NCERT Solutions Ganita Prakash Chapter 5 Figure it Out — Prime Factorisation

Book page 120 Updated on2026-09-05

Q1.

Find the prime factorisations of the following numbers: 64, 104, 105, 243, 320, 141, 1728, 729, 1024, 1331, 1000.

Answer

Keep dividing by the smallest prime that works, until only primes are left.

Number	How it breaks up	Prime factorisation
64	64 = 2 × 32 = 2 × 2 × 16 = …	2 × 2 × 2 × 2 × 2 × 2
104	104 = 8 × 13	2 × 2 × 2 × 13
105	105 = 5 × 21 = 5 × 3 × 7	3 × 5 × 7
243	243 = 3 × 81 = 3 × 3 × 27 = …	3 × 3 × 3 × 3 × 3
320	320 = 64 × 5	2 × 2 × 2 × 2 × 2 × 2 × 5
141	1 + 4 + 1 = 6, so 3 divides it: 141 = 3 × 47	3 × 47
1728	1728 = 64 × 27	2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3
729	729 = 27 × 27	3 × 3 × 3 × 3 × 3 × 3
1024	1024 = 2 × 512 = 2 × 2 × 256 = …	2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
1331	1331 = 11 × 121 = 11 × 11 × 11	11 × 11 × 11
1000	1000 = 10 × 10 × 10 = (2 × 5) × (2 × 5) × (2 × 5)	2 × 2 × 2 × 5 × 5 × 5

Checks: 2⁶ = 64 ✔ 8 × 13 = 104 ✔ 3 × 5 × 7 = 105 ✔ 3⁵ = 243 ✔
64 × 5 = 320 ✔ 3 × 47 = 141 ✔ 64 × 27 = 1728 ✔ 3⁶ = 729 ✔
2¹⁰ = 1024 ✔ 11 × 11 × 11 = 1331 ✔ 8 × 125 = 1000 ✔

Tip for 141: a number is divisible by 3 when its digits add up to a multiple of 3. Here 1 + 4 + 1 = 6, so 141 = 3 × 47, and 47 is prime.

Q2.

The prime factorisation of a number has one 2, two 3s, and one 11. What is the number?

Answer

Just multiply the primes as many times as they occur.

Number = 2 × 3 × 3 × 11
= 2 × 9 × 11
= 18 × 11
= 198

Answer: 198.

Check: 198 = 2 × 99 = 2 × 9 × 11 = 2 × 3 × 3 × 11 ✔ — one 2, two 3s and one 11, exactly as asked.

Q3.

Find three prime numbers, all less than 30, whose product is 1955.

Answer

Break 1955 down step by step. It ends in 5, so start with 5.

1955 ÷ 5 = 391
Now split 391. It is odd, 3 + 9 + 1 = 13 (not a multiple of 3), does not end in 0 or 5.
Try 7 → no. 11 → no. 13 → no. 17 → 391 ÷ 17 = 23 ✔
And 23 is prime.

1955 = 5 × 17 × 23

All three — 5, 17 and 23 — are prime and all are less than 30. ✔

Check: 17 × 23 = 391, and 391 × 5 = 1955 ✔

Q4.

Find the prime factorisation of these numbers without multiplying first. a. 56 × 25 b. 108 × 75 c. 1000 × 81

Answer

Factorise each part separately and then write the two lists side by side — no big multiplication needed.

a. 56 × 25
56 = 2 × 2 × 2 × 7 25 = 5 × 5
56 × 25 = 2 × 2 × 2 × 5 × 5 × 7

b. 108 × 75
108 = 2 × 2 × 3 × 3 × 3 75 = 3 × 5 × 5
108 × 75 = 2 × 2 × 3 × 3 × 3 × 3 × 5 × 5

c. 1000 × 81
1000 = 2 × 2 × 2 × 5 × 5 × 5 81 = 3 × 3 × 3 × 3
1000 × 81 = 2 × 2 × 2 × 3 × 3 × 3 × 3 × 5 × 5 × 5

Check if you like: 56 × 25 = 1400, and 8 × 25 × 7 = 1400 ✔; 108 × 75 = 8100, and 4 × 81 × 25 = 8100 ✔; 1000 × 81 = 81000, and 8 × 81 × 125 = 81000 ✔

Q5.

What is the smallest number whose prime factorisation has: a. three different prime numbers? b. four different prime numbers?

Answer

To keep the answer as small as possible, use the smallest primes and use each of them only once.

a. smallest three primes → 2, 3, 5
2 × 3 × 5 = 30

b. smallest four primes → 2, 3, 5, 7
2 × 3 × 5 × 7 = 30 × 7 = 210

Why nothing smaller can work: the number must contain three (or four) different primes. Swapping any of 2, 3, 5, 7 for a bigger prime only makes the product larger, and repeating a prime also makes it larger without adding a new prime. So 30 and 210 are the smallest.

Try This: continue the pattern — the smallest number with five different primes is 2 × 3 × 5 × 7 × 11 = 2310.

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