NCERT Solutions for Class 6 Maths Chapter 05 Prime Time

 

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Question 1. At what number is 'idli-vada' said for the 10th time?
Answer: 'Idli-vada' is said when a number is a multiple of both 3 and 5 (meaning multiples of 15). To find the 10th occurrence of 'idli-vada', we calculate 15 × 10 = 150. So the 10th time 'idli-vada' is said at the number 150.
In simple words: The number is 150. You say 'idli-vada' for every number that both 3 and 5 can divide into evenly.

Exam Tip: Identify what makes a number get each label - 'idli' for multiples of 3, 'vada' for multiples of 5, and 'idli-vada' for multiples of both.

 

Question 2. If the game is played for the numbers from 1 till 90, find out:
(a) How many times would the children say 'idli' (including the times they say 'idli-vada')?
(b) How many times would the children say 'vada' (including the times they say 'idli-vada')?
(c) How many times would the children say 'idli-vada'?
Answer:
(a) 'Idli' is said for multiples of 3. The multiples of 3 up to 90 are: 3, 6, 9, 12, ..., 90. There are 90/3 = 30 multiples of 3. However, 'idli-vada' is said for multiples of both 3 and 5, meaning multiples of 15. The multiples of 15 up to 90 are: 15, 30, 45, 60, 75, 90. So there are 6 multiples of 15. Therefore, the total number of times 'idli' is said (including 'idli-vada') is 30 (for multiples of 3) and no subtraction is needed for 'idli-vada' as they are already counted. Total times 'idli' is said: 30.
(b) 'Vada' is said for multiples of 5. The multiples of 5 up to 90 are: 5, 10, 15, 20, ..., 90. There are 90/5 = 18 multiples of 5. Subtract the 6 times 'idli-vada' is said (for multiples of 15). Total times 'vada' is said: 18.
(c) 'Idli-vada' is said for multiples of 15. The multiples of 15 up to 90 are: 15, 30, 45, 60, 75, 90. So, 'idli-vada' is said 6 times.
In simple words: Count how many numbers are divided by 3 (that is 30), how many by 5 (that is 18), and how many by both 3 and 5 together (that is 6).

Exam Tip: Remember that 'idli-vada' numbers are counted in both 'idli' and 'vada' totals, so do not subtract them from the individual counts.

 

Question 3. What if the game was played till 900? How would your answers change?
Answer: If the game is played till 900: Times 'idli' is said (including 'idli-vada'): There are 900/3 = 300 multiples of 3. Total times 'idli' is said: 300. Times 'vada' is said (including 'idli-vada'): There are 900/5 = 180 multiples of 5. Subtract the 60 times 'idli-vada' is said. Total times 'vada' is said: 180. Times 'idli-vada' is said: 'Idli-vada' is said for multiples of 15, and there are 900/15 = 60 multiples of 15. Total times 'idli-vada' is said: 60.
In simple words: When the game goes up to 900, there are more numbers to count, so all three counts go up - 'idli' becomes 300, 'vada' becomes 180, and 'idli-vada' becomes 60.

Exam Tip: Use division to find how many multiples of each number fit into the total range - this is faster than listing them all out.

 

Question 4. Is this figure somehow related to the 'idli-vada' game?
Answer: Yes, it is related to idli-vada game, if we assume 3 for idli and 5 for vada. The Venn diagram shows the relationship between multiples. The left circle holds multiples of 3, the right circle holds multiples of 5, and the overlapping section in the middle shows common multiples - which are the multiples of 15. This matches the pattern of the idli-vada game, where 'idli' are multiples of 3, 'vada' are multiples of 5, and 'idli-vada' are common multiples of both.
In simple words: The figure shows multiples of 3 on one side, multiples of 5 on the other, and where they overlap are the numbers you say 'idli-vada' for.

Exam Tip: Venn diagrams are a visual way to show which numbers belong to which group and which numbers belong to two groups at the same time.

 

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Question 5. Let us now play the 'idli-vada' game with different pairs of numbers:
(a) 2 and 5,
(b) 3 and 7,
(c) 4 and 6.
We will say 'idli' for multiples of the smaller number, 'vada' for multiples of the larger number and 'idli-vada' for common multiples. Draw a figure similar to Figure 5.1 if the game is played up to 60.
Answer:
(a) 2 and 5
For the pair 2 and 5, we say 'idli' for multiples of 2 (the smaller number) and 'vada' for multiples of 5 (the larger number). The common multiples of 2 and 5 up to 60 are multiples of 10. The Venn diagram shows multiples of 2 in the left circle, multiples of 5 in the right circle, and their common multiples (multiples of 10) in the overlapping region.
(b) 3 and 7
For the pair 3 and 7, we say 'idli' for multiples of 3 and 'vada' for multiples of 7. Since 3 and 7 share no common factors other than 1, they are co-prime. The multiples of 3 up to 60 are placed in the left circle, multiples of 7 in the right circle, and there are no numbers in the overlapping region because 3 and 7 do not share any common multiples up to 60 (the first common multiple would be 21, but we can check if any exist up to 60 - actually 21, 42 are common multiples). The Venn diagram displays this relationship.
(c) 4 and 6
For the pair 4 and 6, we say 'idli' for multiples of 4 and 'vada' for multiples of 6. The common multiples are multiples of 12 (the least common multiple of 4 and 6). The Venn diagram shows multiples of 4 in the left circle, multiples of 6 in the right circle, and their common multiples in the overlap.
In simple words: Draw two circles that overlap. One circle gets the multiples of the first number, the other gets multiples of the second number, and where they cross shows numbers that are multiples of both.

Exam Tip: The numbers in the overlap region are always multiples of the least common multiple (LCM) of the two numbers.

 

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Question 6. Which of the following could be the other number: 2, 3, 5, 8, 10?
Answer: We need to pick a number from the list (2, 3, 5, 8, 10) that shares common multiples with 4 in such a way that it would result in 'idli' (from 4) or 'idli-vada' but no 'vada' on its own. The number that fits this criterion is 8, because the multiples of 4 and 8 overlap in such a way that they would result in 'idli-vada' (when they are common multiples) and there would be no unique 'vada' for 8 since every multiple of 8 is also a multiple of 4, meaning you only say 'idli' or 'idli-vada' but not 'vada' alone.
In simple words: Pick the number where all its multiples are also multiples of 4. That number is 8, because every number you can count by 8s is also a number you can count by 4s.

Exam Tip: If one number's multiples are all contained within another number's multiples, the two numbers will not have unique 'vada' entries.

 

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Question 7. What jump size can reach both 15 and 30? There are multiple jump sizes possible. Try to find them all.
Answer: The factors of 15 are: 1, 3, 5, 15. The factors of 30 are: 1, 2, 3, 5, 6, 10, 15, 30. The common factors of 15 and 30 are: 1, 3, 5, 15. Thus, the jump sizes that can land on both 15 and 30 are 1, 3, 5 and 15.
In simple words: A jump size works if it divides both 15 and 30 evenly. The sizes that work are 1, 3, 5, and 15.

Exam Tip: Common factors of two numbers are the numbers that divide both of them without leaving a remainder - these are the jump sizes that will land on both targets.

 

Question 8. Look at the table below. What do you notice?

Answer:
1. The shaded numbers in the table are: 33, 36, 39, 42, 45, 48, 51, 54, 57, 60, 63, 66, 69. These are all multiples of 3. So, the numbers in shaded boxes are multiples of 3.
2. The circled numbers in the table are: 32, 36, 40, 44, 52, 56, 60, 64, 68. These are all multiples of 4. So, the numbers in circles are multiples of 4.
3. The numbers that are both shaded and circled are: 36, 48, 60, 64. These numbers are called common multiples of 12 (3 and 4 both).
In simple words: Shaded boxes show numbers that 3 goes into. Circles show numbers that 4 goes into. Where both appear together, those are the numbers that both 3 and 4 divide into evenly.

Exam Tip: When two different markings (shading and circles) appear on the same number, it shows a common multiple - a number that belongs to both groups.

 

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Question 9. Find all multiples of 40 that lie between 310 and 410.
Answer: The multiples of 40 between 310 and 410 are: 320, 360, 400.
In simple words: List out the numbers you get when you skip-count by 40s, and pick the ones between 310 and 410.

Exam Tip: Divide 310 and 410 by 40 to find which multiples fall in that range - this is faster than listing all multiples.

 

Question 10. Who am I?
(a) I am a number less than 40. One of my factors is 7. The sum of my digits is 8.
(b) I am a number less than 100. Two of my factors are 3 and 5. One of my digits is 1 more than the other.
Answer:
(a) The number is 35 (since 3 + 5 = 8 and 35 is divisible by 7).
(b) The number is 45 (since 4 + 1 = 5 and 45 has factors 3 and 5).
In simple words: Use the clues given to narrow down what the number could be. Check each clue one by one until you find the number that fits all of them.

Exam Tip: Work through the clues in order - first find all numbers that match the first clue, then see which of those also match the second clue.

 

Question 11. A number for which the sum of all its factors is equal to twice the number is called a perfect number. The number 28 is a perfect number. Its factors are 1, 2, 4, 7, 14 and 28. Their sum is 56 which is twice 28. Find a perfect number between 1 and 10.
Answer: The perfect number between 1 and 10 is 6. The factors of 6 are 1, 2, 3 and 6, and their sum is 12, which is twice 6.
In simple words: A perfect number is one where its divisors (except itself) add up to the number itself. For 6, the divisors 1, 2, and 3 add to 6.

Exam Tip: Find all factors first, add them up, then check if the sum equals twice the original number.

 

Question 12. Find the common factors of:
(a) 20 and 28
(b) 35 and 50
(c) 4, 8 and 12
(d) 5, 15 and 25
Answer:
(a) 20 and 28: The common factors are 1, 2 and 4.
(b) 35 and 50: The common factors are 1 and 5.
(c) 4, 8 and 12: The common factors are 1 2 and 4.
(d) 5, 15 and 25: The common factors are 1 and 5.
In simple words: Find all the divisors of each number, then pick out the ones that show up in every list.

Exam Tip: List the factors of each number carefully - even one missed factor can change the list of common factors.

 

Question 13. Find any three numbers that are multiples of 25 but not multiples of 50.
Answer: Three numbers that are multiples of 25 but not multiples of 50 are: 25, 75 and 125.
In simple words: Find numbers that 25 goes into evenly, but 50 does not. Multiples of 25 that are odd when divided by 50 work.

Exam Tip: Multiples of 25 that are not even are not multiples of 50 - so pick the odd multiples of 25.

 

Question 14. If the first time 'idli-vada' is said after the number 50 and the two numbers are smaller than 10, we have to find two numbers whose least common multiple is just above 50. The least common multiple of 6 and 9 is 54, so the two numbers could be 6 and 9. This means the first time they say 'idli-vada' would be at 54, as 54 is the first number that is a multiple of both 6 and 9.
Answer: We need two numbers whose least common multiple is just above 50. The least common multiple of 6 and 9 is 54, so the two numbers could be 6 and 9. This means the first time they say 'idli-vada' would be at 54, as 54 is the first number that is a multiple of both 6 and 9.
In simple words: Find two small numbers whose first common multiple comes right after 50. The numbers 6 and 9 work because their LCM is 54.

Exam Tip: The least common multiple (LCM) is the smallest number that both numbers divide into evenly.

 

Question 15. Anshu and his friends play the 'idli-vada' game with two numbers, which are both smaller than 10. The first time anybody says 'idlivada' is after the number 50. What could the two numbers be which are assigned 'idli' and 'vada'?
Answer: The two numbers could be 6 and 9. The least common multiple of 6 and 9 is 54. Since 54 is greater than 50 and is the first number where both 6 and 9 divide it evenly, the two numbers assigned are 'idli' for 6 and 'vada' for 9 (or vice versa).
In simple words: You need two numbers smaller than 10 whose LCM is just after 50. The numbers 6 and 9 work because LCM(6,9) = 54.

Exam Tip: The LCM must be slightly larger than 50 but still small enough that both numbers are less than 10.

 

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Question 16. How many prime numbers are there from 21 to 30? How many composite numbers are there from 21 to 30?
Answer: Prime numbers are numbers that have only two divisors: 1 and the number itself. The prime numbers between 21 and 30 are: 23 and 29. So, there are 2 prime numbers. Composite numbers are numbers that have more than two divisors. The composite numbers between 21 and 30 are: 22, 24, 25, 26, 27, 28 and 30. So, there are 7 composite numbers.
In simple words: Prime numbers only have two divisors - 1 and itself. Composite numbers can be divided by other numbers too. Between 21 and 30, there are 2 primes and 7 composites.

Exam Tip: Check each number in the range to see if it has exactly two divisors (prime) or more than two (composite).

 

Question 17. We see that 2 is a prime and also an even number. Is there any other even prime?
Answer: No, 2 is the only even prime number. This is because all other even numbers are divisible by 2, meaning they have more than two divisors (1, 2 and the number itself). Hence, they cannot be prime.
In simple words: Every even number except 2 can be divided by 2, so it has at least three divisors - 1, 2, and itself. That makes them composite, not prime.

Exam Tip: Use the divisibility rule for 2 - if a number ends in 0, 2, 4, 6, or 8, it is even and therefore not prime (unless it is 2 itself).

 

Question 18. Look at the list of primes till 100. What is the smallest difference between two successive primes? What is the largest difference?
Answer: The list of prime numbers up to 100 is: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89 and 97. The smallest difference between two successive primes is 1, which occurs between 2 and 3. The largest difference between two successive primes up to 100 is 8, which occurs between 89 and 97.
In simple words: Look at each pair of primes in order and subtract to find the gap. The smallest gap is 1 (between 2 and 3), and the largest gap up to 100 is 8 (between 89 and 97).

Exam Tip: When comparing consecutive primes, subtract the smaller from the larger to find the difference.

 

Question 19. Are there an equal number of primes occurring in every row in the table on the previous page? Which decades have the least number of primes? Which have the most number of primes?
Answer: No, there are not an equal number of primes in every row. Some rows contain more primes than others. The decades with the least number of primes are 40 - 49 and 90 - 99, each containing only 2 primes. The decades with the most number of primes are 0 - 9 and 30 - 39, each containing 4 primes.
In simple words: When you group primes by tens (like 0-9, 10-19, etc.), some groups have more primes than others. The 40s and 90s have only 2 primes each, while the 0s and 30s have 4 primes each.

Exam Tip: Count the primes in each row or decade carefully to make sure you do not miss any or count duplicates.

 

Question 20. Which of the following numbers are prime? 23, 51, 37, 26
Answer: The prime numbers from the list are 23 and 37. 23 is prime (it has no divisors other than 1 and 23). 51 is not prime (it is divisible by 1, 3 and 17). 37 is prime (it has no divisors other than 1 and 37). 26 is not prime (it is divisible by 1, 2 and 13).
In simple words: A prime has exactly two divisors. Check each number: 23 is prime, 51 is not, 37 is prime, and 26 is not.

Exam Tip: To check if a number is prime, try to divide it by numbers up to its square root - if none divide evenly, it is prime.

 

Question 21. Write three pairs of prime numbers less than 20 whose sum is a multiple of 5.
Answer: Three pairs of prime numbers less than 20 whose sum is a multiple of 5:
2 + 3 = 5
7 + 3 = 10
13 + 2 = 15
In simple words: Find pairs of primes (like 2 and 3, 7 and 3, 13 and 2) that add up to make a number that 5 divides into evenly.

Exam Tip: List all primes less than 20 first (2, 3, 5, 7, 11, 13, 17, 19), then pick pairs that sum to 5, 10, 15, 20, etc.

 

Question 22. The numbers 13 and 31 are prime numbers. Both these numbers have same digits 1 and 3. Find such pairs of prime numbers up to 100.
Answer: Such prime numbers up to 100 are: 13 and 31 (already given in question). 17 and 71. 37 and 73.
In simple words: Find two prime numbers where one is the reverse of the other - like 13 and 31, or 17 and 71.

Exam Tip: Look for primes where the digits are reversed - both must be prime themselves.

 

Question 23. Find seven consecutive composite numbers between 1 and 100.
Answer: The seven consecutive composite numbers between 1 and 100 are: 90, 91, 92, 93, 94, 95, and 96.
In simple words: Look for seven numbers in a row that are all composite (not prime). The numbers 90 through 96 all work.

Exam Tip: After you find one composite number, check if the next few numbers are also composite - this helps you find consecutive composites.

 

Question 24. Twin primes are pairs of primes having a difference of 2. For example, 3 and 5 are twin primes. So are 17 and 19. Find the other twin primes between 1 and 100.
Answer: Twin primes between 1 and 100: (3, 5), (5, 7), (11, 13), (17, 19), (29, 31), (41, 43), (59, 61) and (71, 73).
In simple words: Find pairs of primes that are exactly 2 apart - like 3 and 5, or 11 and 13.

Exam Tip: Look at the prime list and find pairs with a difference of exactly 2.

 

Question 25. Identify whether each statement is true or false. Explain.
(a) There is no prime number whose units digit is 4.
(b) A product of primes can also be prime.
(c) Prime numbers do not have any factors.
(d) All even numbers are composite numbers.
(e) 2 is a prime and so is the next number, 3. For every other prime, the next number is composite.
Answer:
(a) There is no prime number whose units digit is 4. - True. Because prime numbers cannot end in 4. Any number ending in 4 is divisible by 2, making it composite.
(b) A product of primes can also be prime. - False. Because the product of two primes will always have more than two divisors, so it cannot be prime.
(c) Prime numbers do not have any factors. - False. Because prime numbers do have factors - exactly two: 1 and the number itself.
(d) All even numbers are composite numbers. - False. Because the number 2 is an even prime, so not all even numbers are composite.
(e) 2 is a prime and so is the next number, 3. For every other prime, the next number is composite. - True. Because except for 2 and 3, all other primes have a composite number immediately after them.
In simple words: A number ending in 4 cannot be prime. Multiplying two primes makes something composite. Primes have exactly two divisors. The number 2 is even and prime. After 3, every next number after a prime tends to be composite.

Exam Tip: For true or false questions, give a clear reason - either show why it must be true, or give a counter-example to show it is false.

 

Question 26. Which of the following numbers is the product of exactly three distinct prime numbers: 45, 60, 91, 105, 330?
Answer: To find numbers that are the product of exactly three distinct prime numbers, we check: 60 = 2² × 3 × 5 (product of three distinct primes). 105 = 3 × 5 × 7 (product of three distinct primes). 330 = 2 × 3 × 5 × 11 (product of four distinct primes, so it doesn't qualify). The numbers are 60 and 105.
In simple words: Break each number into its prime factors. Count how many different primes are in each. Only 60 and 105 have exactly three different primes multiplied together.

Exam Tip: Find the prime factorization of each number carefully - count how many distinct primes appear in the factorization.

 

Question 27. How many three-digit prime numbers can you make using each of 2, 4 and 5 once?
Answer: The possible three-digit numbers using 2, 4 and 5 once each are: 245, 254, 425, 452, 524, 542. Checking each: 245, 254, 425, 452, 524, 542 - none of these are prime because they all have multiple factors or are even. Therefore, no three-digit prime numbers can be made using 2, 4 and 5 once.
In simple words: You can arrange 2, 4, and 5 in six different ways to make six different numbers. Check if each is prime. None of them are prime.

Exam Tip: To check if a number is prime, make sure it has only two divisors: 1 and itself. Numbers ending in 4, 2, or 5 (except 5 itself) are usually composite.

 

Question 28. Observe that 3 is a prime number, and 2 × 3 + 1 = 7 is also a prime. Are there other primes for which doubling and adding 1 gives another prime? Find at least five such examples.
Answer: Yes, here are such examples where doubling a prime and adding 1 gives another prime:
2 × 2 + 1 = 5 (both 2 and 5 are prime).
2 × 3 + 1 = 7 (both 3 and 7 are prime).
2 × 5 + 1 = 11 (both 5 and 11 are prime).
2 × 11 + 1 = 23 (both 11 and 23 are prime).
2 × 11 + 1 = 23 (both 11 and 23 are prime).
In simple words: Take a prime, double it, add 1. If the result is also prime, you have found an example. The primes 2, 3, 5, and 11 all work this way.

Exam Tip: Write out the formula (2 × p + 1) for different primes p and check if the result is prime.

 

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Question 29. Where should Grumpy place the treasures so that Jumpy cannot reach both the treasures?
Answer: To prevent Jumpy from reaching both treasures, Grumpy should place the treasures on two numbers that do not share any common factors (other than 1). These numbers are called co-prime. For example, Grumpy could place the treasures on 4 and 9, as they are co-prime (they don't have any common factors except 1). This would mean that Jumpy cannot reach both numbers with a single jump size, except for a jump size of 1, which is not allowed by the game rules. This strategy ensures that Jumpy will not be able to land on both numbers using any jump size other than 1.
In simple words: Place the treasures on two numbers that share no common divisors except 1. Then no jump size (other than 1) can land on both.

Exam Tip: Two numbers are co-prime if their greatest common factor (GCF) is 1 - this guarantees they share no common divisors.

 

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Question 30. Which of the following pairs of numbers are co-prime?
(a) 18 and 35
(b) 15 and 37
(c) 30 and 415
(d) 17 and 69
(e) 81 and 18
Answer:
(a) 18 and 35: Factors of 18: 1, 2, 3, 6, 9, 18 and Factors of 35: 1, 5, 7, 35. Common factor: 1. Co-prime: Yes.
(b) 15 and 37: Factors of 15: 1, 3, 5, 15 and Factors of 37: 1, 37. Common factor: 1. Co-prime: Yes.
(c) 30 and 415: Factors of 30: 1, 2, 3, 5, 6, 10, 15, 30 and Factors of 415: 1, 5, 83, 415. Common factor: 1, 5. Co-prime: No (They share a common factor of 5).
(d) 17 and 69: Factors of 17: 1, 17 and Factors of 69: 1, 3, 23, 69. Common factor: 1. Co-prime: Yes.
(e) 81 and 18: Factors of 81: 1, 3, 9, 27, 81 and Factors of 18: 1, 2, 3, 6, 9, 18. Common factors: 1, 3, 9. Co-prime: No (They share common factors of 3 and 9).
Final Result:
18 and 35 - Co-prime
15 and 37 - Co-prime
30 and 415 - Not co-prime
17 and 69 - Co-prime
81 and 18 - Not co-prime
In simple words: Two numbers are co-prime if the only number that divides both of them is 1. Find all divisors of each number, then check if they share any divisor besides 1.

Exam Tip: If two numbers share a common factor besides 1, they are not co-prime. Make sure to list all factors carefully.

 

Question 31. While playing the 'idli-vada' game with different number pairs, Anshu observed something interesting!
(a) Sometimes the first common multiple was the same as the product of the two numbers.
(b) At other times the first common multiple was less than the product of the two numbers.
Give examples for each of the above. How is it related to the number pair being co-prime?
Answer:
(a) Sometimes the first common multiple was the same as the product of the two numbers. This happens when the two numbers are co-prime, meaning they have no common factors other than 1. For example: Consider the pair 3 and 5. The product of 3 and 5 is 3 × 5 = 15. The least common multiple (LCM) of 3 and 5 is also 15. Since 3 and 5 are co-prime, their LCM is equal to their product.
(b) The first common multiple less than the product of the two numbers: This occurs when the two numbers are not co-prime (they share a common factor greater than 1). In such cases, the LCM is less than the product of the two numbers. For example: Consider the pair 4 and 6. The product of 4 and 6 is 4 × 6 = 24. The LCM of 4 and 6 is 12 (not 24) because they share a common factor (2). Since 4 and 6 are not co-prime, the LCM is less than their product.
Relation to Co-Prime Numbers: Co-prime numbers: When two numbers are co-prime (like 3 and 5), their LCM is the product of the two numbers because they do not share any common factors. Non co-prime numbers: When two numbers are not co-prime (like 4 and 6), their LCM is less than their product because the common factor allows for a smaller multiple.
So, if the numbers are co-prime, their LCM equals the product. If they are not co-prime, the LCM is less than the product.
In simple words: When two numbers share no common factors (co-prime), their smallest common multiple is their product. When they do share a factor, their smallest common multiple is smaller than their product.

Exam Tip: Use the relationship: If numbers are co-prime, LCM = product. If not, LCM - product because of shared factors.

 

Question 32. But where did Anshu go wrong?
Answer: To determine if two numbers are co-prime, all their prime factors must be examined carefully - doing a prime factorization of both numbers ensures that no common prime factors are overlooked. Anshu's mistake was incomplete factorization, which led to a wrong conclusion.
In simple words: When checking if two numbers are co-prime, break down each number into its prime factors completely. Anshu did not do this fully, so he missed a shared factor.

Exam Tip: Always do complete prime factorization - do not stop early, as you might miss a common factor that appears later in the breakdown.

 

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Question 33. Co-prime Art
Observe the following thread art. The first diagram has 12 pegs and the thread is tied to every fourth peg (we say that the thread-gap is 4). The second diagram has 13 pegs and the thread-gap is 3. What about the other diagrams? Observe these pictures, share and discuss your findings in class.
Make such pictures for the following:
(a) 15 pegs, thread-gap of 10
(b) 10 pegs, thread-gap of 7
(c) 14 pegs, thread-gap of 6
(d) 8 pegs, thread-gap of 3
Answer:
Third diagram: 16 pegs with a thread-gap of 6.
Fourth diagram: 24 pegs with a thread-gap of 6.

When you look at the diagrams, the key observation is whether the number of pegs and the thread-gap are co-prime (share no common factors other than 1). If they are co-prime, the thread will visit every peg and form a complete star pattern. If they are not co-prime, the thread will miss some pegs and create smaller designs within the circle.
In simple words: When the number of pegs and the gap size are co-prime, the thread touches every peg. When they are not co-prime, the thread skips some pegs.

Exam Tip: Check if the number of pegs and the thread-gap are co-prime - if they are, the thread will visit all pegs; if not, it will create a smaller pattern.

 

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Question 34. But where did Anshu go wrong?
Answer: To check if two numbers are co-prime, you must consider all their factors, especially by doing a prime factorisation of both numbers to make sure that no common prime factors exist. Anshu's mistake was incomplete factorisation, leading to an incorrect conclusion.
In simple words: Break both numbers down into their prime factors completely. Anshu only partially factorized one or both numbers, missing a shared prime factor.

Exam Tip: Always perform complete prime factorization - stopping too early means you might miss a shared factor that shows they are not co-prime.

 

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Question 35. Find the prime factorizations of the following numbers: 64, 104, 105, 243, 320, 141, 1728, 729, 1024, 1331, 1000.
Answer:
Prime factorizations:
64 = 2⁶
105 = 3 × 5 × 7
320 = 2⁶ × 5
1728 = 2⁶ × 3³
1024 = 2¹⁰
1000 = 2³ × 5³
104 = 2³ × 13
243 = 3⁵
141 = 3 × 47
729 = 3⁶
1331 = 11³
In simple words: Break each number into primes by dividing repeatedly. Keep dividing until you cannot divide anymore, then write as powers of the prime factors.

Exam Tip: Work systematically by dividing by 2, then 3, then 5, then 7, and so on. Stop when you reach 1.

 

Question 36. The prime factorisation of a number has one 2, two 3s, and one 11. What is the number?
Answer: The required number where prime factorization with one 2, two 3s and one 11: 2 × 3² × 11 = 198. The number is 198.
In simple words: Multiply together one 2, two 3s, and one 11 to get your answer: 2 × 3 × 3 × 11 = 198.

Exam Tip: If you know the prime factors, just multiply them together - remember to count duplicates correctly.

 

Question 37. Find the prime factorisation of these numbers without multiplying first:
(a) 56 × 25
(b) 108 × 75
(c) 1000 × 81
Answer:
(a) 56 × 25: First, find prime factorizations: 56 = 2³ × 7 and 25 = 5². So the prime factorization of 56 × 25 = 2³ × 7 × 5².
(b) 108 × 75: First, find prime factorizations: 108 = 2² × 3³ and 75 = 3 × 5². So the prime factorization of 108 × 75 = 2² × 3³ × 3 × 5² = 2² × 3⁴ × 5².
(c) 1000 × 81: First, find prime factorizations: 1000 = 2³ × 5³ and 81 = 3⁴. So the prime factorization of 1000 × 81 = 2³ × 3⁴ × 5³.
In simple words: Do not multiply first. Instead, break each number into primes separately, then combine all the prime factors together.

Exam Tip: When multiplying factored numbers, combine like primes by adding their exponents.

 

Question 38. Smallest numbers with prime factorisation:
(a) What is the smallest number whose prime factorisation has: three different prime numbers?
(b) What is the smallest number whose prime factorisation has: four different prime numbers?
Answer:
(a) Three different prime numbers: 2 × 3 × 5 = 30
(b) Four different prime numbers: 2 × 3 × 5 × 7 = 210
In simple words: Use the smallest primes available and multiply each once. For three primes, multiply 2, 3, and 5. For four primes, multiply 2, 3, 5, and 7.

Exam Tip: To get the smallest number with a given prime factorization, use the smallest primes (2, 3, 5, 7, ...) and use each only once.

 

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Question 39. Are the following pairs of numbers co-prime? Guess first and then use prime factorisation to verify your answer.
(a) 30 and 45
(b) 57 and 85
(c) 121 and 1331
(d) 343 and 216
Answer:
(a) 30 and 45: 30 = 2 × 3 × 5 and 45 = 3² × 5. Not co-prime (common factors: 3 and 5).
(b) 57 and 85: 57 = 3 × 19 and 85 = 5 × 17. Co-prime (no common factors).
(c) 121 and 1331: 121 = 11² and 1331 = 11³. Not co-prime (common factor: 11).
(d) 343 and 216: 343 = 7³ and 216 = 2³ × 3³. Co-prime (no common factors).
In simple words: Break each number into its prime factors. If both numbers share a prime factor, they are not co-prime. If they share no prime factors, they are co-prime.

Exam Tip: Write out the complete prime factorization of each number, then compare to see if any prime factor appears in both.

 

Question 40. Is the first number divisible by the second? Use prime factorisation.
(a) 225 and 27
(b) 96 and 24
(c) 343 and 17
(d) 999 and 99
Answer:
(a) 225 and 27: 225 = 5² × 3² and 27 = 3³. No, 225 is not divisible by 27 (225 has only two 3's, while 27 has three 3's).
(b) 96 and 24: 96 = 2⁵ × 3 and 24 = 2³ × 3. Yes, 96 is divisible by 24.
(c) 343 and 17: 343 = 7³ and 17 is a prime number. No, 343 is not divisible by 17.
(d) 999 and 99: 999 = 3³ × 37 and 99 = 3² × 11. No, 999 is not divisible by 99 (since 999 has 37 as a factor and 99 has 11 as a factor).
In simple words: For the first number to divide by the second, all of the second number's prime factors must appear in the first number with equal or higher powers.

Exam Tip: Compare the prime factorizations - the first number must have every prime factor of the second number, with at least as many copies.

 

Question 41. First number: 2 × 3 × 7 and second number: 3 × 7 × 11. Are they co-prime? Does one of them divide the other?
Answer: First number: 2 × 3 × 7. Second number: 3 × 7 × 11. They are not co-prime because they have common factors (3 and 7). Neither number divides the other because neither contains all the prime factors of the other.
In simple words: They are not co-prime because both have 3 and 7. The first has 2 which the second lacks, and the second has 11 which the first lacks, so neither divides the other.

Exam Tip: If two numbers share common factors but neither has all the factors of the other, they are neither co-prime nor does one divide the other.

 

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Question 42. Consider this statement: Numbers that are divisible by 10 are those that end with '0'. Do you agree?
Answer: Yes, I agree. Numbers that are divisible by 10 always end with a '0'. This is because 10 is a multiple of both 2 and 5, and for a number to be divisible by 10, it must have both 2 and 5 as factors. When a number ends in 0, it indicates that the number can be divided by 10 without leaving a remainder. For example: 20, 30, 100 and 150 all end in 0 and they are divisible by 10.
In simple words: Any number ending in 0 can be divided by 10 evenly. Any number that 10 divides into will always end in 0.

Exam Tip: Divisibility by 10 is simple - just check the last digit. If it is 0, the number is divisible by 10.

 

Question 43. Consider this statement: Numbers that are divisible by 5 are those that end with either a '0' or a '5'. Do you agree?
Answer: Yes, I agree. Numbers that are divisible by 5 always end with either '0' or '5'. This is because the divisibility rule for 5 states that a number must have 5 as one of its factors. For this to happen, the last digit of the number must be either 0 or 5. For example: Numbers like 15, 25, 40 and 55 are all divisible by 5 because they end in 5 or 0.
In simple words: If a number ends in 0 or 5, you can divide it by 5 evenly. Numbers like 15, 25, 40 and 55 are all divisible by 5.

Exam Tip: For divisibility by 5, check only the last digit - if it is 0 or 5, the number is divisible by 5.

 

Question 44. Consider this statement: Numbers that are divisible by 2 are those that end with '0', '2', '4', '6' or '8'. Do you agree? What are all the multiples of 2 between 399 and 411?
Answer: Yes, I agree. Numbers that are divisible by 2 are those that end with '0', '2', '4', '6' or '8'. This is because for a number to be divisible by 2, it must be an even number, and all even numbers end in these digits. The multiples of 2 between 399 and 411 are: 400, 402, 404, 406, 408 and 410. These numbers end in 0, 2, 4, 6 or 8, confirming that they are divisible by 2.
In simple words: If a number ends in 0, 2, 4, 6, or 8, it is divisible by 2 (it is even). Between 399 and 411, those numbers are 400, 402, 404, 406, 408, and 410.

Exam Tip: Divisibility by 2 is easy - just check if the number is even (ends in 0, 2, 4, 6, or 8).

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Question 1. Find numbers between 330 and 340 that are divisible by 4. Also, find numbers between 1730 and 1740, and 2030 and 2040, that are divisible by 4. What do you observe?
Answer: Numbers that can be divided evenly by 4 are:
Between 330 and 340: 332, 336 and 340.
Between 1730 and 1740: 1732, 1736 and 1740.
Between 2030 and 2040: 2032, 2036 and 2040.
Observation: Across these different ranges, the numbers divisible by 4 appear at regular intervals, separated by 4.
In simple words: When you look for numbers that 4 can divide evenly, they always come spaced exactly 4 apart, no matter which range you check.

Exam Tip: The pattern of divisibility remains consistent regardless of the number range - this is a key observation to mention.

 

Question 2. Is 8536 divisible by 4?
Answer: Yes, 8536 can be divided evenly by 4. To confirm this, take the last two digits (36) and check if they divide evenly by 4. Since 36 is divisible by 4, the whole number 8536 is also divisible by 4.
In simple words: Look only at the last two digits. If those two digits divide evenly by 4, then the whole number does too.

Exam Tip: Always use the divisibility rule for 4 (check only the last two digits) to save time rather than performing full division.

 

Question 3. Consider these statements:
(a) Only the last two digits matter when deciding if a given number is divisible by 4.
(b) If the number formed by the last two digits is divisible by 4, then the original number is divisible by 4.
(c) If the original number is divisible by 4, then the number formed by the last two digits is divisible by 4.
Do you agree? Why or why not?
Answer:
(a) Only the last two digits matter when deciding if a given number is divisible by 4. True - When checking whether 4 divides a number evenly, you need only examine these final two digits. The earlier digits have no impact on this divisibility.
(b) If the number formed by the last two digits is divisible by 4, then the original number is divisible by 4. True - This is the key rule for divisibility by 4. When the last two digits form a number that 4 divides, the entire number is also divisible by 4.
(c) If the original number is divisible by 4, then the number formed by the last two digits is divisible by 4. True - If a number can be divided evenly by 4, then its last two digits will always form a number that 4 can divide as well.
In simple words: All three statements are correct. The last two digits decide everything about divisibility by 4.

Exam Tip: Remember that all three conditions are equivalent - any one of them can be used to establish the divisibility rule for 4.

 

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Question 4. Find numbers between 120 and 140 that are divisible by 8. Also, find numbers between 1120 and 1140, and 3120 and 3140, that are divisible by 8. What do you observe?
Answer: Numbers that can be divided evenly by 8 are:
Between 120 and 140: 120, 128 and 136.
Between 1120 and 1140: 1120, 1128 and 1136.
Between 3120 and 3140: 3120, 3128 and 3136.
Observation: The numbers divisible by 8 follow a consistent pattern across these different ranges, appearing at intervals of 8.
In simple words: Numbers that 8 divides evenly keep showing up in the same pattern, spaced 8 apart, in every range.

Exam Tip: The spacing between multiples of 8 remains constant across all number ranges - use this observation to predict other multiples.

 

Question 5. Change the last two digits of 8560 so that the resulting number is a multiple of 8.
Answer: The last two digits of 8560 are "60", and 60 cannot be divided evenly by 8. By changing the final two digits to "64", we get a number divisible by 8. The new number is 8564, which is a multiple of 8.
In simple words: Swap out "60" with "64" to make 8564, which 8 can divide evenly.

Exam Tip: When changing digits to meet divisibility requirements, always verify your answer by checking the divisibility rule.

 

Question 6. Consider this statement:
(a) Only the last three digits matter when deciding if a given number is divisible by 8.
(b) If the number formed by the last three digits is divisible by 8, then the original number is divisible by 8.
(c) If the original number is divisible by 8, then the number formed by the last three digits is divisible by 8.
Do you agree? Why or why not?
Answer:
(a) Only the last three digits matter when deciding if a given number is divisible by 8. False - Because only the final three digits are checked when testing divisibility by 8, as we work with powers of 2.
(b) If the number formed by the last three digits is divisible by 8, then the original number is divisible by 8. True - If the last three digits form a number that 8 divides, then the whole number is divisible by 8. If these three final digits divide evenly by 8, the entire number will as well.
(c) If the original number is divisible by 8, then the number formed by the last three digits is divisible by 8. True - If a number can be divided evenly by 8, then its last three digits will always form a number that 8 can divide.
In simple words: For divisibility by 8, the last three digits tell you everything you need to know.

Exam Tip: The divisibility rule for 8 depends on checking the last three digits, not just two - this is because 8 is a higher power of 2 than 4.

 

Page 125 - Figure it Out

 

Question 7. 2024 is a leap year (as February has 29 days). Leap years occur in the years that are multiples of 4, except for those years that are evenly divisible by 100 but not 400.
(a) From the year you were born till now, which years were leap years?
(b) From the year 2024 till 2099, how many leap years are there?
Answer:
(a) Leap years occur every 4 years (except for years divisible by 100 but not 400). A person born in 2011 would have experienced leap years in 2012, 2016 and 2020 from 2011 to 2024.
(b) Starting from 2024, leap years happen every 4 years: 2024, 2028, 2032, 2036, 2040, 2044, 2048, 2052, 2056, 2060, 2064, 2068, 2072, 2076, 2080, 2084, 2088, 2092 and 2096. There are 19 leap years between 2024 and 2099.
In simple words: Leap years come every 4 years. Count them out and you'll find 19 of them from 2024 to 2099.

Exam Tip: Remember the exception rule: years divisible by 100 are NOT leap years unless they are also divisible by 400 (e.g., 1900 was not a leap year, but 2000 was).

 

Question 8. Find the largest and smallest 4-digit numbers that are divisible by 4 and are also palindromes.
Answer: A palindrome is a number that reads the same forwards and backwards. The smallest 4-digit palindrome divisible by 4 is 1221. The largest 4-digit palindrome divisible by 4 is 8888.
In simple words: A palindrome looks the same when you read it both ways. Find the smallest and largest ones among 4-digit numbers where 4 divides evenly.

Exam Tip: Check the last two digits of palindromes to verify divisibility by 4 - in a 4-digit palindrome ABBA, check if BA is divisible by 4.

 

Question 9. Explore and find out if each statement is always true, sometimes true or never true. You can give examples to support your reasoning
(a) Sum of two even numbers gives a multiple of 4.
(b) Sum of two odd numbers gives a multiple of 4.
Answer:
(a) Sum of two even numbers gives a multiple of 4. Sometimes true - While both even numbers are multiples of 2, they may not both be multiples of 4, so their sum is not always divisible by 4. For instance, 2 + 6 = 8 (which is a multiple of 4), but 2 + 12 = 14 (which is not a multiple of 4).
(b) Sum of two odd numbers gives a multiple of 4. Always true - Any two odd numbers always add to an even number that 4 can divide. For example, 3 + 5 = 8 and 9 + 7 = 16.
In simple words: When you add two odd numbers, you always get a result that 4 divides evenly. But adding two even numbers doesn't always work that way.

Exam Tip: Use concrete examples to test whether statements hold in all cases, some cases, or never - this systematic approach prevents errors.

 

Question 10. Find the remainders obtained when each of the following numbers are divided by
(i) 10,
(ii) 5,
(iii) 2
Numbers: 78, 99, 173, 572, 980, 1111, 2345
Answer:
Remainders of 78:
(i) 78 ÷ 10 = remainder 8
(ii) 78 ÷ 5 = remainder 3
(iii) 78 ÷ 2 = remainder 0

Remainders of 99:
(i) 99 ÷ 10 = remainder 9
(ii) 99 ÷ 5 = remainder 4
(iii) 99 ÷ 2 = remainder 1

Remainders of 173:
(i) 173 ÷ 10 = remainder 3
(ii) 173 ÷ 5 = remainder 3
(iii) 173 ÷ 2 = remainder 1

Remainders of 572:
(i) 572 ÷ 10 = remainder 2
(ii) 572 ÷ 5 = remainder 2
(iii) 572 ÷ 2 = remainder 0

Remainders of 980:
(i) 980 ÷ 10 = remainder 0
(ii) 980 ÷ 5 = remainder 0
(iii) 980 ÷ 2 = remainder 0

Remainders of 1111:
(i) 1111 ÷ 10 = remainder 1
(ii) 1111 ÷ 5 = remainder 1
(iii) 1111 ÷ 2 = remainder 1

Remainders of 2345:
(i) 2345 ÷ 10 = remainder 5
(ii) 2345 ÷ 5 = remainder 0
(iii) 2345 ÷ 2 = remainder 1
In simple words: When you divide these numbers, the leftover part (remainder) depends only on the last digit of the number.

Exam Tip: Notice that the remainder when dividing by 10, 5, or 2 always depends only on the last digit - use this shortcut to avoid full division calculations.

 

Question 11. The teacher asked if 14560 is divisible by all of 2, 4, 5, 8 and 10. Guna checked for divisibility of 14560 by only two of these numbers and then declared that it was also divisible by all of them. What could those two numbers be?
Answer: The two key numbers to check would be 4 and 5. If a number is divisible by 4 and 5, it must also be divisible by 2, 8, and 10. This is because 4 and 5 are factors that, when combined properly, ensure divisibility by all the others on the list.
In simple words: Check if 4 divides the number and if 5 divides it. If both do, then you know all five numbers will divide it.

Exam Tip: Understand the relationships between divisors - knowing divisibility by certain key numbers lets you deduce divisibility by others.

 

Question 12. Which of the following numbers are divisible by all of 2, 4, 5, 8 and 10: 572, 2352, 5600, 6000, 77622160.
Answer: The numbers that meet the conditions of divisibility by 2, 4, 5, 8, and 10 are: 5600, 6000 and 77622160. These three numbers satisfy all the divisibility requirements because they meet the conditions needed for each of the five divisors.
In simple words: Only 5600, 6000 and 77622160 can be divided evenly by all five of these numbers.

Exam Tip: To check divisibility by all these numbers at once, verify divisibility by 8 and 5 - if both work, all five divisors will divide the number.

 

Question 13. Write two numbers whose product is 10,000. The two numbers should not have 0 as the units digit.
Answer: Two numbers whose product equals 10,000 and which do not end in 0 are 25 and 400. When you multiply 25 × 400, you get 10,000. Both numbers meet the requirement of not having 0 as the units digit (25 ends in 5, and 400 ends in 0 - actually, 400 does end in 0, so this example doesn't work). A better pair would be 25 and 400, but since 400 ends in 0, try 50 and 200 (but these also end in 0). The correct answer is: there is no valid pair, as any two factors of 10,000 that multiply to give 10,000 will have at least one number ending in 0 or 5.
In simple words: Find two numbers that multiply to make 10,000 but neither can end in 0. This is actually impossible, since 10,000 = 10^4, which always needs factors of 2 and 5.

Exam Tip: Recognize that 10,000 = 2^4 × 5^4, so any factorization will include at least one factor containing 5, which means at least one number must end in 0 or 5.