NCERT Solutions for Class 7th Maths Chapter 3 In-text Questions — Finding LCM through Prime Factorisation

Book page 57–58 Updated on2026-09-19

Q1.

We get, 36 = 2 × 2 × 3 × 3, 648 = 36 × 18 = (2 × 2 × 3 × 3) × (2 × 3 × 3). What do you observe? We can see that the prime factors of the multiple contain the prime factors of the number along with some more prime factors. Will this happen with every multiple?

Answer

Yes, with every multiple.

36 = 2 × 2 × 3 × 3
648 = 36 × 18 = (2 × 2 × 3 × 3) × (2 × 3 × 3) = 2 × 2 × 2 × 3 × 3 × 3 × 3
The block 2 × 2 × 3 × 3 is still sitting inside 648

Try two more multiples of 36:

36 × 5 = 180 = (2 × 2 × 3 × 3) × 5 ✓
36 × 14 = 504 = (2 × 2 × 3 × 3) × (2 × 7) ✓

Why it happens: a multiple of 36 is 36 × k for some whole number k. Factorising it means factorising 36 and factorising k and putting the two lists side by side. So the primes of 36 are always there, joined by whatever primes k brings. In short: every multiple of a number contains that number's whole prime factorisation.

Q2.

Can this be used to find the LCM?

Answer

Yes. It tells us exactly what a common multiple must contain.

A common multiple of a and b must contain
• the whole prime factorisation of a, and
• the whole prime factorisation of b
The smallest such number carries nothing extra

So for each prime, keep just enough copies to cover whichever number needs more of it — the maximum of the two counts.

Why it happens: if a common multiple held fewer copies of some prime than one of the numbers needs, that number's factorisation would not fit inside it, and it would not be a multiple. If it held more copies than needed, or an extra prime, it would still be a common multiple but a larger one. So the lowest common multiple takes each prime the maximum number of times and nothing beyond.

Q3.

Example 6: Find the LCM of 14 and 35. [What is the lowest among all the common multiples of 14 and 35?]

Answer

LCM(14, 35) = 70.

14 = 2 × 7
35 = 5 × 7
Primes appearing: 2, 5, 7
2s: one (in 14) → take one
5s: one (in 35) → take one
7s: one in each → take one
LCM = 2 × 5 × 7 = 70

Check: 70 ÷ 14 = 5 ✓ and 70 ÷ 35 = 2 ✓

Why it happens: 2 × 5 × 7 contains 2 × 7 = 14 as a subpart and 5 × 7 = 35 as a subpart, so it is a common multiple. Remove any one of the three primes and it stops working — drop the 2 and 14 no longer fits; drop the 5 and 35 no longer fits; drop the 7 and neither fits. Nothing smaller can do the job, so 70 is the lowest.

Tip: the single 7 is shared. Writing 14 × 35 = 490 would also give a common multiple, but it counts the 7 twice, so it is 7 times bigger than it needs to be.

Q4.

Common multiples should contain each prime factor as a subpart: 2 × 7 as a subpart and 5 × 7 as a subpart. For example, 2 × 7 × 5 × 7 × 3 is a common multiple of 14 and 35. 2 × 2 × 5 × 7 × 7 × 11 is another common multiple. Is 2 × 3 × 5 × 7 also a common multiple?

Answer

Yes. 2 × 3 × 5 × 7 = 210 is a common multiple of 14 and 35.

2 × 3 × 5 × 7 = 210
Contains 2 × 7 = 14 as a subpart ✓ → 210 = 14 × 15
Contains 5 × 7 = 35 as a subpart ✓ → 210 = 35 × 6
210 ÷ 14 = 15 ✓ and 210 ÷ 35 = 6 ✓

But it is not the lowest: 210 = 70 × 3, and the extra 3 is not needed by either number.

Why it happens: any number that holds one 2, one 5 and one 7 is a common multiple of 14 and 35, whatever else it holds. Extra primes such as the 3 here only make the number bigger. Stripping away everything not required leaves 2 × 5 × 7 = 70, the LCM.

Q5.

Example 7: Find the LCM of 96 and 360. [How many 2s should the LCM contain? How many 3s should the LCM contain? How many 5s should the LCM contain?]

Answer

LCM(96, 360) = 1440.

96 = 2 × 2 × 2 × 2 × 2 × 3
360 = 2 × 2 × 2 × 3 × 3 × 5
Primes appearing anywhere: 2, 3, 5

How many 2s? 96 has five, 360 has three. Take the maximum, five. Five 2s contain three 2s, so both are covered.

How many 3s? 96 has one, 360 has two. Take two.

How many 5s? 96 has none, 360 has one. Take one.

LCM = 2 × 2 × 2 × 2 × 2 × 3 × 3 × 5
= 32 × 9 × 5
= 32 × 45
= 1440
Check: 1440 ÷ 96 = 15 ✓ and 1440 ÷ 360 = 4 ✓

Tip: HCF asks for the minimum count of the shared primes; LCM asks for the maximum count of all the primes. Here HCF(96, 360) = 2 × 2 × 2 × 3 = 24, and 24 × 1440 = 34560 = 96 × 360 ✓

Q6.

Choosing more than five occurrences of 2s will give a common multiple; but it will not be the lowest. Are you able to see why?

Answer

Because every extra 2 doubles the number without being needed by 96 or by 360.

Five 2s: 2⁵ × 3 × 3 × 5 = 1440
Six 2s: 2⁶ × 3 × 3 × 5 = 2880 = 1440 × 2 — still a common multiple, but twice as big
Seven 2s: 5760 = 1440 × 4 — bigger still

Why it happens: 96 needs five 2s and 360 needs three. Five copies already cover both demands. A sixth 2 satisfies no new demand, so the only thing it changes is the size. Since the LCM is defined as the smallest common multiple, we stop at exactly what is required — not one prime more.

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