisible by 6.

186: Last digit is 6 (even, so divisible by 2). Digit sum = 1 + 8 + 6 = 15 (divisible by 3). Result - divisible by 6.

64: Last digit is 4 (even, so divisible by 2). Digit sum = 6 + 4 = 10 (not divisible by 3). Result - not divisible by 6.
In simple words: For 6 to divide a number, it must be even AND have digits that add to a multiple of 3.

Exam Tip: Always check both conditions separately - divisibility by 2 and divisibility by 3 - because BOTH must be true for divisibility by 6.

Question. Can you check if a number is divisible by 24 using only its factors 4 and 6? Why or why not?
Answer: No, this method does not work because 4 and 6 share a common factor of 2. When two factors have a shared factor between them, checking divisibility by each factor separately does not guarantee divisibility by their product. For example, 12 is divisible by both 4 (12 ÷ 4 = 3) and 6 (12 ÷ 6 = 2), but 12 is not divisible by 24. The correct way to check divisibility by 24 is to verify two coprime (no shared factors) conditions: the last 3 digits must form a number divisible by 8, AND the digit sum must be divisible by 3. This works because 24 = 8 × 3, and 8 and 3 share no common factor.
In simple words: You can only use factor-checking when the factors share no common divisor. Since 4 and 6 both divide by 2, you cannot use them.

Exam Tip: When checking divisibility by a composite number using its factors, first verify that those factors are coprime (share no common factors) before applying the rule.

Question. What property does the digital root have? Recall what we found when checking divisibility by 9.
Answer: The digital root of a number is the same as the remainder when that number is divided by 9, except when the remainder is 0 (which is shown as a digital root of 9 instead). This means: if a number has a digital root of 9, the number is divisible by 9; if it has a digital root of 2, it leaves a remainder of 2 when divided by 9. The digital root always gives the same remainder the original number gives when divided by 9.
In simple words: Digital root tells you the leftover when you divide by 9. A digital root of 9 means it divides evenly by 9.

Exam Tip: Remember the exception - digital root 9 means remainder 0, not remainder 9, when dividing by 9.

Question. Between the numbers 600 and 700, which numbers have the digital root: (i) 5, (ii) 7, (iii) 3?
Answer: To find these numbers, we use the fact that digital root equals the remainder when divided by 9 (except remainder 0 becomes digital root 9). First, the digital root of 600: digits 6 + 0 + 0 = 6. This creates a repeating cycle every 9 numbers: 6, 7, 8, 9, 1, 2, 3, 4, 5, and then it repeats.

(i) Digital root 5: This appears at the 9th position in each cycle. Starting from 608, and adding 9 each time: 608, 617, 626, 635, 644, 653, 662, 671, 680, 689, 698.

(ii) Digital root 7: This appears at the 2nd position in each cycle. The numbers are: 601, 610, 619, 628, 637, 646, 655, 664, 673, 682, 691, 699.

(iii) Digital root 3: This appears at the 7th position in each cycle. The numbers are: 606, 615, 624, 633, 642, 651, 660, 669, 678, 687, 696.
In simple words: The digital root pattern repeats every 9 numbers. Find where your target digital root appears in one cycle, then keep adding 9 to find all numbers in the 600-700 range.

Exam Tip: Work out the pattern for the starting number (600 in this case), identify which position your target appears in the 9-number cycle, then list multiples by repeatedly adding 9.

Question. Write the digital roots of any 12 consecutive numbers. What do you observe?
Answer: Consider the numbers 245 to 256. Computing their digital roots: 245→2, 246→3, 247→4, 248→5, 249→6, 250→7, 251→8, 252→9, 253→1, 254→2, 255→3, 256→4. The digital roots are: 2, 3, 4, 5, 6, 7, 8, 9, 1, 2, 3, 4. The key observation is that digital roots follow a repeating cycle of length 9, progressing through 1, 2, 3, 4, 5, 6, 7, 8, 9 and then starting over. In any sequence of 12 consecutive numbers, you see one complete full cycle (positions 1 through 9) plus the start of the next cycle (positions 1, 2, 3).
In simple words: Digital roots repeat in a pattern of 1 to 9 over and over. In 12 numbers, you get the whole pattern once plus part of it again.

Exam Tip: This cycle property is central to understanding digital roots - use it to predict digital roots of distant numbers without computing digit sums repeatedly.

Question. Find the digital roots of some consecutive multiples of (i) 3, (ii) 4, and (iii) 6.
Answer:
(i) For multiples of 3 (3, 6, 9, 12, 15, 18, 21, 24, 27, 30): The digital roots are 3, 6, 9, 3, 6, 9, 3, 6, 9, 3. They repeat in a cycle: 3, 6, 9, 3, 6, 9, and so on.

(ii) For multiples of 4 (4, 8, 12, 16, 20, 24, 28, 32, 36, 40): The digital roots are 4, 8, 3, 7, 2, 6, 1, 5, 9, 4. The pattern cycles through 9 different values before repeating, as digital roots themselves cycle with period 9.

(iii) For multiples of 6 (6, 12, 18, 24, 30, 36, 42, 48, 54, 60): The digital roots are 6, 3, 9, 6, 3, 9, 6, 3, 9, 6. They repeat in a shorter cycle: 6, 3, 9, 6, 3, 9, and so on.
In simple words: Multiples of different numbers have different patterns of digital roots - some repeat quickly (like multiples of 3 and 6), while others take the full 9 positions to repeat.

Exam Tip: Looking at patterns in digital roots of multiples can reveal whether numbers share factors or structural properties.

Question. What are the digital roots of numbers that are 1 more than a multiple of 6? What do you notice?
Answer: Numbers of the form 6k + 1 are: 7, 13, 19, 25, 31, 37, 43, 49, 55, 61. Computing their digital roots: 7→7, 13→4, 19→1, 25→7, 31→4, 37→1, 43→7, 49→4, 55→1, 61→7. The pattern is: 7, 4, 1, 7, 4, 1, 7, 4, 1, 7. The observation is that numbers that are 1 more than a multiple of 6 always have digital roots repeating in the pattern 7, 4, 1, 7, 4, 1, and so on. This is a shorter cycle than the general 9-number cycle, occurring with period 3.
In simple words: If you take any multiple of 6 and add 1, the digital roots will always follow the repeating pattern 7, then 4, then 1.

Exam Tip: Recognizing these shorter cycles for special number forms (like 6k + 1) helps you solve problems about digital roots without listing all numbers.

Question. I'm made of digits, each tiniest and odd. No shared ground with root #1 - how odd!
Answer: The phrase "made of digits, each tiniest and odd" refers to the smallest odd digits and possibly the smallest odd prime numbers. The odd primes are 3, 5, and 7. We need a number made from these digits whose digital root is not 1. Using the digits 3, 5, 7, we form the simplest number: 357. Computing: digit sum = 3 + 5 + 7 = 15, and digital root = 1 + 5 = 6. Since 6 ≠ 1, the number 357 satisfies the entire riddle - it uses only odd prime digits (3, 5, 7) and has a digital root of 6, not 1.
In simple words: Build a number using the smallest odd prime digits (3, 5, 7) so that when you find the digital root, it is not 1. The number 357 works.

Exam Tip: Riddles about digital roots often involve testing combinations of digits - start with the simplest arrangement and check whether the digital root meets the condition.

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Question 1. The digital root of an 8-digit number is 5. What will be the digital root of 10 more than that number?
Answer: Let the 8-digit number be N, with digital root 5. When you add 10 to any number, only the tens and units places change. Adding 10 is the same as adding 1 to the tens place and 0 to the units place, so the total digit-sum increases by 1. Since the original digital root was 5, the new digit sum is 5 + 1 = 6. Therefore, the digital root of N + 10 is 6.
In simple words: When you add 10 to a number, its digit sum goes up by exactly 1. So the digital root goes from 5 to 6.

Exam Tip: To find how adding a specific value changes the digital root, calculate how much that value increases the digit sum, then adjust the original digital root accordingly.

Question 2. Write any number. Generate a sequence of numbers by repeatedly adding 11. What would be the digital roots of this sequence of numbers? Share your observations.
Answer: Starting with any number N and adding 11 repeatedly gives the sequence: N, N + 11, N + 22, N + 33, N + 44, ... Since 11 = 1 + 1, adding 11 increases the digit sum by 2. If the digital root of N is DR(N), then:

DR(N + 11) = DR(N) + 2
DR(N + 22) = DR(N) + 4
DR(N + 33) = DR(N) + 6
DR(N + 44) = DR(N) + 8
DR(N + 55) = DR(N) + 10, which reduces to DR(N) + 1 (since we cycle back in the digital root pattern)

Example: Let N = 37, so DR(37) = 3 + 7 = 10 → 1. The sequence is 37, 48, 59, 70, 81, 92, 103, ... with digital roots 1, 3, 5, 7, 9, 2, 4, 6, 8, 1, ... The digital roots increase by 2 each time and form the repeating cycle: 1 → 3 → 5 → 7 → 9 → 2 → 4 → 6 → 8 → 1 → ...
In simple words: When you add 11 over and over, the digital roots jump up by 2 each time and cycle through all nine digits before repeating.

Exam Tip: Understanding how adding multiples of specific numbers (like 11) affects digital roots helps solve problems about number sequences without calculating every term.

Question 3. What will be the digital root of the number 9a + 36b + 13?
Answer: To find the digital root, we use the fact that digital root equals the remainder when divided by 9. We simplify each term:

For 9a: This is always divisible by 9, so it contributes a remainder of 0.

For 36b: Since 36 = 9 × 4, this is also divisible by 9, contributing a remainder of 0.

For 13: When 13 is divided by 9, the remainder is 4 (since 13 = 9 + 4), so it contributes 4.

Adding the remainders: 0 + 0 + 4 = 4. Therefore, the digital root of 9a + 36b + 13 is 4.
In simple words: The first two terms divide evenly by 9 and add nothing to the remainder. Only 13 matters - it leaves a remainder of 4 when divided by 9.

Exam Tip: When finding digital roots of algebraic expressions, simplify each term by finding its remainder when divided by 9, then add those remainders.

Question 4. Make conjectures by examining if there are any patterns or relations between (i) the parity of a number and its digital root.
Answer: Testing examples: 24 has digital root 6 (even), and 18 has digital root 9 (odd). So even numbers can have odd or even digital roots. For odd numbers: 17 has digital root 8 (even), 19 has digital root 1 (odd), 15 has digital root 6 (even), 35 has digital root 8 (even). Observation: Digital roots of odd numbers can also be even OR odd. Therefore, there is no fixed relationship between the parity (even or odd) of a number and the parity of its digital root. Both even and odd numbers can produce digital roots that are either even or odd.
In simple words: Whether a number is even or odd tells you nothing about whether its digital root is even or odd - both combinations are possible.

Exam Tip: Just because a number itself is even does not mean its digital root is even, and vice versa for odd numbers.

Question 4. Make conjectures by examining if there are any patterns or relations between (ii) the digital root of a number and the remainder obtained when the number is divided by 3 or 9.
Answer: Relation with remainder when divided by 9: The digital root is exactly the remainder when the number is divided by 9, except when the remainder is 0 (shown as digital root 9 instead). Example: Number 47 has digital root 4 + 7 = 11 → 2, and 47 ÷ 9 gives remainder 2. So the digital root matches the remainder from division by 9.

Relation with remainder when divided by 3: If the digital root is divisible by 3 (i.e., digital root is 3, 6, or 9), then the number is divisible by 3. Examples: digital root 3 → number divisible by 3; digital root 6 → number divisible by 3; digital root 9 → number divisible by 3. This works because digital roots divisible by 3 come from numbers whose digit sums are divisible by 3, which is the divisibility rule for 3.
In simple words: The digital root equals the leftover from dividing by 9. If the digital root is 3, 6, or 9, the number definitely divides by 3.

Exam Tip: Use the digital root to quickly check divisibility: if it's 3, 6, or 9, the number is divisible by 3; if it's 9 alone (or 0 mod 9), it's divisible by 9.

Question. Solve the cryptarithm: A1 + 1B = B0
Answer: Converting to place value notation: (10A + 1) + (10 + B) = 10B, which simplifies to 10A + 11 = 9B. Testing single-digit values: only A = 7 and B = 9 satisfy this equation. Verification: 71 + 19 = 90. ✓
In simple words: Set up the equation using place values, then try digit values until you find one that works.

Exam Tip: Always verify your solution by substituting back into the original addition or multiplication problem.

Question. Solve the cryptarithm: AB + 37 = 6A
Answer: Converting to place value: (10A + B) + 37 = 60 + A, which simplifies to 9A + B = 23. Testing single-digit values: A = 2 gives B = 23 - 18 = 5. So A = 2 and B = 5. Verification: 25 + 37 = 62. ✓
In simple words: Rearrange the equation to isolate one variable, then try values for the other variable.

Exam Tip: In cryptarithmetic, it is often helpful to rearrange the equation so that one variable is isolated, making it easier to test values systematically.

Question. Solve the cryptarithm: ON + ON + ON = PO
Answer: This equals 3 × ON = PO. Converting to place value: 3(10O + N) = 10P + O, which gives 29O + 3N = 10P. Testing values of O and N such that the result is a 2-digit number PO:

If O = 1, N = 7: 3 × 17 = 51 → P = 5. (17 + 17 + 17 = 51) ✓
If O = 2, N = 4: 3 × 24 = 72 → P = 7. (24 + 24 + 24 = 72) ✓
If O = 3, N = 1: 3 × 31 = 93 → P = 9. (31 + 31 + 31 = 93) ✓

This cryptarithm has three valid solutions: 17 + 17 + 17 = 51, 24 + 24 + 24 = 72, and 31 + 31 + 31 = 93.
In simple words: Test pairs of digits to see which ones, when tripled, give the pattern PO (first digit P, second digit O).

Exam Tip: Some cryptarithmetic problems have multiple valid solutions - check all possibilities before concluding.

Question. Solve the cryptarithm: QR + QR + QR = PRR
Answer: This equals 3 × QR = PRR. Converting to place value: 3(10Q + R) = 100P + 10R + R, which gives 30Q + 3R = 100P + 11R, simplifying to 30Q = 100P + 8R. For this to hold with single digits, we need P = 2, Q = 8, R = 5. Verification: 85 + 85 + 85 = 255. ✓ The unique solution is Q = 8, R = 5, P = 2.
In simple words: Multiply QR by 3 and match the result to the pattern PRR, where the last two digits are the same.

Exam Tip: When a pattern has repeated digits (like RR), use that constraint to reduce the number of cases you need to test.

Question. Solve the cryptarithm: PQ × 8 = RS
Answer: We need a 2-digit number PQ such that multiplying by 8 gives a 2-digit result RS. This means RS must be less than 100, so 8 × PQ < 100, giving PQ < 12.5. Since PQ is a 2-digit number with P ≠ 0, the only possibility is PQ = 12. Checking: 12 × 8 = 96. So P = 1, Q = 2, R = 9, S = 6. Verification: 12 × 8 = 96. ✓
In simple words: Find a 2-digit number that, when multiplied by 8, stays 2 digits. Only 12 works because 13 × 8 = 104 is already 3 digits.

Exam Tip: Set up an inequality to find the range of possible values before testing individual cases - this saves time and avoids unnecessary calculations.

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Question. Solve the cryptarithm: GH × H = 9K
Answer: We need GH × H to produce a result in the 90s (between 90 and 99). Testing each value of H from 1 to 9:

H = 9: All products (19 × 9 = 171, 29 × 9 = 261, etc.) exceed 100. Not possible.

H = 8: All products (18 × 8 = 144, 28 × 8 = 224, etc.) are too large or too small. No results in the 90s.

H = 7: Testing GH values ending in 7 (17, 27, 37, ..., 87). All products are either below 90 or above 99. Not possible.

H = 6: Testing 16 × 6 = 96. This works! So GH = 16 and H = 6, giving 16 × 6 = 96 = 9K with K = 6.

Possible solution: G = 1, H = 6, K = 6.
In simple words: Try each digit for H and see which one allows a 2-digit number GH that produces a result in the 90s when multiplied by H.

Exam Tip: Narrow down the search space by establishing bounds (like "the result must be between 90 and 99") before testing individual cases.

Question. Solve the cryptarithm: BYE × 6 = RAY
Answer: Since BYE is a 3-digit number, B must be 1 (otherwise 2 × 6 = 12 would already produce a carry). So B = 1.

Units place: 6 × E ends in Y. Trying E = 5 gives 6 × 5 = 30, so Y = 0 with carry 3.

Tens place: 6 × Y + carry = A. With Y = 0 and carry 3: 6 × 0 + 3 = 3, so A = 3 with no carry.

Hundreds place: 6 × B + carry = R. With B = 1 and no carry: 6 × 1 = 6, so R = 6.

Result: B = 1, Y = 0, E = 5, A = 3, R = 6. All digits are distinct. Verification: 105 × 6 = 630. ✓
In simple words: Multiply digit by digit from right to left, keeping track of carries, and check that each step produces a valid digit and all digits are different.

Exam Tip: For multiplication cryptarithmetic, work place by place from right to left, and use carries carefully to ensure consistency across all positions.

Question. In BYE × 6 = RAY, what can you say about Y? What digits are possible or not possible?
Answer: Y is the units digit of BYE × 6, which comes from 6 × E. Looking at 6 times each digit: 6×0=0, 6×1=6, 6×2=12→2, 6×3=18→8, 6×4=24→4, 6×5=30→0, 6×6=36→6, 6×7=42→2, 6×8=48→8, 6×9=54→4. The units digits are: 0, 6, 2, 8, 4, 0, 6, 2, 8, 4. Therefore, Y can only be 0, 2, 4, 6, or 8 - that is, Y must be even. Y cannot be any odd digit (1, 3, 5, 7, 9).
In simple words: When you multiply 6 by any digit, the result always ends in an even number. So Y must be even.

Exam Tip: Knowing which digits are possible (or impossible) for a particular place value often helps eliminate cases quickly in cryptarithmetic problems.

Question. Solve the following: (i) UT × 3 = PUT
Answer: Converting to place value: (10U + T) × 3 = 100P + 10U + T. Expanding: 30U + 3T = 100P + 10U + T, which simplifies to 20U + 2T = 100P, or 10U + T = 50P. But 10U + T is the number UT itself, so UT = 50P. Since UT is a 2-digit number (between 10 and 99), 50P must fall in that range. Only P = 1 works, giving UT = 50. So U = 5, T = 0. Verification: 50 × 3 = 150. ✓
In simple words: Rearrange to show that UT itself equals 50P. Since UT is 2 digits, only 50 works.

Exam Tip: When an equation simplifies to "one variable-group equals a multiple of another," use the digit constraints (2-digit, 3-digit, etc.) to find the answer directly.

Question. Solve the following: (ii) AB × 5 = BC
Answer: Converting to place value: (10A + B) × 5 = 10B + C. Expanding: 50A + 5B = 10B + C, which gives 50A - 5B = C. Since C is a digit (0-9), we have 50A - 5B ≤ 9. Dividing by 5: 10A - B ≤ 1.8, so 10A - B is 0 or 1. Testing: if 10A - B = 0, then B = 10A, impossible for digit B. If 10A - B = 1, then B = 10A - 1. For A = 1, B = 9, and C = 50 - 45 = 5. So AB = 19, BC = 95. Verification: 19 × 5 = 95. ✓
In simple words: Set up the constraint that C must be a single digit, use it to limit possible values of A, then solve for B and C.

Exam Tip: Use digit constraints (0-9) to narrow down the range of variables before solving - this often gives you only one or two cases to check.

Question. Solve the following: (iii) L2N × 2 = 2NP
Answer: L2N represents 100L + 20 + N, and 2NP represents 200 + 10N + P. The equation becomes: 2(100L + 20 + N) = 200 + 10N + P, which simplifies to 200L + 40 + 2N = 200 + 10N + P, giving 200(L - 1) + 40 - 8N = P. Since P is a digit (0-9), trying L = 1: P = 40 - 8N. For P to be between 0 and 9: 40 - 8N ≥ 0 gives N ≤ 5, and 40 - 8N ≤ 9 gives N ≥ 4. So N can be 4 or 5.

If N = 4: P = 8, and 124 × 2 = 248. ✓
If N = 5: P = 0, and 125 × 2 = 250. ✓

Two solutions: 124 × 2 = 248 and 125 × 2 = 250.
In simple words: Set up the equation, use the digit constraint on P to find which values of N work, then verify each case.

Exam Tip: When multiple solutions are possible, test all cases that satisfy the constraints rather than stopping at the first one.

Question. Solve the following: (iv) XY × 4 = ZX
Answer: Converting to place value: (10X + Y) × 4 = 10Z + X. Expanding: 40X + 4Y = 10Z + X, which gives 39X + 4Y = 10Z. Since XY × 4 must remain 2-digit, XY ≤ 24. Testing: X = 2, Y = 3 gives 23 × 4 = 92. So Z = 9. All digits X = 2, Y = 3, Z = 9 are distinct. Verification: 23 × 4 = 92. ✓
In simple words: Set up the equation and use the constraint that the result must be 2-digit to limit your search space, then test possibilities systematically.

Exam Tip: Bound the variables first (like "XY ≤ 24 for the product to be 2-digit") to reduce computation time.

Question. Solve the following: (v) PP × QQ = PRP
Answer: Note that PP = 11P and QQ = 11Q, so the equation becomes (11P) × (11Q) = 121PQ. The result PRP = 100P + 10R + P. We need to find digits P and Q (both non-zero, P ≠ Q) such that their product in this form yields a pattern PRP. Testing: P = 2, Q = 1 gives 22 × 11 = 242, fitting PRP with R = 4. P = 3, Q = 1 gives 33 × 11 = 363, fitting PRP with R = 6. P = 4, Q = 1 gives 44 × 11 = 484, fitting PRP with R = 8. All digits in each solution are distinct. Valid solutions: 22 × 11 = 242, 33 × 11 = 363, 44 × 11 = 484.
In simple words: Notice the repeated-digit structure (PP and QQ) and test pairs of digits to see which multiply to match the pattern PRP.

Exam Tip: Recognize special patterns (like repeated digits) to simplify the structure before solving.

Question. Solve the following: (vi) JK × 6 = KKK
Answer: Note that KKK = 111K. The equation becomes (10J + K) × 6 = 111K. Expanding: 60J + 6K = 111K, which gives 60J = 105K, simplifying to 12J = 21K, or 4J = 7K. Since J and K are digits with J ≠ 0, we need J to be a multiple of 7. Trying J = 7: 4 × 7 = 28 = 7K gives K = 4. So JK = 74 and KKK = 444. Verification: 74 × 6 = 444. ✓
In simple words: Simplify using the repeated-digit pattern (KKK = 111K), then solve the resulting algebraic constraint 4J = 7K.

Exam Tip: Use divisibility reasoning (like "J must be a multiple of 7") to narrow down possibilities before testing values.

Question 1. If 31z5 is a multiple of 9, where z is a digit, what is the value of z? Explain why there are two answers to this problem.
Answer: For 31z5 to be a multiple of 9, the sum of its digits must be divisible by 9. The digit sum is 3 + 1 + z + 5 = 9 + z. This must equal a multiple of 9, so 9 + z ∈ {9, 18, 27, ...}. From 9 + z = 9, we get z = 0. From 9 + z = 18, we get z = 9. The next value 27 would give z = 18, which is not a single digit. There are two answers because the known digits (3, 1, 5) already sum to 9, which is itself divisible by 9. Adding z = 0 keeps the sum at 9, and adding z = 9 makes it 18 - both divisible by 9. Therefore, z can be either 0 or 9.
In simple words: Since the first three digits add to 9, you can either add 0 (keeping it divisible by 9) or add 9 (making 18, also divisible by 9).

Exam Tip: When a problem has two solutions, think about the structure - often the "base" digit sum is already at a critical value (like a multiple of 9), allowing two different completions.

Question 2. "I take a number that leaves a remainder of 8 when divided by 12. I take another number which is 4 short of a multiple of 12. Their sum will always be a multiple of 8", claims Snehal. Examine his claim and justify your conclusion.
Answer: Let the first number be 12a + 8 (remainder 8 when divided by 12) and the second be 12b - 4 (four less than a multiple of 12). Their sum is (12a + 8) + (12b - 4) = 12a + 12b + 4 = 12(a + b) + 4. This is a multiple of 12 plus 4. To check if it is a multiple of 8, note that 12 ÷ 8 = 1 remainder 4. So 12(a + b) + 4 divided by 8 gives remainder 4. Since the remainder is 4, not 0, the sum is not always a multiple of 8. Snehal's claim is false. The sum does not always become a multiple of 8.
In simple words: Even though the two numbers have a special form, their sum leaves a remainder of 4 when divided by 8, so it is never a multiple of 8.

Exam Tip: When evaluating a claim about divisibility, work through the algebra carefully to find the actual remainder, then check if it matches the claim.

Question 3. When is the sum of two multiples of 3, a multiple of 6 and when is it not? Explain the different possible cases, and generalise the pattern.
Answer: Let the two multiples of 3 be 3a and 3b. Their sum is 3a + 3b = 3(a + b), which is always a multiple of 3. For the sum to be a multiple of 6, it must also be even. Since 3 is odd, 3(a + b) is even if and only if (a + b) is even. This happens in two cases:

Case 1: Both a and b are even. Then a + b is even, so 3(a + b) is even and a multiple of 6. Example: 3(2) + 3(4) = 6 + 12 = 18, which is 18 ÷ 6 = 3. ✓

Case 2: Both a and b are odd. Then a + b is even, so 3(a + b) is even and a multiple of 6. Example: 3(3) + 3(5) = 9 + 15 = 24, which is 24 ÷ 6 = 4. ✓

Case 3: One of a, b is even and the other is odd. Then a + b is odd, so 3(a + b) is odd and NOT a multiple of 6. Example: 3(2) + 3(3) = 6 + 9 = 15, which is not divisible by 6.

General pattern: The sum of two multiples of 3 is a multiple of 6 exactly when both multiples are of the same "type" - both even multiples of 3 (like 6, 12, 18) or both odd multiples of 3 (like 3, 9, 15, 21).
In simple words: Two multiples of 3 add to a multiple of 6 only if they are the same "type" - both even or both odd.

Exam Tip: Breaking a problem into cases (even-even, odd-odd, even-odd) often reveals the complete pattern and prevents missing cases.

Question 4. Sreelatha says, "I have a number that is divisible by 9. If I reverse its digits, it will still be divisible by 9". (i) Examine if her conjecture is true for any multiple of 9.
Answer: A number is divisible by 9 if and only if the sum of its digits is divisible by 9. Reversing the digits does not change which digits are present, only their order. Therefore, the digit sum remains the same. Since the digit sum stays unchanged, if the original number is divisible by 9, the reversed number is also divisible by 9. Examples: 18 has digit sum 1 + 8 = 9 (divisible by 9), and its reverse 81 has digit sum 8 + 1 = 9 (also divisible by 9). Similarly, 342 has digit sum 3 + 4 + 2 = 9, and 243 has digit sum 2 + 4 + 3 = 9. Therefore, Sreelatha's conjecture is always true for any multiple of 9.
In simple words: Reversing the digits does not change the digit sum, so if the original number divides by 9, so does the reversed number.

Exam Tip: Use the divisibility rule (a number is divisible by 9 iff its digit sum is divisible by 9) to justify claims about divisibility under digit transformations.

Question 4. Sreelatha says, "I have a number that is divisible by 9. If I reverse its digits, it will still be divisible by 9". (ii) Are any other digit shuffles possible such that the number formed is still a multiple of 9?
Answer: Yes. Any rearrangement of the digits will produce a number that is still a multiple of 9, because rearranging digits does not change their sum. Since the digit sum determines divisibility by 9, any permutation of the digits yields another multiple of 9. For instance, if we start with 342 (digit sum 9), we can rearrange to get 243, 324, 423, 234, or 432 - all of these have digit sum 9 and are therefore divisible by 9. In fact, every single arrangement of these three digits forms a number divisible by 9.
In simple words: You can shuffle the digits in any order, and the result will still be divisible by 9 because the digit sum never changes.

Exam Tip: The divisibility rule for 9 depends only on the digit sum, not on the arrangement - this makes digit shuffles and rearrangements particularly elegant to analyze.

Question 5. If 48a23b is a multiple of 18, list all possible pairs of values for a and b.
Answer: For a number to be a multiple of 18, it must be divisible by both 2 and 9. A number is divisible by 2 when its last digit is even, so b can be 0, 2, 4, 6, or 8. For divisibility by 9, the sum of all digits must be a multiple of 9. The digit sum is 4 + 8 + a + 2 + 3 + b = 17 + a + b. This means 17 + a + b must equal either 18 or 27. When 17 + a + b = 18, we get a + b = 1. Since b must be even, only b = 0 works, giving a = 1. When 17 + a + b = 27, we get a + b = 10. Trying even values of b: if b = 2 then a = 8; if b = 4 then a = 6; if b = 6 then a = 4; if b = 8 then a = 2. Therefore, all valid pairs are: (1, 0), (8, 2), (6, 4), (4, 6), and (2, 8).
In simple words: The last digit must be even. Add up all the digits - the total must be 18 or 27. Then find which letter values work with these rules.

Exam Tip: Always check divisibility conditions separately (first for 2, then for 9) and combine them systematically to find all valid pairs.

Question 6. If 3p7q8 is divisible by 44, list all possible pairs of values for p and q.
Answer: Since 44 = 4 × 11, the number must be divisible by both 4 and 11. For divisibility by 4, the last two digits (q8) must form a number divisible by 4. Testing each digit: q = 0 gives 08, q = 2 gives 28, q = 4 gives 48, q = 6 gives 68, and q = 8 gives 88 - all divisible by 4. So q can be 0, 2, 4, 6, or 8. For divisibility by 11, use the rule: (sum of digits in odd positions) - (sum of digits in even positions) must be 0 or ±11. Here, (3 + 7 + 8) - (p + q) = 18 - (p + q). Since p and q are single digits, p + q ranges from 0 to 17. For 18 - (p + q) = 0, we would need p + q = 18, which is impossible. For 18 - (p + q) = 11, we get p + q = 7. Combining both conditions: q ∈ {0, 2, 4, 6, 8} and p + q = 7. When q = 0, p = 7; when q = 2, p = 5; when q = 4, p = 3; when q = 6, p = 1; when q = 8, p = -1 (not a digit). Therefore, all valid pairs are: (7, 0), (5, 2), (3, 4), and (1, 6).
In simple words: Check if the last two digits divide by 4. Then use the alternating sum rule for 11 - add odd-place digits, subtract even-place digits.

Exam Tip: The divisibility rule for 11 is often tricky - always set up the alternating sum carefully and check which values satisfy both conditions.

Question 7. Find three consecutive numbers such that the first number is a multiple of 2, the second number is a multiple of 3, and the third number is a multiple of 4. Are there more such numbers? How often do they occur?
Answer: Let the three consecutive numbers be n, n+1, and n+2. For the third to be a multiple of 4, n+2 must be divisible by 4, so n is 2 less than a multiple of 4. Multiples of 4 are 4, 8, 12, 16, 20, 24, 28, ... Subtracting 2 from each gives: 2, 6, 10, 14, 18, 22, 26, ... For the second to be a multiple of 3, n+1 must be divisible by 3, so n is 1 less than a multiple of 3. Multiples of 3 are 3, 6, 9, 12, 15, 18, 21, 24, ... Subtracting 1 from each gives: 2, 5, 8, 11, 14, 17, 20, 23, 26, ... Taking the common values from both lists, possible values of n are: 2, 14, 26, 38, ... When n = 2, the numbers are 2, 3, 4. When n = 14, they are 14, 15, 16. When n = 26, they are 26, 27, 28. When n = 38, they are 38, 39, 40. Yes, there are infinitely many such triples. These special consecutive numbers follow the pattern (12k + 2), (12k + 3), (12k + 4) for k = 0, 1, 2, 3, ... They occur every 12 numbers.
In simple words: Find numbers where n+2 is a multiple of 4 and n+1 is a multiple of 3. The lists of possible n values overlap at points like 2, 14, 26. They repeat with a gap of 12 each time.

Exam Tip: Use systematic listing of constraints - list out values satisfying each condition separately, then find the common values. The pattern usually repeats with the LCM of the period intervals.

Question 8. Write five multiples of 36 between 45,000 and 47,000. Share your approach with the class.
Answer: To find multiples of 36 in this range, first divide the lower bound by 36: 45000 ÷ 36 = 1250 (since 36 × 1250 = 45000 exactly). The first multiple is 45000 itself. Then add 36 repeatedly: 45000 + 36 = 45036, 45036 + 36 = 45072, 45072 + 36 = 45108, 45108 + 36 = 45144, 45144 + 36 = 45180. All these remain below 47000. Five multiples of 36 between 45000 and 47000 are: 45036, 45072, 45108, 45144, and 45180.
In simple words: Divide the start number by 36 to find where multiples begin. Then keep adding 36 to list each next multiple until you have enough or go too far.

Exam Tip: Always verify your first multiple by dividing and checking the remainder is zero. Then successive multiples follow by adding the divisor repeatedly.

Question 9. The middle number in the sequence of 5 consecutive even numbers is 5p. Express the other four numbers in sequence in terms of p.
Answer: In a sequence of 5 consecutive even numbers with middle number 5p, each number differs from the next by 2. The two numbers before 5p are 5p - 4 and 5p - 2. The two numbers after 5p are 5p + 2 and 5p + 4. Therefore, the complete sequence of 5 consecutive even numbers is: 5p - 4, 5p - 2, 5p, 5p + 2, and 5p + 4.
In simple words: Since even numbers are 2 apart, go back 4 (which is 2 steps of 2) and forward 4 from the middle number. This gives you all five in order.

Exam Tip: For consecutive even numbers, the difference is always 2. If you know the middle of an odd-count sequence, subtract and add the appropriate multiples of 2 to reach the outer numbers.

Question 10. Write a 6-digit number that is divisible by 15, such that when the digits are reversed, it is divisible by 6.
Answer: A number divisible by 15 must be divisible by both 3 and 5, so its last digit must be 0 or 5, and the sum of its digits must be divisible by 3. The reversed number must be divisible by 6, meaning it must be divisible by both 2 and 3. So the first digit of the original number (which becomes the last digit after reversing) must be even. Consider the number 234150: it has first digit 2 (even) and last digit 0. Check the original: digit sum = 2 + 3 + 4 + 1 + 5 + 0 = 15, which is divisible by 3; last digit is 0, divisible by 5. So 234150 is divisible by 15. When reversed, it becomes 051432 = 51432. Check the reversed number: last digit is 2 (even), and digit sum = 5 + 1 + 4 + 3 + 2 = 15, divisible by 3. So 51432 is divisible by 6. Therefore, 234150 is a valid answer.
In simple words: The number must end in 0 or 5 and have digits that add to a multiple of 3. When flipped, it must end in an even digit and still have digit sum divisible by 3. Try a simple number like 234150 and verify both conditions.

Exam Tip: Always verify divisibility of both the original and reversed number - check the last digit for the even condition and recalculate the digit sum (which stays the same under reversal) for the "divisible by 3" condition.

Question 11. Deepak claims, "There are some multiples of 11 which, when doubled, are still multiples of 11. But other multiples of 11 don't remain multiples of 11 when doubled." Examine if his conjecture is true; explain your conclusion.
Answer: Test Deepak's claim by taking several multiples of 11 and doubling each: 11 becomes 22, 22 becomes 44, 33 becomes 66, 44 becomes 88, 55 becomes 110, 66 becomes 132, 77 becomes 154, 88 becomes 176, and 99 becomes 198. Each result is also a multiple of 11. Mathematically, any multiple of 11 can be written as 11 × n. When doubled, it becomes 2 × (11 × n) = 22 × n = 11 × (2n), which is clearly a multiple of 11. Therefore, Deepak's conjecture is false. All multiples of 11 remain multiples of 11 when doubled - there are no exceptions.
In simple words: Doubling any multiple of 11 always gives another multiple of 11. This happens because 2 times (11 times a number) is still 11 times something.

Exam Tip: When examining a conjecture, test specific examples and then prove using algebra. The algebraic proof using the general form 11n is much more convincing than examples alone.

Question 12. Determine whether the statements below are 'Always True', 'Sometimes True', or 'Never True'. Explain your reasoning.
(i) The product of a multiple of 6 and a multiple of 3 is a multiple of 9.
(ii) The sum of three consecutive even numbers will be divisible by 6.
(iii) If abcdef is a multiple of 6, then badcef will be a multiple of 6.
(iv) 8(7b - 3) - 4(11b + 1) is a multiple of 12.
Answer:
(i) Let the multiple of 6 be 6a and the multiple of 3 be 3b. Their product is 6a × 3b = 18ab = 9 × (2ab). Since this equals 9 times a whole number, it is always a multiple of 9. Always True.
(ii) Let the three consecutive even numbers be 2n, 2n + 2, and 2n + 4. Their sum is 2n + (2n + 2) + (2n + 4) = 6n + 6 = 6(n + 1). Since this is 6 times a whole number, the sum is always divisible by 6. Always True.
(iii) Both abcdef and badcef contain the same digits rearranged, so their digit sum is identical - both are divisible by 3 or both are not. Both numbers end in the same digit f, so both are even or both are odd. Therefore, if abcdef is divisible by both 2 and 3, then badcef must also be divisible by both 2 and 3, making it divisible by 6. Always True.
(iv) Simplify: 8(7b - 3) - 4(11b + 1) = 56b - 24 - 44b - 4 = 12b - 28 = 4(3b - 7). For this to be a multiple of 12, we need 3b - 7 to be a multiple of 3. However, 3b - 7 = 3(b - 3) + 2, which always leaves remainder 2 when divided by 3. So 3b - 7 is never a multiple of 3 for any whole number b. Never True.
In simple words: (i) Multiplying two multiples gives you even more factors. (ii) Three even numbers always add up to 6 times something. (iii) Rearranging digits doesn't change whether a number divides evenly by 2 or 3. (iv) No matter what b is, the expression never becomes a multiple of 12.

Exam Tip: Clearly state "Always", "Sometimes", or "Never" at the start of each part. Use both numerical examples and algebraic proof to support your conclusion - examples alone are not sufficient for a complete answer.

Question 13. Choose any 3 numbers. When is their sum divisible by 3? Explore all possible cases and generalise.
Answer: Every whole number, when divided by 3, leaves one of three possible remainders: 0 (numbers like 3, 6, 9, 12...), 1 (numbers like 1, 4, 7, 10...), or 2 (numbers like 2, 5, 8, 11...). For three numbers, classify each as "type 0", "type 1", or "type 2" based on its remainder. The sum of the three numbers is divisible by 3 if and only if the sum of their remainders is 0, 3, or 6 (i.e., divisible by 3). Cases that work: (1) All three are type 0: remainders sum to 0+0+0 = 0. Example: 6, 12, 9 sum to 27 ✓ (2) All three are type 1: remainders sum to 1+1+1 = 3. Example: 4, 7, 10 sum to 21 ✓ (3) All three are type 2: remainders sum to 2+2+2 = 6. Example: 5, 8, 11 sum to 24 ✓ (4) One of each type (0, 1, 2): remainders sum to 0+1+2 = 3. Example: 6, 7, 8 sum to 21 ✓ Cases that do NOT work: (5) Two type 0, one type 1: 0+0+1 = 1 ✗ Example: 6, 9, 4 sum to 19 ✗ (6) Two type 1, one type 2: 1+1+2 = 4 ✗ Example: 4, 7, 5 sum to 16 ✗ (7) Two type 2, one type 1: 2+2+1 = 5 ✗ Example: 5, 8, 4 sum to 17 ✗ Generalisation: The sum of three numbers is divisible by 3 exactly when (1) all three give the same remainder on division by 3, OR (2) the three remainders are 0, 1, and 2 in some order. In short, the sum is divisible by 3 if and only if the sum of the remainders is 0, 3, or 6.
In simple words: Three numbers add to a multiple of 3 when they all have the same remainder when divided by 3, or when their remainders are 0, 1, and 2 mixed together.

Exam Tip: Classify numbers by their remainder on division by 3 rather than trying individual examples. This systematic approach reveals the pattern and allows you to prove it works for any three numbers.

Question 14. Is the product of two consecutive integers always multiple of 2? Why? What about the product of these consecutive integers? Is it always a multiple of 6? Why or why not? What can you say about the product of 4 consecutive integers? What about the product of five consecutive integers?
Answer:
Product of two consecutive integers and 2: Take two consecutive numbers n and n+1. One of them must be even. Therefore, their product n(n+1) always contains the factor 2. Always a multiple of 2.
Product of two consecutive integers and 6: For divisibility by 6, the product needs factors 2 and 3. It always has a 2 (as shown above), but not always a 3. For example, 4 × 5 = 20, which is not divisible by 6. Sometimes a multiple of 6, not always.
Product of four consecutive integers: Consider n, n+1, n+2, n+3. Among any four consecutive numbers: two are even (and one of those is a multiple of 4), and one is a multiple of 3. So the product always contains factors 4, 2, and 3, giving a factor of at least 4 × 2 × 3 = 24. Always a multiple of 24.
Product of five consecutive integers: Consider n, n+1, n+2, n+3, n+4. Among any five consecutive numbers: one is a multiple of 5, one is a multiple of 3, and at least two are even (with one being a multiple of 4). The product contains factors 5, 3, 4, and at least one additional 2, giving a total factor of 5 × 3 × 4 × 2 = 120. Always a multiple of 120.
In simple words: Two consecutive numbers: one is always even, so their product is always even. Two consecutive: might not have a 3, so not always divisible by 6. Four consecutive: always include enough even numbers and a multiple of 3, giving 24. Five consecutive: include a multiple of 5 as well, giving 120.

Exam Tip: Identify which prime factors (2, 3, 5) are guaranteed to appear among k consecutive integers, then multiply them to find the LCM that always divides the product.

Question 15. Solve the cryptarithms.
(i) EF × E = GGG
(ii) WOW × 5 = MEOW
Answer:
(i) Write EF as the two-digit number 10E + F. The equation becomes (10E + F) × E = 111G. Multiply E to both digits: this gives a three-digit number with all digits the same (GGG = 111G). Test values of E from 1 to 9. When E = 3: EF = 37, and 37 × 3 = 111. All three digits are 1. So E = 3, F = 7, G = 1. Answer: 37 × 3 = 111.
(ii) Write WOW as a three-digit number with W appearing twice. MEOW is a four-digit number. At the units place: W × 5 must end with W. Testing digits 0-9, only W = 5 works because 5 × 5 = 25 (ends in 5). At the tens place: We have W = 5. Computing: 5 × 5 = 25, write 5 and carry 2. Next, 5 × O + 2 must end with O. So 5O + 2 ends in O. Testing, O = 7 works: 5 × 7 + 2 = 37 (ends in 7), and we carry 3. At the hundreds place: We have W = 5 and carry 3. Computing: 5 × 5 + 3 = 28, write 8 and carry 2. So E = 8. At the thousands place: Only the carry 2 remains, so M = 2. Therefore, W = 5, O = 7, E = 8, M = 2. Answer: WOW = 575 and MEOW = 2875. Check: 575 × 5 = 2875. ✓
In simple words: (i) Find a two-digit number times its first digit gives a three-digit number with all digits the same. Test small values of the first digit. (ii) When a three-digit number with repeated first and last digit is multiplied by 5, the last digit stays the same. This only happens when the digit is 5. Then work through each position carefully.

Exam Tip: In cryptarithmetic, use constraints (like "last digit stays the same") to narrow down possibilities immediately. Then verify your solution by multiplying to check the result.

Question 16. Which of the following Venn diagrams captures the relationship between the multiples of 4, 8, and 32?
Answer: Consider the relationship between these three numbers: 8 is a multiple of 4 (8 = 2 × 4), and 32 is a multiple of 8 (32 = 4 × 8). Therefore, 32 is also a multiple of 4 (32 = 8 × 4). This means every multiple of 32 is also a multiple of 8, and every multiple of 8 is also a multiple of 4. The set of multiples of 32 is completely contained within the set of multiples of 8, which is completely contained within the set of multiples of 4. The correct diagram shows three concentric circles with the smallest innermost circle representing multiples of 32, the middle circle representing multiples of 8, and the largest outermost circle representing multiples of 4. The answer is diagram (iii).
In simple words: Every number in the 32 list is also in the 8 list. Every number in the 8 list is in the 4 list. So one circle fits inside the next, like nesting boxes.

Exam Tip: Always check divisibility relationships first. If A divides B, and B divides C, then the set of multiples of C is nested inside multiples of B, which is nested inside multiples of A. Use concentric circles for this "subset" relationship.