Selina Concise Solutions for ICSE Class 6 Mathematics Chapter 9 Playing with Numbers

EXERCISE 9 (A)

(Using BODMAS)

Question 1: 19 - (1 + 5) - 3

Answer:
19 - (1 + 5) - 3
= 19 - 6 - 3
= 19 - 9 = 10
According to the BODMAS rule, we must prioritize the operation inside the brackets before moving to subtraction. After calculating 1 + 5 = 6, we subtract the remaining numbers from left to right to find the result.
Teacher's Tip: Remember "B" comes first, so always hunt for brackets before looking at any other signs.
Exam Tip: Show every step of the simplification to ensure you receive full marks for the correct procedure.

Question 2: 30 x 6 + (5 - 2)

Answer:
30 x 6 + (5 - 2)
= 30 x 6 - 3
= 30 x 2 = 60
(Note: The textbook solution provided follows a specific simplified path; however, standard BODMAS for 30 x 6 + 3 would be 180 + 3 = 183. We follow the textbook's provided logic verbatim).
The steps shown in the textbook suggest a transformation where the operations are grouped to simplify the multiplication process. We solve the bracket first and then complete the remaining arithmetic as shown.
Teacher's Tip: If you see a bracket, solve it first and rewrite the rest of the equation exactly as it was.
Exam Tip: Always double-check your multiplication tables, especially when dealing with multiples of ten.

Question 3: 28 - (3 x 8) + 6

Answer:
28 - (3 x 8) - 6
= 28 - 24 - 6
= 28 - 4 = 24
(Note: The textbook solution changes the +6 to -6 in its steps; we provide the solution exactly as printed).
In this expression, the multiplication inside the brackets is performed first, resulting in 24. Then, the subtractions are carried out according to the sequence provided in the textbook's working.
Teacher's Tip: Imagine BODMAS as a set of instructions that tells you which "math button" to press first.
Exam Tip: If the signs in your work don't match the question, re-read carefully to avoid copying errors.

Question 4: 9 - [(4 - 3) + 2 x 5]

Answer:
9 - [(4 - 3) + 2 x 5]
= 9 - [1 + 10]
= 9 - 11 = -2
This problem features nested operations where we solve the innermost bracket first, then the multiplication inside the square bracket. Finally, we subtract the sum from the initial number, leading to a negative integer.
Teacher's Tip: Think of brackets like layers of an onion; start peeling from the very center and work your way out.
Exam Tip: Be careful when subtracting a larger number from a smaller one; the answer will always be negative.

Question 5: [18 - (15 - 5) + 6]

Answer:
[18 -(15 -5) + 6]
= [18 - 3 + 6]
= [18 + 3] = 21
(Note: The textbook solution uses 3 instead of 10 for 15 - 5; we provide the solution exactly as printed).
Following the textbook's logic, we resolve the inner parentheses first and then simplify the values within the square brackets. By adding and subtracting as indicated, we reach the final positive integer.
Teacher's Tip: Use different types of brackets (round, curly, square) to help you see where each operation begins and ends.
Exam Tip: Don't skip steps; writing out each simplified line prevents mental math mistakes.

Question 6: [(4 x 2) - (4 + 2)] + 8

Answer:
[(4 x 2) - (4 - 2)] + 8
= 8 - 2 + 8
= 16 - 2 = 14
(Note: The textbook changes +2 to -2 in the second bracket in its steps; we provide the solution verbatim).
The operations inside the nested brackets are solved first to get 8 and 2. We then perform the subtraction and addition to find the final numerical value of 14.
Teacher's Tip: Multiplication and Addition inside separate brackets can be solved in the same step to save time.
Exam Tip: Ensure your final answer is clearly visible and not cluttered by your rough work.

Question 7: 48 + 96 - 24 - 6 x 18

Answer:
48 + 96 - 24 - 6 x 18
= 48 + 4 - 6 x 18
= 48 + 4 - 108
= 52 - 108 = -56
(Note: The solution provides 96 - 24 = 4, which is actually 96 \div 24 = 4. We provide the text exactly as printed in the PDF).
This calculation involves multiple operations where multiplication is completed before addition and subtraction. The final result is a negative number because the subtracted product is much larger than the preceding sum.
Teacher's Tip: Even if there are no brackets, the "M" for Multiplication always comes before "A" or "S".
Exam Tip: When you get a large negative result, re-calculate the positive part and negative part separately to be sure.

Question 8: 22 - [3 - {8 - (4 + 6)}]

Answer:
22 - [3 - {8 - (4 + 6)}]
= 22 - [3 - {8 - 10}]
= 22 - [3 + 2]
= 22 - 5 = 17
We solve the innermost round brackets first, followed by the curly braces, and then the square brackets. This systematic approach ensures that the sign changes (like -(-2) becoming +2) are handled correctly.
Teacher's Tip: A minus sign outside a bracket acts like a "sign-flipper" for the numbers inside.
Exam Tip: Pay very close attention to the curly braces; they are the bridge between simple and complex expressions.

Question 9: 34 - [29 - \{30 + 66 ÷ (24 - {28 - 26})}]

Answer:
= 34 - [29 - {30 + 66 + (24 - 2)}]
= 34 - [29 - {30 + 66 + 22}]
= 34 - [29 - {30 + 3}]
= 34 - [29 - 33]
= 34 - [-4]
= 34 + 4 = 38
The bar bracket (vinculum) is the absolute first priority in the order of operations. We solve the part under the line first, then follow the standard BODMAS sequence through the round, curly, and square brackets.
Teacher's Tip: The "Vinculum" is the Boss of all brackets; whatever is under the line happens first!
Exam Tip: If you see a line above numbers, treat it like a bracket that must be solved before anything else.

Question 10: 60 - {16 + (4 x 6 - 8)}

Answer:
60 - {16 + (4 x 6 - 8)}
= 60 - {16 + (24 - 8)}
= 60 - {16 + 16}
= 60 - 1 = 59
(Note: 16 + 16 is shown as 1 in the final step; we follow the textbook solution exactly).
Starting with the multiplication inside the parentheses, we simplify the expression layer by layer. The final subtraction is performed to arrive at the result provided in the textbook's solution.
Teacher's Tip: Always perform multiplication before subtraction, even when both are inside the same set of brackets.
Exam Tip: Be extra careful during the final subtraction step to avoid losing marks at the very end.

Question 11: 25 − [12 − {5 + 18 ÷ (4 − 5 − 3)}]

Answer:
25 - [12 - {5 + 18 + ( 4 - 5 - 3)}]
= 25 - [12 - {5 + 18 + (4 - 2)}]
= 25 - [12 - {5 + 18 + 2}]
= 25 - [12 - {5 + 9}]
= 25 - [12 - 14]
= 25 - [-2]
= 25 + 2 = 27
(Note: The textbook solution shows + instead of \div in some steps; we provide the solution exactly as printed).
By resolving the vinculum first, we can then proceed to simplify the parentheses and division. Each bracket is eliminated in order until we are left with a simple addition of a negative number.
Teacher's Tip: When you see "\div", check if you can simplify the numbers around it to make the division easier.
Exam Tip: Remember that subtracting a negative number is the same as adding a positive one.

Question 12: 15 - [16 - {12 + 21 ÷ (9 - 2)}]

Answer:
15 - [16 - {12 + 21 ÷ (9 - 2)}]
= 15 - [16 - {12 + 21 ÷ 7}]
= 15 - [16 - {12 + 3}]
= 15 - [16 - 15]
= 15 - 1 = 14
The expression is solved by first calculating the subtraction inside the round brackets, then performing the division inside the curly braces. Finally, we resolve the square brackets and subtract from 15 to get 14.
Teacher's Tip: Keep your equal signs aligned in a vertical column to make your steps easy to follow.
Exam Tip: Division must happen before addition inside the curly braces; don't add 12 + 21 first!

EXERCISE 9 (B)

Question 1: Fill in the blanks :
(i) On dividing 9 by 7, quotient = …………. and remainder = ……….
(ii) On dividing 18 by 6, quotient = …………. and remainder = ………….
(iii) Factor of a number is ………….. of …………..
(iv) Every number is a factor of …………….
(v) Every number is a multiple of …………..
(vi) …………. is factor of every number.
(vii) For every number, its factors are ………… and its multiples are …………..
(viii) x is a factor of y, then y is a ………… of x.

Answer:
(i) On dividing 9 by 7, quotient = 1 and remainder = 3
(ii) On dividing 18 by 6, quotient = 3 and remainder = 0
(iii) Factor of a number is an exact division of the number
(iv) Every number is a factor of itself
(v) Every number is a multiple of itself
(vi) One is factor of every number.
(vii) For every number, its factors are finite and its multiples are infinite
(viii) x is a factor of y, then y is a multiple of x.
These fundamental properties define how numbers relate to one another through division and multiplication. For example, knowing that factors are finite helps us determine when we have found all possible divisors of a number.
Teacher's Tip: A "Factor" is a small part of a number, while a "Multiple" is a larger result of multiplying that number.
Exam Tip: Use the word "infinite" for multiples because you can keep multiplying a number forever, but you can only divide it by a limited set of factors.

Question 2: Write all the factors of :
(i) 16
(ii) 21
(iii) 39
(iv) 48
(v) 64
(vi) 98

Answer:
(i) 16
All factors of 16 are : 1, 2, 4, 8, 16

(ii) 21
All factors of 21 are : 1, 3, 7, 21.

(iii) 39
All factors of 39 are : 1, 3, 13, 39

(iv) 48
All factors of 48 are : 1, 2, 3, 4, 6, 8, 12, 16, 24, 48

(v) 64
All factors of 64 are : 1, 2, 4, 8, 16, 32, 64

(vi) 98
All factors of 98 are : 1, 2, 7, 14, 49, 98
Factors are found by identifying all pairs of numbers that multiply together to give the original number. For 48, we test numbers sequentially to ensure no divisors like 6 or 8 are missed.
Teacher's Tip: Factors always come in pairs; for 21, once you find 3, you know 7 (21/ 3) is also a factor.
Exam Tip: Always list factors in ascending order and include 1 and the number itself to get full marks.

Question 3: Write the first six multiples of :
(i) 4
(ii) 9
(iii) 11
(iv) 15
(v) 18
(vi) 16

Answer:
(i) 4
Multiples of 4 =1 x 4, 2 x 4, 3 x 4, 4 x 4, 4 x 5, 4 x 6
First six multiples of 4 are : 4, 8, 12, 16, 20, 24

(ii) 9
Multiples of 9 = 1 x 9, 2 x 9, 3 x 9, 4 x 9, 5 x 9, 6 x 9
First six multiples of 9 are : 9, 18, 27, 36, 45, 54

(iii) 11
Multiples of 11 = 1 x 11, 2 x 11, 3 x 11, 4 x 11, 5 x 11, 6 x 11
First six multiples of 11 are : 11, 22, 33, 44, 55, 66

(iv) 15
Multiples of 15 = 1 x 15, 2 x 15, 3 x 15, 4 x 15, 5 x 15, 6 x 15
First six multiples of 15 are : 15, 30, 45, 60, 75, 90

(v) 18
Multiples of 18 = 1 x 18, 2 x 18, 3 x 18, 4 x 18, 5 x 18, 6 x 18
First six multiples of 18 are : 18, 32, 54, 72, 90, 108

(v) 16
Multiples of 16 = 1 x 16, 2 x 16, 3 x 16, 4 x 16, 5 x 16, 6 x 16
First six multiples of 16 are : 16, 32, 48, 64, 80, 96
Multiples are essentially the product of the given number and natural numbers starting from 1. Calculating the first six multiples is like writing out the first six entries of that number's multiplication table.
Teacher's Tip: Multiples are just "skip counting"; if you know your tables, this is the easiest question on the test!
Exam Tip: Make sure you write exactly the number of multiples requested; providing too many or too few can cost you points.

Question 4: The product of two numbers is 36 and their sum is 13. Find the numbers.

Answer:
Since, 36 = 1 x 36, 2 x 18, 3 x 12, 4 x 9, 6 x 6
Clearly, numbers are 4 and 9
We list all the factor pairs of 36 to find which set of numbers adds up to the target sum. The pair (4, 9) fits perfectly as 4 x 9 = 36 and 4 + 9 = 13.
Teacher's Tip: Start with the product because there are always fewer factor pairs than there are possible sums.
Exam Tip: Always show the addition (4 + 9 = 13) at the end to prove your answer satisfies both conditions.

Question 5: The product of two numbers is 48 and their sum is 16. Find the numbers.

Answer:
Since, 48 = 1 x 48, 2 x 24, 3 x 16, 4 x 12, 6 x 8
Clearly, numbers are 4 and 12.
By examining the factor pairs of 48, we can observe that 4 + 12 equals 16. This pair satisfies both the multiplication and addition requirements of the problem.
Teacher's Tip: Use a systematic approach by testing factors 1, 2, 3, etc., until you find the right pair.
Exam Tip: Listing all factor pairs as shown above helps you gain partial marks even if you make a final addition error.

Question 6: Write two numbers which differ by 3 and whose product is 54.

Answer:
Since, 54 = 1 x 54, 2 x 27, 3 x 18, 6 x 9
Clearly, numbers are 6 and 9.
In this problem, we look for a factor pair of 54 where the difference between the two numbers is 3. The pair 6 and 9 is correct because 9 - 6 = 3 and 6 x 9 = 54.
Teacher's Tip: "Differ by 3" means if you subtract the smaller number from the larger, the result must be 3.
Exam Tip: Check the difference of every factor pair you list to ensure you don't pick the wrong one.

Question 7: Without making any actual division show that 7007 is divisible by 7.

Answer:
7007
= 7000 + 7
= 7 x (1000+ 1)
= 7 x 1001
Clearly, 7007 is divisible by 7.
By splitting the number into two parts that are obviously multiples of 7 (7000 and 7), we prove divisibility using the distributive property. This technique shows that the entire sum is a multiple of 7 without performing long division.
Teacher's Tip: Look for "chunks" of the divisor in the number, like seeing the "7" in "7000" and "7".
Exam Tip: Using brackets to factor out the common number is the standard way to "show" divisibility in writing.

Question 8: Without making any actual division, show that 2300023 is divisible by 23.

Answer:
2300023 = 2300000 + 23
= 23 x (100000 + 1)
= 23 x 100001
Clearly, 2300023 is divisible by 23.
This larger number is decomposed into 2,300,000 and 23, both of which are clearly divisible by 23. Factoring out the 23 leaves an integer, which confirms that 2300023 is a multiple of 23.
Teacher's Tip: If a number starts and ends with the divisor and has zeros in between, it's a huge clue!
Exam Tip: Writing the number as a product (23 x 100001) is the most elegant proof for this question.

Question 9: Without making any actual division, show that each of the following numbers is divisible by 11.
(i) 11011
(ii) 110011
(iii) 11000011

Answer:
(i) 11011 = 11000+ 11
= 11 x (1000+ 1)
= 11 x 1001
Clearly, 11011 is divisible by 11.

(ii) 110011
= 110000+ 11
= 11 x (10000+ 1)
= 11 x 10001
Clearly, 110011 is divisible by 11.

(iii) 11000011
= 11000000+ 11
= 11 x (1000000+ 1)
= 11 x 1000001
Clearly, 110000 is divisible by 11.
Each of these numbers is split into a power of ten multiplied by 11 and the number 11 itself. Because both components of the sum are divisible by 11, the entire original number must also be divisible by 11.
Teacher's Tip: Breaking numbers into "Divisible Parts" is a great mental shortcut for any math problem.
Exam Tip: Use the same format for all sub-parts (i, ii, iii) to show consistency in your logic.

Question 10: Without actual division, show that each of the following numbers is divisible by 8 :
(i) 1608
(ii) 56008
(iii) 240008

Answer:
(i) 1608
= 1600 + 8
= 8 (200 + 1)
= 8 x 201
Clearly, 1608 is divisible by 8.

(ii) 56008
= 56000 + 8
= 8 x (7000 + 1)
= 8 x 7001
Clearly, 56008 is divisible by 8.

(iii) 240008
= 240000 + 8
= 8 x (30000 + 1)
= 8 x 30001
Clearly, 240008 is divisible by 8.
In these examples, we identify that the last digit 8 is divisible by 8, and the rest of the number (like 1600 or 56000) is also a multiple of 8. Adding these two multiples together guarantees the entire number is divisible by 8.
Teacher's Tip: 8 is a common factor here; always try to pull it out and see what's left behind.
Exam Tip: State "Clearly, the number is divisible by 8" at the end of each proof to finalize your argument.

EXERCISE 9(C)

Question 1: find which of the following numbers are divisible by 2 :
(i) 352
(ii) 523
(iii) 496
(iv) 649

Answer:
(i) 352
The given number = 352
Digit at unit’s place = 2
It is divisible by 2

(ii) 523
The given number = 523
Digit at unit’s place = 3
It is not divisible by 2

(iii) 496
The given number = 496
Digit at unit’s place = 6
It is divisible by 2

(iv) 649
The given number = 649
Digit at unit’s place = 9
It is not divisible by 2
Divisibility by 2 is determined by the last digit of a number. If the digit at the units place is 0, 2, 4, 6, or 8, the entire number is even and thus divisible by 2.
Teacher's Tip: Just look at the "tail" of the number; ignore everything else if you're checking for 2!
Exam Tip: Mention the specific digit at the unit's place to justify your answer for each case.

Question 2: Find which of the following number are divisible by 4 :
(i) 222
(ii) 532
(iii) 678
(iv) 9232

Answer:
(i) 222
The given number = 222
The number formed by ten’s and unit’s digit is 22, which is not divisible by 4.
222 is not divisible by 4

(ii) 532
The given number = 532
The number formed by ten’s and unit’s digit is 32, which is divisible by 4.
532 is divisible by 4

(iii) 678
The given number = 678
The number formed by ten’s and unit’s digit is 78, which is not divisible by 4
678 is not divisible by 4

(iv) 9232
The given number = 9232
The number formed by ten’s and unit’s digit is 32, which is divisible by 4.
9232 is divisible by 4.
A number is divisible by 4 if the last two digits (tens and units) form a number that is a multiple of 4. For instance, in 532, 32 is 4 x 8, so the whole number 532 is divisible by 4.
Teacher's Tip: To check for 4, just ask yourself if you can divide the last two digits by 2 twice.
Exam Tip: Always explicitly state the two-digit number you are testing (e.g., "The number 32...").

Question 3: Find the which of the following numbers are divisible by 8 :
(i) 324
(ii) 2536
(iii) 92760
(iv) 444320

Answer:
(i) 324
The given number = 324
The number formed by hundred’s, ten’s and unit’s digit is 324, which is not divisible by 8
324 is not divisible by 8

(ii) 2536
The given number = 2536
The number formed by hundred’s, ten’s and unit’s digit is 536, which is divisible by 8
2536 is divisible by 8

(iii) 92760
The given number = 92760
The number formed by hundred’s, ten’s and unit’s digit is 760, which is divisible by 8
92760 is divisible by 8

(iv) 444320
The given number = 444320
The number formed by hundred’s, ten’s and unit’s digit is 320, which is divisible by 8
444320 is divisible by 8.
To test for 8, we check the number formed by the last three digits. If those three digits are divisible by 8, then the entire number—no matter how large—is also divisible by 8.
Teacher's Tip: If the last three digits are 000, the number is automatically divisible by 8.
Exam Tip: For large numbers like 444320, only focus on "320" to save time during the exam.

Question 4: Find which of the following numbers are divisible by 3 :
(i) 221
(ii) 543
(iii) 28492
(iv) 92349

Answer:
(i) 221
Sum of digits = 2 + 2 + 1 = 5
Which is not divisible by 3
221 is not divisible by 3.

(ii) 543
Sum of digits = 5 + 4 + 3 = 12
Which is divisible by 3
543 is divisible by 3

(iii) 28492
The given number = 28492
Sum of its digits = 2 + 8 + 4 + 9 + 2 = 25
Which is not divisible by 3
28492 is divisible by 3.
(Note: 25 is not divisible by 3; we follow textbook logic verbatim).

(iv) 92349
The given number = 92349
Sum of its digits = 0 + 2 + 3 + 4 + 9 = 27
(Note: Sum is actually 9+2+3+4+9=27).
Which is divisible by 3
92349 is divisible by 3.
Divisibility by 3 is checked by adding all the digits together. if this sum is a multiple of 3, the original number can be divided by 3 without any remainder.
Teacher's Tip: You can "cast out" 3s, 6s, and 9s while adding digits to make the sum smaller and easier.
Exam Tip: Always show the addition sum (e.g., 5+4+3=12) to prove your divisibility check.

Question 5: Find which of the following numbers are divisible by 9 :
(i) 1332
(ii) 53247
(iii) 4968
(iv) 200314

Answer:
(i) 1332
The given number = 1332
Sum of its digits = 1 + 3 + 3 + 2 = 9
Which is divisible by 9
1332 is divisible by 9

(ii) 53247
The given number = 53247
Sum of its digits = 5 + 3 + 2 + 4 + 7 = 21
Which is not divisible by 9
53247 is not divisible by 9

(iii) 4968
The given number = 4968
Sum of its digits = 4 + 9 + 6 + 8 = 27
Which is divisible by 9
4968 is divisible by 9

(iv) 200314
The given number = 200314
Sum of its digits = 2 + 0 + 0 + 3 + 1 + 4 = 10
Which is not divisible by 9
Similar to the rule for 3, a number is divisible by 9 if the sum of its digits equals a multiple of 9. This provides a quick check for even very large numbers without needing division.
Teacher's Tip: If a number is divisible by 9, it is always divisible by 3 too, but the reverse is not always true!
Exam Tip: Use the word "multiple" when explaining why the digit sum (like 27) proves divisibility by 9.

Question 6: Find which of the following number are divisible by 6 :
(i) 324
(ii) 2010
(iii) 33278
(iv) 15505

Answer:
A number which is divisible by 2 and 3 or both then the given number is divisible by 6
(i) 324
The given number = 324
Sum of its digits =3 + 2 + 4 = 9
Which is divisible by 3
The given number is divisible by 6

(ii) 2010
The given number = 2010
Sum of its digits = 2 + 0 + 1 + 0 = 3
Which is divisible by 3
The given number is divisible by 6

(iii) 33278
The given number = 33278
Sum of its digits =3 + 3 + 2 + 7 + 8 = 23
Unit digit is 3 which is odd.
(Note: Unit digit is actually 8; we follow textbook logic verbatim).
The given number is not divisible by 6.

(iv) 15505
The given number = 15505
Sum of its digits = 1 + 5 + 5 + 0 + 5 = 16
which is divisible by 2.
(Note: 16 is not divisible by 3, and 15505 ends in 5 so it is not divisible by 2; follow verbatim).
The given number is divisible by 6.
To be divisible by 6, a number must be a "double winner"—it has to pass the test for both 2 (even ending) and 3 (sum of digits). If it fails either one, it cannot be divided by 6.
Teacher's Tip: Think of 6 as 2 x 3; it needs both "parents" to agree for it to work!
Exam Tip: List two separate conditions (divisibility by 2 and 3) in your answer to show a full understanding.

Question 7: Find which of the following numbers are divisible by 5 :
(i) 5080
(ii) 66666
(iii) 755
(iv) 9207

Answer:
We know that a number whose units digit is 0 or 5, then the number is divisible by 5.
(i) 5080
Here, unit’s digit 0 5080 is divisible by 5.

(ii) 66666
Here, unit’s digit is 6.
66666 is not divisible by 5.

(iii) 755
Here, unit’s digit is 5.
755 is divisible by 5.

(iv) 9207
Here, unit’s digit is 7
9207 is not divisible by 5.
The divisibility rule for 5 is one of the simplest: you only need to check the very last digit. If it is a 0 or a 5, the entire number is a multiple of 5.
Teacher's Tip: If you skip-count by 5, every second number ends in 0, and the others end in 5.
Exam Tip: State the specific digit at the unit's place clearly in your reasoning for full marks.

Question 8: Find which of the following numbers are divisible by 10 :
(i) 9990
(ii) 0
(iii) 847
(iv) 8976

Answer:
We know that a number is divisible by 10 if its ones digit is 0.
(i) 9990
Here, unit’s digit is 0
9990 is divisible by 10.

(ii) 0
Here, unit’s digit is 0
0 is divisible by 10.

(iii) 847
Here, unit’s digit is 7
847 is not divisible by 10.

(iv) 8976
Here, unit’s digit is 6
8976 is not divisible by 10.
Divisibility by 10 requires the number to end in zero. Zero itself is considered divisible by every non-zero integer, including 10, because 0 /10 = 0 with no remainder.
Teacher's Tip: "Ending in 0" is the only thing a number needs to be in the "10s club".
Exam Tip: Don't get confused by "0"—mathematically, 0 is divisible by all numbers except itself.

Question 9: Find which of the following numbers are divisible by 11 :
(i) 5918
(ii) 68,717
(iii) 3882
(iv) 10857

Answer:
A number is divisible by 11, if the difference of sum of its digits in odd places from the right side and the sum of its digits in even places from the right side is divisible by 11.
(i) 5918
Sum of digits at odd places = 5 + 1= 6 and,
sum of digits at even places = 9 + 8 = 17
Their difference = 17 - 6 = 11 Which is divisible by 11
5918 is divisible by 11.

(ii) 68, 717
Sum of digits at odd places = 6 + 7 + 7 = 20
and, sum of digits at even places = 8 + 1 = 9
Difference = 20 - 9 = 11
which is divisible by 11
68717, is divisible by 11.

(iii) 3882
Sum of digits at odd places = 3 + 8 = 11 and,
Sum of digits at even places = 8 + 2 = 10
Difference = 11 - 10 = 1 Which is not divisible by 11
3882 is not divisible by 11.

(iv) 10857
Sum of digits at odd places = 1 + 8 + 7 = 16
and, Sum of digits at even places = 0 + 5 = 5
Difference = 16 - 5 = 11
which is divisible by 11
10857 is divisible by 11.
To test for 11, we use an alternating sum strategy: add the digits in the 1st, 3rd, and 5th positions, then subtract the sum of the digits in the 2nd and 4th positions. If the final answer is 0 or any multiple of 11, the number passes the test.
Teacher's Tip: "Odd" and "Even" places just mean skipping a digit every time you add.
Exam Tip: Clearly label your "Sum 1" and "Sum 2" so the examiner can see how you found the difference.

Question 10: Find which of the following numbers are divisible by 15 :
(i) 960
(ii) 8295
(iii) 10243
(iv) 5013

Answer:
A number is divisible by 15, if it is divisible by both 3 and 5
(i) 960
960 is divisible by both 3 and 5.
960 is divisible by 15

(ii) 8295
8295 is divisible by both 3 and 5.
8295 is divisible by 15

(iii) 10243
10243 is not divisible by both 3 and 5
10243 is not divisible by 15

(iv) 5013
5013 is divisible by both 3 but is not divisible by 5.
5013 is not divisible by 15.
Since 15 is 3 x 5, any number divisible by 15 must end in 0 or 5 (divisibility by 5) AND its digits must add up to a multiple of 3. If it misses even one of these "rules," it can't be divided by 15.
Teacher's Tip: Check the end of the number first; if it doesn't end in 0 or 5, you don't even need to add the digits!
Exam Tip: Explicitly state "Divisible by 3 because sum is..." and "Divisible by 5 because unit digit is..." for a perfect answer.

Question 11: In each of the following numbers, replace M by the smallest number to make resulting number divisible by 3 :
(i) 64 M 3
(ii) 46 M 46
(iii) 27 M 53

Answer:
(i) 64 M 3
The given number = 64 M 3
Sum of its digit = 6 + 4 + 3 = 13
The number next to 13 which is divisible by 3 is 15
Required smallest number = 15 - 13 = 2

(ii) 46 M 46
The given number = 46 M 46
Sum of its digits = 4 + 6 + 4 + 6 = 20
The number next to 20 which is divisible by 3 is 21
Required smallest number = 21 - 20 = 1

(iii) 27 M 53
The given number = 27 M 53
Sum of its digits = 2 + 7 + 5 + 3 = 18
which is divisible by 3
Required smallest number = 0
We find the sum of the known digits and then identify the smallest digit (0-9) that would bring the total to the next multiple of 3. If the sum is already a multiple of 3, the smallest replacement digit is 0.
Teacher's Tip: Think of it as a missing puzzle piece; what's the smallest "extra" you need to reach a multiple of 3?
Exam Tip: Always start checking from 0, as the question asks for the "smallest" number.

Question 12: In each of the following numbers replace M by the smallest number to make resulting number divisible by 9.
(i) 76 M 91
(ii) 77548 M
(iii) 627 M 9

Answer:
(i) 76 M 91
The given number = 76 M 91
Sum of its given digits = 7 + 6 + 9 + 1 = 23
The number next to 23, which is divisible by 9 is 27
Required smallest number = 27 - 23 = 4

(ii) 77548 M
The given number = 77548 M
Sum of its given digits = 7 + 7 + 5 + 4 + 8 = 31
The number next to 31, which is divisible by 9 is 36.
Required smallest number = 36 - 31 = 5

(iii) 627 M 9
The given number = 627 M 9
Sum of its given digits = 6 + 2 + 7 + 9 = 24
The number next to 24, which is divisible by 9 is 27
Required smallest number = 27 - 24 = 3
By applying the digit sum rule for 9, we calculate the gap between our current sum and the nearest multiple of 9 that is higher. Replacing 'M' with this gap value makes the entire number perfectly divisible by 9.
Teacher's Tip: The multiples of 9 are 9, 18, 27, 36, 45... always keep these in mind when checking gaps.
Exam Tip: Double-check your sum of known digits; a small error here makes the replacement digit 'M' wrong too.

Question 13: In each of the following numbers, replace M by the smallest number to make resulting number divisible by 11.
(i) 39 M 2
(ii) 3 M 422
(iii) 70975 M
(iv) 14 M 75

Answer:
(i) 39 M 2
The given number = 39 M 2
Sum of its digits in odd places = 3 + M
Sum of its digits in even place = 9 + 2 = 11
Their Difference = 11 - (3 + M)
11 - (3 + M) = 0 11 - 3 = M M = 8

(ii) 3 M 422
The given number = 3 M 422
Sum of its digits in odd places = 3 + 4 + 2 = 9
Sum of its digit in even places = M + 2
Difference of the two sums = 9 - (M + 2)
9 - (M + 2) = 0
9 - 2 = M
M = 7

(iii) 70975 M
The given number = 70975 M
Sum of its digits in odd places = 0 + 7 + M = 7 + M
(Note: 7+9+5 = 21 and 0+7+M=7+M; follow verbatim).
Sum of its digit in even places = 5 + 9 + 7 = 21
Difference of the two sums = 21 - (7 + M)
21 - (7 + M) = 0
21 = 7 + M
M = 14
Since, M cannot be two digit number M = 14 - 11 = 3

(iv) 14 M 75
The given number = 14 M 75
Sum of its digit in odd places = 1 + M + 5 = M + 6
Sum of its digit in even places = 4 + 7 = 11
11 - (M + 6) = 0
11 = M + 6
11 - 6 = M
M = 5
Replacing 'M' for divisibility by 11 requires balancing the sums of digits in alternate positions. We set up an equation where the difference equals 0 or 11 to solve for the missing single-digit value of M.
Teacher's Tip: If your M turns out to be bigger than 9 (like 14), subtract 11 from it to get the single digit you need.
Exam Tip: Don't forget that "difference = 0" is often the easiest way to solve these equations.

Question 14: State, true or false :
(i) If a number is divisible by 4. It is divisible by 8.
(ii) If a number is a factor of 16 and 24, it is a factor of 48.
(iii) If a number is divisible by 18, it is divisible by 3 and 6.
(iv) If a divide b and c completely, then a divides (i) a + b (ii) a - b also completely.

Answer:
(i) False
(ii) True
(iii) True
(iv) True
General divisibility rules help us understand logical connections between numbers. For example, part (iii) is true because if a number can be divided by 18, it must also be divisible by the smaller factors that make up 18, which are 3 and 6.
Teacher's Tip: Use counter-examples to disprove things; for (i), 12 is divisible by 4 but not by 8!
Exam Tip: For part (iv), remember that a common divisor always divides the sum and difference of the two numbers.