Chapter 3 exercise answers: Finding Common Ground

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(a) 90

1. 90 = 2 × 3 × 3 × 5.
2. List every piece you can cut from this row: nothing, one prime, two primes, and so on.

Answer1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 90

(b) 105

1. 105 = 3 × 5 × 7.
2. Every combination of these three primes gives a factor.

Answer1, 3, 5, 7, 15, 21, 35, 105

(c) 132

1. 132 = 2 × 2 × 3 × 11.
2. Combine the primes in every possible way.

Answer1, 2, 3, 4, 6, 11, 12, 22, 33, 44, 66, 132

(d) 360 (this number has 24 factors)

1. 360 = 2 × 2 × 2 × 3 × 3 × 5.
2. Adding 1 to each prime's count (3+1, 2+1, 1+1) and multiplying gives 4 × 3 × 2 = 24 factors — matching what the book tells us.

Answer1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 18, 20, 24, 30, 36, 40, 45, 60, 72, 90, 120, 180, 360

(e) 840 (this number has 32 factors)

1. 840 = 2 × 2 × 2 × 3 × 5 × 7.
2. (3+1) × (1+1) × (1+1) × (1+1) = 4 × 2 × 2 × 2 = 32 factors — matching the book.

Answer1, 2, 3, 4, 5, 6, 7, 8, 10, 12, 14, 15, 20, 21, 24, 28, 30, 35, 40, 42, 56, 60, 70, 84, 105, 120, 140, 168, 210, 280, 420, 840

Watch this explained “225, listed properly”, 4:38 into Reading every factor of a number off its prime factorisation · हिंदी में देखें

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(a) 50, 60

1. 50 = 2 × 5 × 5, 60 = 2 × 2 × 3 × 5.
2. Shared pieces: 2, 5, and 2 × 5.

AnswerCommon factors 1, 2, 5, 10. HCF = 10

(b) 140, 275

1. 140 = 2 × 2 × 5 × 7, 275 = 5 × 5 × 11.
2. Only the single 5 is shared.

AnswerCommon factors 1, 5. HCF = 5

(c) 77, 725

1. 77 = 7 × 11, 725 = 5 × 5 × 29.
2. No prime is shared.

AnswerCommon factor 1 only. HCF = 1

(d) 370, 592

1. 370 = 2 × 5 × 37, 592 = 2 × 2 × 2 × 2 × 37.
2. Shared pieces: 2, 37, and 2 × 37.

AnswerCommon factors 1, 2, 37, 74. HCF = 74

(e) 81, 243

1. 81 = 3 × 3 × 3 × 3, 243 = 3 × 3 × 3 × 3 × 3.
2. 81 divides 243 exactly, so every factor of 81 is common.

AnswerCommon factors 1, 3, 9, 27, 81. HCF = 81

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(a) 24, 180

1. 24 = 2 × 2 × 2 × 3, 180 = 2 × 2 × 3 × 3 × 5.
2. Take the smaller count of each shared prime: two 2s and one 3.

AnswerHCF = 12

(b) 42, 75, 24

1. 42 = 2 × 3 × 7, 75 = 3 × 5 × 5, 24 = 2 × 2 × 2 × 3.
2. The only prime common to all three is a single 3.

AnswerHCF = 3

(c) 240, 378

1. 240 = 2 × 2 × 2 × 2 × 3 × 5, 378 = 2 × 3 × 3 × 3 × 7.
2. Shared: one 2 and one 3.

AnswerHCF = 6

(d) 400, 2500

1. 400 = 2 × 2 × 2 × 2 × 5 × 5, 2500 = 2 × 2 × 5 × 5 × 5 × 5.
2. Shared: two 2s and two 5s.

AnswerHCF = 100

(e) 300, 800

1. 300 = 2 × 2 × 3 × 5 × 5, 800 = 2 × 2 × 2 × 2 × 2 × 5 × 5.
2. Shared: two 2s and two 5s.

AnswerHCF = 100

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Open NCERT p. 54One way to think about it

1. 6, 12, 8, 18 are not prime, so this split does not show all the common factors.
2. Break everything into primes instead: 72 = 2 × 2 × 2 × 3 × 3, and 144 = 2 × 2 × 2 × 2 × 3 × 3.
3. Every prime in 72's row also appears in 144's row, so 72 itself is a common factor (144 = 72 × 2).
4. Comparing composite pieces like 6 and 12 against 8 and 18 can hide a shared prime — for example 6 and 8 both hide a 2, and 12 and 18 both hide a 6.

In shortNo. 72 and 144 share far more than 1 — in fact 72 is a common factor of both. Composite factors can hide common primes, so you must break numbers all the way down to primes before deciding what they share.

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(a) 30, 72

1. 30 = 2 × 3 × 5, 72 = 2 × 2 × 2 × 3 × 3.
2. Take the larger count of each prime: three 2s, two 3s, one 5.

AnswerLCM = 360

(b) 36, 54

1. 36 = 2 × 2 × 3 × 3, 54 = 2 × 3 × 3 × 3.
2. Take the larger count of each prime: two 2s, three 3s.

AnswerLCM = 108

(c) 105, 195, 65

1. 105 = 3 × 5 × 7, 195 = 3 × 5 × 13, 65 = 5 × 13.
2. Take every prime that appears anywhere, at its largest count: 3, 5, 7, 13.

AnswerLCM = 1365

(d) 222, 370

1. 222 = 2 × 3 × 37, 370 = 2 × 5 × 37.
2. Take the larger count of each prime: 2, 3, 5, 37.

AnswerLCM = 1110

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Open NCERT p. 59One way to think about it

(a) Two consecutive even numbers

1. Try 6 & 8, 14 & 16: HCF is always 2.
2. Both are 2 × (a number), and those two numbers are consecutive, so they share no further factor.

In shortThe HCF of two consecutive even numbers is always 2.

(b) Two consecutive odd numbers

1. Try 9 & 11, 15 & 17: HCF is always 1.
2. Any number dividing both would divide their difference, which is 2. Since both are odd, 2 cannot be a common factor.

In shortThe HCF of two consecutive odd numbers is always 1.

(c) Two even numbers

1. Try 2 & 4, 4 & 6 (HCF 2), but 8 & 12 gives HCF 4, and 20 & 30 gives HCF 10.

In shortThe HCF of two even numbers is always even, but not always exactly 2 — it depends on the numbers.

(d) Two consecutive numbers

1. Try 8 & 9, 44 & 45: HCF is always 1.
2. A common factor of two consecutive numbers must divide their difference, 1, so it can only be 1.

In shortThe HCF of two consecutive numbers is always 1.

(e) Two co-prime numbers

1. Co-prime means 'shares no common factor except 1', by definition.

In shortThe HCF of two co-prime numbers is always 1.

Watch this explained “Evens, and a wrong turn”, 2:18 into What HCF and LCM do for consecutive, even, and co-prime numbers · हिंदी में देखें

Open NCERT p. 59Checked by computerAnswers can differ: one example

(a) Find more such number pairs

1. This happens whenever one number is a multiple of the other.
2. 4 divides 12 exactly, so 12 already holds every prime that 4 holds.

Answer4 and 12 (LCM = 12)

(b) Describe such number pairs using algebra.

1. Let the smaller number be n, and let the other be a multiple of it, kn.
2. kn already contains every prime of n, so the smallest number containing both rows is kn itself.

AnswerLCM(n, kn) = kn

Watch this explained “The same question, of the LCM”, 4:06 into What HCF and LCM do for consecutive, even, and co-prime numbers · हिंदी में देखें

Open NCERT p. 59One way to think about it

(a) Two multiples of 3

1. Try 6 and 9: LCM 18. Try 6 and 12: LCM 12. Try 9 and 15: LCM 45.
2. Both numbers contain a 3, so every common multiple of them must contain a 3 as well.

In shortThe LCM of two multiples of 3 is always a multiple of 3. There is no fixed rule for its exact value.

(b) Two consecutive even numbers

1. Try 6 and 8: product 48, LCM 24. Try 10 and 12: product 120, LCM 60.
2. Two consecutive even numbers have HCF 2: they share exactly one 2 and nothing else. The product counts that shared 2 twice, but the LCM needs it only once.

In shortThe LCM of two consecutive even numbers is half their product.

(c) Two consecutive numbers

1. Try 8 and 9: LCM 72 = 8 × 9. Try 14 and 15: LCM 210 = 14 × 15.
2. Two consecutive numbers share no factor other than 1, because any common factor would also divide their difference, which is 1. With nothing shared, every prime of both numbers is needed in full.

In shortThe LCM of two consecutive numbers is their product.

(d) Two co-prime numbers

1. Try 4 and 9: LCM 36 = 4 × 9.
2. Co-prime numbers share no prime, so the LCM must hold all the primes of both numbers, with nothing counted twice.

In shortThe LCM of two co-prime numbers is their product.

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1. Row 1 repeats every 6 stars: yellow, green, orange, blue, magenta, gray. Blue is at position 4, then 10, 16, 22, ... (every 6th star from position 4).
2. Row 2 repeats every 4 stars: green, orange, yellow, blue. Blue is at position 4, then 8, 12, 16, ... (every 4th star).
3. Both rows already show blue together at position 4, in the picture.
4. The next position where both lists agree is the next common value after 4: 4 + LCM(6, 4) = 4 + 12 = 16.

AnswerThe blue stars meet again at the 16th star.

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(a) a multiple of 5 × 7 × 7 × 11 × 2?

1. A multiple of 5×7×7×11×2 must contain a 2 and two 7s in its row.
2. 5×7×11×11 has no 2 at all, and only one 7.

AnswerNo

(b) a factor of 5 × 7 × 7 × 11 × 2?

1. A factor of 5×7×7×11×2 can use at most one 11.
2. 5×7×11×11 needs two 11s, which the other row does not have.

AnswerNo

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(a) 3 × 3 × 5 × 7 × 7 and 12 × 7 × 11

1. First break 12 into primes: 12 × 7 × 11 = 2 × 2 × 3 × 7 × 11.
2. Common primes with 3 × 3 × 5 × 7 × 7: one 3 and one 7.
3. For the LCM, take every prime at its largest count: two 2s, two 3s, one 5, two 7s, one 11.

AnswerHCF = 3 × 7. LCM = 2 × 2 × 3 × 3 × 5 × 7 × 7 × 11

(b) 45 and 36

1. 45 = 3 × 3 × 5, 36 = 2 × 2 × 3 × 3.
2. Common primes: two 3s. For the LCM, take two 2s, two 3s, one 5.

AnswerHCF = 3 × 3. LCM = 2 × 2 × 3 × 3 × 5

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Open NCERT p. 63Checked by computerAnswers can differ: one example

1. 66 = 2 × 3 × 11.
2. HCF 1 means the two numbers share no prime. LCM 66 means that between them they use exactly the primes 2, 3 and 11, each once.
3. So split 2, 3 and 11 into two groups with nothing in common. There are four ways: 1 and 66, 2 and 33, 3 and 22, 6 and 11. Any of these pairs is a correct answer.
4. For example, 6 = 2 × 3 and 11 share no prime, so their HCF is 1, and their LCM is 2 × 3 × 11 = 66.

Answer6 and 11 (1 and 66, 2 and 33, 3 and 22 are also correct)

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1. Splitting equally through 3, 5 and 7 gates means the number of cows is a multiple of 3, of 5 and of 7.
2. So it is a multiple of LCM(3, 5, 7) = 105.
3. Multiples of 105 under 200: only 105 itself (the next one, 210, is too big).

Answer105 cows

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1. A cube packs without gaps only if its side divides all three box measurements: 12, 18 and 36.
2. HCF(12, 18, 36) = 6, so any factor of 6 works: 1, 2, 3, 6.
3. Checking the options: 9 does not divide 12; 6, 3 and 2 all divide 12, 18 and 36; 4 does not divide 18.

Answer6 cm, 3 cm and 2 cm cubes all pack without gaps

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1. The largest number that divides both 306 and 36 is their HCF.
2. 306 = 2 × 3 × 3 × 17 and 36 = 2 × 2 × 3 × 3.
3. Take each shared prime at its smaller count: one 2 and two 3s. HCF = 2 × 3 × 3 = 18.
4. The larger options fail: 36 does not divide 306 (306 ÷ 36 = 8.5), and 612 and 360 are bigger than 36, so they cannot divide it.

Answer(c) 18

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1. Being divisible by 3, 4, 5 and 7 means the number is a multiple of LCM(3, 4, 5, 7) = 420.
2. Check multiples of 420 one by one for their remainder on dividing by 11: 420, 840, 1260, 1680, 2100, ...
3. 420 mod 11 = 2, 840 mod 11 = 4, 1260 mod 11 = 6, 1680 mod 11 = 8, 2100 mod 11 = 10 — found it.

Answer2100

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1. In this game, children group themselves by the called number; nobody is left out exactly when the total is a multiple of that number.
2. 'No one got out' for 6 and for 9 means the total is a multiple of both 6 and 9, so a multiple of LCM(6, 9) = 18.
3. 'Some got out' for 10 means the total is NOT a multiple of 10.
4. Check the options: 72 (mult. of 18, not of 10) works; 90 is a multiple of 10 too, so fails; 45 is not a multiple of 6; 3 is not a multiple of 6 or 9; 36 (mult. of 18, not of 10) works.

Answer72 or 36

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1. Two different primes share no common factor, so they are co-prime.
2. For co-prime numbers, LCM = m × n exactly — not less, not more.
3. Since m and n are both at least 2, m × n is bigger than each of m and n on its own.

AnswerOnly (c): the LCM is always greater than both numbers

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1. In each round, the dog covers 9 feet while the rabbit covers 7 feet, so the dog gains 9 − 7 = 2 feet every leap.
2. The dog must close a 150 foot head start.
3. Number of leaps needed = 150 ÷ 2 = 75.

Answer75 leaps

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1. This is the LCM of 1, 2, 3, 4, 5, 6, 8, 9, 10.
2. The highest power of 2 needed comes from 8 = 2 × 2 × 2. The highest power of 3 comes from 9 = 3 × 3. The only 5 comes from 5 or 10.
3. LCM = 2 × 2 × 2 × 3 × 3 × 5 = 360.

Answer360

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1. Factorise every denominator: 15 = 3×5, 20 = 2×2×5, 36 = 2×2×3×3, 63 = 3×3×7, 21 = 3×7.
2. The LCM of all five denominators is 2×2×3×3×5×7 = 1260 — this is the efficient common denominator, found without multiplying all five denominators together.
3. Rewrite each fraction with denominator 1260: 8/15 = 672/1260, 1/20 = 63/1260, 7/36 = 245/1260, 11/63 = 220/1260, 1/21 = 60/1260.
4. Add the numerators: 672 + 63 + 245 + 220 + 60 = 1260, giving 1260/1260.

AnswerThe sum is exactly 1. Finding the LCM of the denominators first (instead of just multiplying them all together) keeps the numbers small and efficient.

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