NCERT Solutions for Class 7th Maths Chapter 3 In-text Questions — Factors of a Number Using Prime Factorisation

Book page 51 Updated on2026-09-19

Q1.

Similarly, is 2 × 7 = 14 a factor of 840? Why or why not?

Answer

Yes, 14 is a factor of 840.

840 = 2 × 2 × 2 × 3 × 5 × 7
= (2 × 7) × (2 × 2 × 3 × 5)
= 14 × 60
Check: 14 × 60 = 840 ✓

Why it happens: 14 asks for one 2 and one 7. The factorisation of 840 has three 2s and one 7, so both are available. The group can be lifted out, and 2 × 2 × 3 × 5 = 60 is what remains.

Q2.

Is 2 × 2 × 2 a factor of 840? Why or why not?

Answer

Yes, 2 × 2 × 2 = 8 is a factor of 840.

840 = (2 × 2 × 2) × (3 × 5 × 7)
= 8 × 105
Check: 8 × 105 = 840 ✓

Why it happens: 840 contains exactly three 2s, and 8 asks for exactly three 2s. There are just enough, so 8 fits.

Check it yourself: is 16 = 2 × 2 × 2 × 2 a factor of 840? No — that would need four 2s and 840 has only three. Indeed 840 ÷ 16 = 52.5.

Q3.

Is 3 × 3 × 3 a factor of 840? Why or why not?

Answer

No, 3 × 3 × 3 = 27 is not a factor of 840.

840 = 2 × 2 × 2 × 3 × 5 × 7
Number of 3s in 840 = only one
27 needs three 3s
Check: 840 ÷ 27 = 31.11… — not a whole number ✗

Why it happens: a factor cannot ask for more copies of a prime than the number actually has. 840 has just one 3, so 3 is a factor and 9 = 3 × 3 is not, let alone 27.

Q4.

Can we use this idea to list down all the possible factors of a number using just its prime factors?

Answer

Yes. Every factor is a subpart of the prime factorisation, so listing all the subparts lists all the factors.

Take 30 = 2 × 3 × 5.

No prime at all: 1
One prime at a time: 2, 3, 5
Two primes together: 2 × 3 = 6, 2 × 5 = 10, 3 × 5 = 15
All three: 2 × 3 × 5 = 30
Factors of 30 = 1, 2, 3, 5, 6, 10, 15, 30

Why it happens: a factor of 30 can only be built out of the primes inside 30. Choosing which of those primes to keep and which to drop produces every factor exactly once — and choosing none of them leaves 1, which is why 1 is always a factor.

Q5.

Find the factors of 225 using prime factorisation.

Answer

First factorise 225 by the division method.

5 | 225
5 | 45
3 |   9
3 |   3
     1
225 = 3 × 3 × 5 × 5

Now form every subpart, systematically:

No prime: 1
One prime: 3, 5
Two primes: 3 × 3 = 9, 5 × 5 = 25, 3 × 5 = 15
Three primes: 3 × 3 × 5 = 45, 3 × 5 × 5 = 75
Four primes: 3 × 3 × 5 × 5 = 225

Factors of 225 = 1, 3, 5, 9, 15, 25, 45, 75, 225 — nine factors in all.

Tip: 225 = 3² × 5². You may take 0, 1 or 2 threes (3 ways) and 0, 1 or 2 fives (3 ways), so there are 3 × 3 = 9 factors. That count matches the list.

Q6.

Check that all the factors of 225 occur in this list.

Answer

Test every number from 1 to 225 that could divide it — or simply pair the factors up.

Pair	Product
1 × 225	225 ✓
3 × 75	225 ✓
5 × 45	225 ✓
9 × 25	225 ✓
15 × 15	225 ✓

The nine factors fall into four pairs plus the middle one, 15. Nothing is missing, and nothing extra has crept in.

Why it happens: factors come in pairs that multiply to the number, and the two members of a pair sit on either side of √225 = 15. So it is enough to test 1 to 15: only 1, 3, 5, 9 and 15 divide 225, and each brings its partner. That gives exactly nine factors.

Was this helpful?