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NCERT Solutions for Class 7th Maths Chapter 3 In-text Questions — Finding the HCF of Numbers Using Prime Factorisation

Book page 51–53 Updated on2026-09-19

Q1.
After observing a few prime factorisations, Anshu claims “The larger a number is, the longer its prime factorisation will be”. What do you think of Anshu’s claim?
Answer

Anshu's claim is false. One counterexample is enough to settle it.

96 = 2 × 2 × 2 × 2 × 2 × 3 → 6 prime factors
121 = 11 × 11 → 2 prime factors
121 is the larger number, but its factorisation is shorter

More counterexamples: 128 = 2⁷ has seven prime factors, while the much larger 1009 is prime and has just one.

Why it happens: the length of a prime factorisation depends on how small the primes are, not on how big the number is. A number built from many 2s grows slowly and collects a long factorisation. A number that is a product of two large primes is huge but has only two factors in its list.
Did you know? A claim made without proof is called a conjecture. Anshu's claim is a conjecture, and we disproved it by finding a counterexample — a single case where it fails. To prove a conjecture true you must cover every case; to prove it false one example is enough.
Q2.
Example 1: Find the common factors, and the HCF of 45 and 75. [So the common factors should be subparts of both the factorisations. Can you write them down?]
Answer

The common factors are 1, 3, 5 and 15, and the HCF is 15.

45 = 3 × 3 × 5
75 = 3 × 5 × 5

A common factor must be a subpart of both lists. Compare them:

SubpartInside 3 × 3 × 5?Inside 3 × 5 × 5?Common factor?
3yesyesyes
5yesyesyes
3 × 5 = 15yesyesyes
3 × 3 = 9yesno (only one 3)no
5 × 5 = 25no (only one 5)yesno

Adding 1, the common factors are 1, 3, 5, 15. The highest is 15.

HCF(45, 75) = 15
Why it happens: 45 has two 3s and one 5; 75 has one 3 and two 5s. A shared subpart can use at most one 3 (that is all 75 has) and at most one 5 (that is all 45 has). So the biggest shared subpart is 3 × 5 = 15.
Q3.
Example 2: Find the common factors, and the HCF of 112 and 84.
Answer

The common factors are 1, 2, 4, 7, 14 and 28, and the HCF is 28.

112 = 2 × 2 × 2 × 2 × 7
84 = 2 × 2 × 3 × 7

Common primes: 2 (112 has four, 84 has two → at most two) and 7 (one each).

Common subpartValue
22
77
2 × 24
2 × 714
2 × 2 × 728
Highest common subpart = 2 × 2 × 7 = 28
Check: 112 ÷ 28 = 4 ✓ and 84 ÷ 28 = 3 ✓
Why it happens: 84 has only two 2s, so no common factor may use more than two. 84 has one 3 but 112 has none, so 3 can never appear. Both have exactly one 7. Taking the most that is allowed of each shared prime — two 2s and one 7 — gives the highest common factor.
Q4.
Example 3: Find the common factors and the HCF of 96 and 275.
Answer

The only common factor is 1, so the HCF is 1.

96 = 2 × 2 × 2 × 2 × 2 × 3
275 = 5 × 5 × 11
Primes in 96: 2, 3
Primes in 275: 5, 11
Shared primes: none

With no prime in common, no subpart other than the empty one is shared. So 1 is the only common factor, and it is also the HCF.

Why it happens: any common factor bigger than 1 would have to contain at least one prime, and that prime would have to sit in both factorisations. Here the two lists of primes do not overlap at all, so nothing bigger than 1 can be common.
Did you know? Two numbers whose HCF is 1 are called co-prime. Neither number need be prime — 96 and 275 are both composite, yet they are co-prime.
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