Chapter 5 exercise answers: Prime Time

Open NCERT p. 108Matches NCERT’s answer

1. 'Idli-vada' is said on a number that is a multiple of both 3 and 5.
2. The numbers that are multiples of both 3 and 5 are 15, 30, 45, 60, ... — they go up in steps of 15.
3. So the 10th such number is 15 × 10.
4. 15 × 10 = 150.

Answer150

Watch this explained “Counting to 90 — and 900”, 4:29 into Common multiples: why the first "idli-vada" is 15 · हिंदी में देखें

Open NCERT p. 108Matches NCERT’s answer

(a) How many times would the children say ‘idli’…

1. 'Idli' is said on every multiple of 3, including the numbers where 'idli-vada' is said.
2. From 1 to 90, the multiples of 3 are 3, 6, 9, ..., 90.
3. Number of multiples of 3 up to 90 = 90 ÷ 3 = 30.

Answer30

(b) How many times would the children say ‘vada’…

1. 'Vada' is said on every multiple of 5, including the numbers where 'idli-vada' is said.
2. Number of multiples of 5 up to 90 = 90 ÷ 5 = 18.

Answer18

(c) How many times would the children say ‘idli-vada’?

1. 'Idli-vada' is said on numbers that are multiples of both 3 and 5, i.e. multiples of 15.
2. Number of multiples of 15 up to 90 = 90 ÷ 15 = 6.

Answer6

Watch this explained “Counting to 90 — and 900”, 4:29 into Common multiples: why the first "idli-vada" is 15 · हिंदी में देखें

Open NCERT p. 108Matches NCERT’s answer

1. The same idea from question 2 works, just with 900 in place of 90.
2. Multiples of 3 up to 900: 900 ÷ 3 = 300 times, so 'idli' is said 300 times.
3. Multiples of 5 up to 900: 900 ÷ 5 = 180 times, so 'vada' is said 180 times.
4. Multiples of 15 up to 900: 900 ÷ 15 = 60 times, so 'idli-vada' is said 60 times.
5. Each answer is exactly 10 times the answer for 1 to 90, because 900 is 10 times 90.

Answer'idli' 300 times, 'vada' 180 times, 'idli-vada' 60 times.

Watch this explained “Counting to 90 — and 900”, 4:29 into Common multiples: why the first "idli-vada" is 15 · हिंदी में देखें

Open NCERT p. 108One way to think about it

1. The figure has two circles: one for multiples of 3, one for multiples of 5, drawn up to 30.
2. Every number that is only a multiple of 3 (like 3, 6, 9, 12, 18, 21, 24, 27) goes in the left circle only — these are the 'idli' numbers.
3. Every number that is only a multiple of 5 (like 5, 10, 20, 25) goes in the right circle only — these are the 'vada' numbers.
4. Numbers that are multiples of both 3 and 5 (15 and 30) go in the overlap — these are the 'idli-vada' numbers.
5. So the figure is exactly a picture of the game: 'idli' numbers, 'vada' numbers, and the overlap of 'idli-vada' numbers.
6. If the game were played till 60, the left circle would hold 3, 6, 9, 12, 18, 21, 24, 27, 33, 36, 39, 42, 48, 51, 54, 57; the right circle would hold 5, 10, 20, 25, 35, 40, 50, 55; and the overlap would hold 15, 30, 45, 60.

In shortYes — the figure is a picture of the game itself: the left circle is the 'idli' numbers (multiples of 3), the right circle is the 'vada' numbers (multiples of 5), and the overlap is the 'idli-vada' numbers (multiples of both, i.e. multiples of 15). Playing till 60 just means listing more numbers in each part, using the same rule.

Watch this explained “Naming: common multiples”, 2:57 into Common multiples: why the first "idli-vada" is 15 · हिंदी में देखें

Open NCERT p. 110Matches NCERT’s answer

1. 310 ÷ 40 = 7.75, so the first multiple of 40 after 310 is 40 × 8 = 320.
2. Keep adding 40: 320, 360, 400.
3. The next one, 400 + 40 = 440, is past 410, so we stop at 400.

Answer320, 360, 400

Watch this explained “Running the machine backwards”, 7:09 into Common factors: which jump sizes can land on a number · हिंदी में देखें

Open NCERT p. 111Matches NCERT’s answer

(a) …less than 40… factors is 7… sum of my digits is 8.

1. The multiples of 7 less than 40 are 7, 14, 21, 28, 35.
2. Check the digit sum of each: 7→7, 14→5, 21→3, 28→10, 35→8.
3. Only 35 has digit sum 8.

Answer35

(b) …less than 100… 3 and 5… is 1 more than the other.

1. Having 3 and 5 both as factors means the number is a multiple of 15.
2. The multiples of 15 less than 100 are 15, 30, 45, 60, 75, 90.
3. Check which one has digits that differ by 1: 15(4), 30(3), 45(1), 60(6), 75(2), 90(9).
4. Only 45 has digits 4 and 5, which differ by 1.

Answer45

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Open NCERT p. 111Matches NCERT’s answer

1. Check numbers between 1 and 10 by listing their factors and adding them up.
2. 6 has factors 1, 2, 3, 6. Their sum is 1 + 2 + 3 + 6 = 12.
3. 12 is twice 6, so 6 is a perfect number.

Answer6

Watch this explained “Perfect numbers”, 8:57 into Common factors: which jump sizes can land on a number · हिंदी में देखें

Open NCERT p. 111Matches NCERT’s answer

(a) 20 and 28

1. Factors of 20: 1, 2, 4, 5, 10, 20.
2. Factors of 28: 1, 2, 4, 7, 14, 28.
3. Numbers in both lists: 1, 2, 4.

Answer1, 2, 4

(b) 35 and 50

1. Factors of 35: 1, 5, 7, 35.
2. Factors of 50: 1, 2, 5, 10, 25, 50.
3. Numbers in both lists: 1, 5.

Answer1, 5

(c) 4, 8 and 12

1. Factors of 4: 1, 2, 4. Factors of 8: 1, 2, 4, 8. Factors of 12: 1, 2, 3, 4, 6, 12.
2. Numbers common to all three lists: 1, 2, 4.

Answer1, 2, 4

(d) 5, 15 and 25

1. Factors of 5: 1, 5. Factors of 15: 1, 3, 5, 15. Factors of 25: 1, 5, 25.
2. Numbers common to all three lists: 1, 5.

Answer1, 5

Watch this explained “Both lists, and the overlap”, 4:00 into Common factors: which jump sizes can land on a number · हिंदी में देखें

Open NCERT p. 111Checked by computerAnswers can differ: one example

1. List multiples of 25: 25, 50, 75, 100, 125, 150, ...
2. Every second one of these (50, 100, 150, ...) is also a multiple of 50, so skip those.
3. The remaining ones — 25, 75, 125, 175, ... — are multiples of 25 but not of 50.
4. Pick any three of them, for example 25, 75 and 125.

Answer25, 75, 125 (any three odd multiples of 25 work)

Watch this explained “Running the machine backwards”, 7:09 into Common factors: which jump sizes can land on a number · हिंदी में देखें

Open NCERT p. 111Checked by computerAnswers can differ: one example

1. 'Idli-vada' is first said at the first number that is a multiple of both numbers. So we need two numbers, both smaller than 10, whose first common multiple is more than 50.
2. The first common multiple can never be more than the product of the two numbers. So the product must be more than 50. With both numbers smaller than 10, the only products above 50 are 6 × 9 = 54, 7 × 8 = 56, 7 × 9 = 63 and 8 × 9 = 72.
3. 6 and 9 do not work: 18 is already a multiple of both, and 18 is not after 50.
4. 7 and 8: multiples of 8 are 8, 16, 24, 32, 40, 48, 56; the first one that is also a multiple of 7 is 56. It is after 50, so this pair works.
5. In the same way, 7 and 9 first meet at 63, and 8 and 9 first meet at 72. So the question has three correct answers: 7 and 8, 7 and 9, or 8 and 9.
6. One answer: 'idli' is said for the smaller number, 7, and 'vada' for the larger number, 8. The first 'idli-vada' is then at 56.

Answeridli = 7, vada = 8 (one of three possible pairs)

Watch the lesson Common multiples: why the first "idli-vada" is 15 · हिंदी में देखें

Open NCERT p. 111Matches NCERT’s answer

1. A jump size lands on a number only when it is a factor of that number.
2. Factors of 28: 1, 2, 4, 7, 14, 28.
3. Factors of 70: 1, 2, 5, 7, 10, 14, 35, 70.
4. Numbers common to both lists: 1, 2, 7, 14.

Answer1, 2, 7, 14

Watch this explained “Both lists, and the overlap”, 4:00 into Common factors: which jump sizes can land on a number · हिंदी में देखें

Open NCERT p. 111Checked by computerAnswers can differ: one example

1. The overlap holds the common multiples of the two numbers: 24, 48 and 72. They go up in steps of 24.
2. Nothing smaller than 24 is in the overlap, so 24 is the first common multiple of the two numbers. (If it were 12, then 12, 36 and 60 would be in the overlap too.)
3. So both numbers are factors of 24, and their first common multiple is exactly 24. Also, both outer regions have numbers to fill in, so neither number can be a multiple of the other.
4. Three pairs fit: 8 and 12, 6 and 8, or 3 and 8. Any one of them is a correct answer. (A pair like 3 and 24 also meets at 24, but then the circle for 24 would have no numbers of its own.)
5. Check 8 and 12: multiples of 8 are 8, 16, 24, 32, 40, 48, 56, 64, 72; multiples of 12 are 12, 24, 36, 48, 60, 72. The common ones are 24, 48 and 72, just as in the diagram.
6. Filling in the numbers up to 72, the largest number shown: Multiples of 8 only: 8, 16, 32, 40, 56, 64. Multiples of 12 only: 12, 36, 60.

AnswerMultiples of 8 and multiples of 12 (one of three possible pairs)

Watch this explained “The same idea in a grid”, 5:37 into Common factors: which jump sizes can land on a number · हिंदी में देखें

Open NCERT p. 111Matches NCERT’s answer

1. The number must be a multiple of 8 and of 9. Multiples of 9: 9, 18, 27, 36, 45, 54, 63, 72. The first one that is also a multiple of 8 is 72, and every common multiple of 8 and 9 is a multiple of 72.
2. The number must also be a multiple of 5. Multiples of 72: 72, 144, 216, 288, 360. A multiple of 5 ends in 0 or 5, and the first of these that does is 360.
3. Now check the other numbers: 360 ÷ 2 = 180, 360 ÷ 3 = 120, 360 ÷ 4 = 90, 360 ÷ 6 = 60, 360 ÷ 10 = 36, and 1 divides every number. So 360 is a multiple of all of 1, 2, 3, 4, 5, 6, 8, 9 and 10.
4. 72, 144, 216 and 288 are the only smaller common multiples of 8 and 9, and none of them is a multiple of 5. So 360 is the smallest such number.

Answer360

Watch this explained “Challenge: one to ten”, 8:48 into Common multiples: why the first "idli-vada" is 15 · हिंदी में देखें

Open NCERT p. 111Matches NCERT’s answer

1. From the previous question, the smallest number that is a multiple of every number from 1 to 10 except 7 is 360.
2. Now 7 must divide the number too, and 360 has no 7 hidden inside it, so multiply by 7.
3. 360 × 7 = 2520.
4. 2520 is divisible by every number from 1 to 10.

Answer2520

Watch this explained “Challenge: one to ten”, 8:48 into Common multiples: why the first "idli-vada" is 15 · हिंदी में देखें

Open NCERT p. 114Matches NCERT’s answer

1. Any even number bigger than 2 can be divided by 2, so 2 is always one of its factors.
2. That means such a number has at least three factors: 1, 2 and itself.
3. A prime number can only have exactly two factors, so no even number bigger than 2 can be prime.
4. So 2 is the only even prime.

AnswerNo

Watch this explained “The only even prime”, 5:13 into Prime and composite: the rectangle test for a number's factors · हिंदी में देखें

Open NCERT p. 114Matches NCERT’s answer

1. The primes up to 100 are 2, 3, 5, 7, 11, 13, ..., 89, 97.
2. Find the gap between each prime and the next one, for example 3 − 2 = 1, 5 − 3 = 2, and so on.
3. The smallest gap is between 2 and 3, which is 1.
4. The largest gap is between 89 and 97, which is 8.

AnswerSmallest difference is 1 (between 2 and 3); largest difference is 8 (between 89 and 97).

Watch this explained “Gaps, neighbours and thinning”, 5:52 into The sieve of Eratosthenes: finding primes by removing multiples · हिंदी में देखें

Open NCERT p. 114Matches NCERT’s answer

1. Count the primes in each row of ten: 1–10 has 4 (2,3,5,7); 11–20 has 4 (11,13,17,19); 21–30 has 2; 31–40 has 2; 41–50 has 3; 51–60 has 2; 61–70 has 2; 71–80 has 3; 81–90 has 2; 91–100 has 1 (97).
2. The counts are not all the same, so the rows do not have an equal number of primes.
3. The smallest count is 1, in the decade 91–100.
4. The largest count is 4, shared by the decades 1–10 and 11–20.

AnswerNo, the rows do not have equal numbers of primes. Least: 91–100 (just 1 prime). Most: 1–10 and 11–20 (4 primes each).

Watch this explained “Gaps, neighbours and thinning”, 5:52 into The sieve of Eratosthenes: finding primes by removing multiples · हिंदी में देखें

Open NCERT p. 114Matches NCERT’s answer

1. 23: 2, 3 and 4 do not divide 23. Since 5 × 5 = 25 is already more than 23, any factor pair of 23 would need a number smaller than 5, and there is none. So 23 has only the factors 1 and 23: it is prime.
2. 51: 51 = 3 × 17, so 51 has the factors 1, 3, 17 and 51. It is not prime.
3. 37: 2, 3, 4, 5 and 6 do not divide 37, and 7 × 7 = 49 is more than 37, so we can stop. 37 has only the factors 1 and 37: it is prime.
4. 26: 26 is even, 26 = 2 × 13. It is not prime.

Answer23 and 37

Watch this explained “Testing a number by hand”, 5:56 into Prime and composite: the rectangle test for a number's factors · हिंदी में देखें

Open NCERT p. 114Checked by computerAnswers can differ: one example

1. Primes less than 20: 2, 3, 5, 7, 11, 13, 17, 19.
2. Try sums of pairs and check if they are multiples of 5: 2 + 3 = 5 ✓; 2 + 13 = 15 ✓; 3 + 7 = 10 ✓.
3. These three pairs all work.

Answer(2, 3), (2, 13), (3, 7) — other correct pairs also exist.

Watch this explained “Prime, and composite”, 3:21 into Prime and composite: the rectangle test for a number's factors · हिंदी में देखें

Open NCERT p. 114Checked by computerReads two ways: both answers shown

1. Reverse the digits of each two-digit prime up to 100 and see if the new number is also prime.
2. 13 ↔ 31, 17 ↔ 71, 37 ↔ 73 and 79 ↔ 97 work: both numbers are prime.
3. Others do not: for example 19 → 91 = 7 × 13, 23 → 32 is even, 29 → 92 is even. 11 reversed is 11 again, so it does not make a pair of two numbers.
4. The question can be read two ways. NCERT's answer key reads it as 'find other pairs like 13 and 31', so we lead with that: 17 and 71, 37 and 73, 79 and 97.
5. Read as 'find all such pairs up to 100', the given pair counts too: 13 and 31, 17 and 71, 37 and 73, 79 and 97.

AnswerRead as pairs other than 13 and 31 (NCERT's key): 17 and 71, 37 and 73, 79 and 97. Read as all such pairs up to 100: 13 and 31, 17 and 71, 37 and 73, 79 and 97.

Watch this explained “Twins, reversals, and a hunt”, 8:22 into Prime and composite: the rectangle test for a number's factors · हिंदी में देखें

Open NCERT p. 114Checked by computerAnswers can differ: one example

1. The primes 89 and 97 are far apart, with a gap of 8, which is the largest gap below 100.
2. That means all the numbers strictly between them are composite: 90, 91, 92, 93, 94, 95, 96.
3. That is exactly seven numbers in a row, all composite (for example, 91 = 7 × 13, which is easy to miss).

Answer90, 91, 92, 93, 94, 95, 96

Watch this explained “Seven composites in a row”, 6:52 into The sieve of Eratosthenes: finding primes by removing multiples · हिंदी में देखें

Open NCERT p. 114Matches NCERT’s answer

1. List all primes up to 100 using the sieve you already made.
2. Look for pairs of primes that are 2 apart.
3. 3 & 5 and 17 & 19 are already given, so skip those.
4. The remaining twin prime pairs below 100 are: 5 & 7, 11 & 13, 29 & 31, 41 & 43, 59 & 61, 71 & 73.

Answer5 & 7, 11 & 13, 29 & 31, 41 & 43, 59 & 61, 71 & 73

Watch this explained “Twins, reversals, and a hunt”, 8:22 into Prime and composite: the rectangle test for a number's factors · हिंदी में देखें

Open NCERT p. 115One way to think about it

(a) There is no prime number whose units digit is 4.

1. A number whose units digit is 4 is even, so 2 is one of its factors.
2. It also has the factors 1 and itself, and it is bigger than 2 (the smallest such number is 4).
3. So it has at least three factors — 1, 2 and itself — and is composite, never prime.

In shortTrue

(b) A product of primes can also be prime.

1. Take any product of two or more primes, for example 2 × 3 = 6.
2. The product has 1 and itself as factors, and also each prime that went into it (for 6: 1, 2, 3 and 6).
3. Each of those primes is bigger than 1 and smaller than the product, so the product has more than two factors.
4. So a product of primes is always composite, never prime.

In shortFalse

(c) Prime numbers do not have any factors.

1. Every prime has exactly two factors: 1 and itself.
2. For example, the factors of 7 are 1 and 7.
3. So primes do have factors — just no factors other than 1 and themselves.

In shortFalse

(d) All even numbers are composite numbers.

1. 2 is even, and its only factors are 1 and 2.
2. So 2 is prime, not composite.
3. One even number that is not composite is enough to make the statement false.

In shortFalse

(e) 2 … 3. For every other prime, the next number is composite.

1. Every prime other than 2 is odd, because every even number other than 2 has 2 as a factor.
2. The number right after an odd prime is even and at least 4 (3 → 4, 5 → 6, 7 → 8, 11 → 12).
3. An even number bigger than 2 has 2 as a factor, so it is composite.
4. So for every prime except 2, the next number is composite.

In shortTrue

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Open NCERT p. 115Matches NCERT’s answer

1. Find the prime factorisation of each number.
2. 45 = 3 × 3 × 5 — only 2 different primes.
3. 60 = 2 × 2 × 3 × 5 — 3 different primes, but 2 is repeated, so it is not simply three primes multiplied once each.
4. 91 = 7 × 13 — only 2 different primes.
5. 105 = 3 × 5 × 7 — exactly three different primes, none repeated.
6. 330 = 2 × 3 × 5 × 11 — 4 different primes.

Answer105

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Open NCERT p. 115Matches NCERT’s answer

1. Every arrangement of the digits 2, 4, 5 ends in one of these three digits.
2. A number ending in 2 or 4 is even, so it can be divided by 2 — not prime.
3. A number ending in 5 can be divided by 5 — not prime.
4. So every single arrangement of 2, 4, 5 fails to be prime.

Answer0 — none of the arrangements is prime

Watch this explained “Twins, reversals, and a hunt”, 8:22 into Prime and composite: the rectangle test for a number's factors · हिंदी में देखें

Open NCERT p. 115Checked by computerAnswers can differ: one example

1. Yes, there are many such primes; any five correct examples answer the question. Here is one set.
2. Take a prime p, work out 2 × p + 1, and check whether the result is prime.
3. p = 5: 2 × 5 + 1 = 11, which is prime.
4. p = 11: 2 × 11 + 1 = 23, which is prime.
5. p = 23: 2 × 23 + 1 = 47, which is prime.
6. p = 29: 2 × 29 + 1 = 59, which is prime.
7. p = 41: 2 × 41 + 1 = 83, which is prime.
8. Other primes that also work include 2, 53, 83 and 89. Not every prime works: for p = 7, 2 × 7 + 1 = 15 = 3 × 5.

AnswerYes — for example 5, 11, 23, 29 and 41 (2p + 1 = 11, 23, 47, 59 and 83, all prime)

Watch this explained “Testing a number by hand”, 5:56 into Prime and composite: the rectangle test for a number's factors · हिंदी में देखें

Open NCERT p. 120Matches NCERT’s answer

1. Keep dividing each number by the smallest prime that divides it, until only 1 is left.
2. 64 = 2⁶
3. 104 = 2³ × 13
4. 105 = 3 × 5 × 7
5. 243 = 3⁵
6. 320 = 2⁶ × 5
7. 141 = 3 × 47
8. 1728 = 2⁶ × 3³
9. 729 = 3⁶
10. 1024 = 2¹⁰
11. 1331 = 11³
12. 1000 = 2³ × 5³

Answer64=2⁶; 104=2³×13; 105=3×5×7; 243=3⁵; 320=2⁶×5; 141=3×47; 1728=2⁶×3³; 729=3⁶; 1024=2¹⁰; 1331=11³; 1000=2³×5³

Watch this explained “Breaking 56 all the way down”, 2:06 into Prime factorisation, and why there is only one of them per number · हिंदी में देखें

Open NCERT p. 120Matches NCERT’s answer

1. Multiply the given prime factors together: 2 × 3 × 3 × 11.
2. 2 × 3 = 6
3. 6 × 3 = 18
4. 18 × 11 = 198

Answer198

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Open NCERT p. 120Matches NCERT’s answer

1. 1955 ends in 5, so 5 divides it: 1955 ÷ 5 = 391.
2. Now break 391 by trying primes in order: 2, 3, 5, 7, 11 and 13 all leave a remainder (for example 391 = 7 × 55 + 6, 391 = 11 × 35 + 6, 391 = 13 × 30 + 1).
3. The next prime, 17, works: 391 ÷ 17 = 23.
4. 23 is prime, so 1955 = 5 × 17 × 23.
5. Check: 5 × 17 = 85 and 85 × 23 = 1955. All three primes, 5, 17 and 23, are less than 30.

Answer5, 17 and 23

Watch this explained “Breaking 56 all the way down”, 2:06 into Prime factorisation, and why there is only one of them per number · हिंदी में देखें

Open NCERT p. 120Matches NCERT’s answer

(a) 56 × 25

1. 56 = 2 × 2 × 2 × 7
2. 25 = 5 × 5
3. Put both lists together: 2 × 2 × 2 × 5 × 5 × 7

Answer2³ × 5² × 7

(b) 108 × 75

1. 108 = 2 × 2 × 3 × 3 × 3
2. 75 = 3 × 5 × 5
3. Put both lists together: 2 × 2 × 3 × 3 × 3 × 3 × 5 × 5

Answer2² × 3⁴ × 5²

(c) 1000 × 81

1. 1000 = 2 × 2 × 2 × 5 × 5 × 5
2. 81 = 3 × 3 × 3 × 3
3. Put both lists together: 2 × 2 × 2 × 3 × 3 × 3 × 3 × 5 × 5 × 5

Answer2³ × 3⁴ × 5³

Watch this explained “Factorising without multiplying”, 8:21 into Prime factorisation, and why there is only one of them per number · हिंदी में देखें

Open NCERT p. 120Matches NCERT’s answer

(a) three different prime numbers?

1. To keep the number as small as possible, use the three smallest primes: 2, 3 and 5.
2. Multiply them once each: 2 × 3 × 5 = 30.

Answer30

(b) four different prime numbers?

1. Use the four smallest primes: 2, 3, 5 and 7.
2. Multiply them once each: 2 × 3 × 5 × 7 = 210.

Answer210

Watch this explained “Two names, carefully distinguished”, 3:19 into Prime factorisation, and why there is only one of them per number · हिंदी में देखें

Open NCERT p. 122Matches NCERT’s answer

(a) 30 and 45

1. 30 = 2 × 3 × 5
2. 45 = 3 × 3 × 5
3. They share the primes 3 and 5.

AnswerNo

(b) 57 and 85

1. 57 = 3 × 19
2. 85 = 5 × 17
3. They share no prime factor.

AnswerYes

(c) 121 and 1331

1. 121 = 11 × 11
2. 1331 = 11 × 11 × 11
3. They both have 11 as a prime factor.

AnswerNo

(d) 343 and 216

1. 343 = 7 × 7 × 7
2. 216 = 2 × 2 × 2 × 3 × 3 × 3
3. They share no prime factor.

AnswerYes

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Open NCERT p. 122Matches NCERT’s answer

(a) 225 and 27

1. 225 = 3 × 3 × 5 × 5
2. 27 = 3 × 3 × 3
3. 27's factorisation needs three 3s, but 225 only has two.

AnswerNo

(b) 96 and 24

1. 96 = 2 × 2 × 2 × 2 × 2 × 3
2. 24 = 2 × 2 × 2 × 3
3. Every prime in 24's list is inside 96's list.

AnswerYes

(c) 343 and 17

1. 343 = 7 × 7 × 7
2. 17 is prime and does not appear anywhere in 343's factorisation.

AnswerNo

(d) 999 and 99

1. 999 = 3 × 3 × 3 × 37
2. 99 = 3 × 3 × 11
3. 99's factorisation needs an 11, but 999 has none.

AnswerNo

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Open NCERT p. 122Matches NCERT’s answer

1. First number = 2 × 3 × 7 = 42. Second number = 3 × 7 × 11 = 231.
2. Both lists share the primes 3 and 7, so they are not co-prime.
3. 42's list (2, 3, 7) is not fully inside 231's list (missing the 2), and 231's list (3, 7, 11) is not fully inside 42's list (missing the 11).
4. So neither number divides the other.

AnswerNot co-prime; neither number divides the other

Watch this explained “Two tests, and a claim about primes”, 7:15 into Reading co-primality and divisibility straight off the prime factorisation · हिंदी में देखें

Open NCERT p. 122Matches NCERT’s answer

1. A prime number's only factors are 1 and itself.
2. Take any two different primes — they cannot share any factor bigger than 1, because neither one's only other factor (itself) equals the other prime.
3. So their only common factor is 1, which is exactly what co-prime means.

AnswerYes, Guna is right

Watch this explained “Two tests, and a claim about primes”, 7:15 into Reading co-primality and divisibility straight off the prime factorisation · हिंदी में देखें

Open NCERT p. 125One way to think about it

(a) From the year you were born till now, … leap years?

1. The answer depends on the year you were born, so every student's list is different.
2. Rule: a year is a leap year if it is a multiple of 4, except years divisible by 100 but not by 400.
3. Example: someone born in 2014. The multiples of 4 from 2014 to 2026 are 2016, 2020 and 2024.
4. None of them is divisible by 100, so all three are leap years.

In shortIt depends on your birth year. For example, for someone born in 2014: 2016, 2020 and 2024.

(b) From the year 2024 till 2099, how many leap years are there?

1. The multiples of 4 from 2024 to 2099 start at 2024 and end at 2096 (2100 is outside the range).
2. Count them: (2096 − 2024) ÷ 4 + 1 = 72 ÷ 4 + 1 = 18 + 1 = 19.
3. No year from 2024 to 2099 is divisible by 100, so none of them has to be removed.

In short19 leap years

Watch this explained “Leap years, palindromes, and a product with no zero”, 9:02 into Divisibility tests for 10, 5, 2, 4 and 8: why only the last digits matter · हिंदी में देखें

Open NCERT p. 125Checked by computer

1. A 4-digit palindrome has the form abba, for example 2112: its last two digits are b then a.
2. Only the last two digits decide whether a number is divisible by 4.
3. Largest: with a = 9 the number ends in 9, which is odd, so no b works. With a = 8, try b = 9: 8998 ends in 98, and 98 ÷ 4 leaves remainder 2 — no. b = 8: 8888 ends in 88, and 88 ÷ 4 = 22 — yes.
4. Smallest: with a = 1 the number ends in 1, which is odd, so no b works. With a = 2, try b = 0: 2002 ends in 02, and 2 is not divisible by 4 — no. b = 1: 2112 ends in 12, and 12 ÷ 4 = 3 — yes.
5. The answer key at the back of the book prints 9999 and 1001, but both are odd numbers, so neither can be divided by 4; the largest is 8888 (8888 = 4 × 2222) and the smallest is 2112 (2112 = 4 × 528).

AnswerLargest: 8888; smallest: 2112

Watch this explained “Leap years, palindromes, and a product with no zero”, 9:02 into Divisibility tests for 10, 5, 2, 4 and 8: why only the last digits matter · हिंदी में देखें

Open NCERT p. 125Matches NCERT’s answer

(a) Sum of two even numbers gives a multiple of 4.

1. 2 + 2 = 4, which is a multiple of 4.
2. 2 + 4 = 6, which is not a multiple of 4.
3. So it works sometimes, not always.

AnswerSometimes true

(b) Sum of two odd numbers gives a multiple of 4.

1. 1 + 3 = 4, which is a multiple of 4.
2. 1 + 5 = 6, which is not a multiple of 4.
3. So it works sometimes, not always.

AnswerSometimes true

Open NCERT p. 126Matches NCERT’s answer

1. Dividing by 10: the remainder is the last digit of the number.
2. Dividing by 5: the part of the number before the last digit is a whole number of tens, and 10 is a multiple of 5, so only the last digit matters. The remainder is the last digit if it is 0 to 4, and the last digit minus 5 if it is 5 to 9.
3. Dividing by 2: the remainder is 0 if the last digit is even and 1 if it is odd.

4. Number	÷10	÷5	÷2
   78	8	3	0
   99	9	4	1
   173	3	3	1
   572	2	2	0
   980	0	0	0
   1111	1	1	1
   2345	5	0	1

Answer÷10: 8, 9, 3, 2, 0, 1, 5 · ÷5: 3, 4, 3, 2, 0, 1, 0 · ÷2: 0, 1, 1, 0, 0, 1, 1

Watch the lesson Divisibility tests for 10, 5, 2, 4 and 8: why only the last digits matter · हिंदी में देखें

Open NCERT p. 126Checked by computerAnswers can differ: one example

1. This question has two correct answers; here is one, and the other is given at the end.
2. If a number is divisible by 8, it is also divisible by 4 and by 2, because 4 and 2 both divide 8.
3. If a number is divisible by 5, it ends in 0 or 5. If it is also divisible by 8 it is even, so it must end in 0 — which means it is divisible by 10.
4. So checking 8 and 5 is enough to be sure about all of 2, 4, 5, 8 and 10.
5. For 14560: its last three digits are 560, and 560 ÷ 8 = 70, so 8 divides it; it ends in 0, so 5 divides it.
6. 8 and 10 also works: 10 covers 5 and 2, and 8 covers 4. No other pair works — for example, 4 and 10 both divide 20, but 8 does not.

Answer8 and 5 (8 and 10 also works)

Watch this explained “Two checks that buy the other three”, 7:04 into Divisibility tests for 10, 5, 2, 4 and 8: why only the last digits matter · हिंदी में देखें

Open NCERT p. 126Matches NCERT’s answer

1. A number is divisible by all of 2, 4, 5, 8 and 10 exactly when it is divisible by 40 (their LCM).
2. 572: last three digits 572 ÷ 8 leaves a remainder, and it doesn't end in 0 — fails.
3. 2352: does not end in 0 — fails.
4. 5600: ends in 0, and 600 ÷ 8 = 75 exactly — passes.
5. 6000: ends in 0, and 000 ÷ 8 = 0 exactly — passes.
6. 77622160: ends in 0, and 160 ÷ 8 = 20 exactly — passes.

Answer5600, 6000 and 77622160

Watch this explained “Two checks that buy the other three”, 7:04 into Divisibility tests for 10, 5, 2, 4 and 8: why only the last digits matter · हिंदी में देखें

Open NCERT p. 126Checked by computerAnswers can differ: one example

1. 10000 = 2 × 2 × 2 × 2 × 5 × 5 × 5 × 5.
2. A number ends in 0 only if it has both a 2 and a 5 among its factors.
3. So give all the 2s to one number and all the 5s to the other: 2×2×2×2 = 16 and 5×5×5×5 = 625.
4. Check: 16 × 625 = 10000, and neither 16 nor 625 ends in 0.

Answer16 and 625

Watch this explained “Leap years, palindromes, and a product with no zero”, 9:02 into Divisibility tests for 10, 5, 2, 4 and 8: why only the last digits matter · हिंदी में देखें