NCERT Solutions for Class 7th Maths Chapter 3 Figure it Out — Factors of a Number Using Prime Factorisation

Book page 51 Updated on2026-09-19

Q1.
List all the factors of the following numbers: (a) 90 (b) 105 (c) 132 (d) 360 (this number has 24 factors) (e) 840 (this number has 32 factors)
Answer

Factorise each number into primes, then build every subpart.

(a) 90 = 2 × 3 × 3 × 5
(b) 105 = 3 × 5 × 7
(c) 132 = 2 × 2 × 3 × 11
(d) 360 = 2 × 2 × 2 × 3 × 3 × 5
(e) 840 = 2 × 2 × 2 × 3 × 5 × 7
NumberAll factorsHow many
(a)901, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 9012
(b)1051, 3, 5, 7, 15, 21, 35, 1058
(c)1321, 2, 3, 4, 6, 11, 12, 22, 33, 44, 66, 13212
(d)3601, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 18, 20, 24, 30, 36, 40, 45, 60, 72, 90, 120, 180, 36024
(e)8401, 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, 84032
Tip — count before you list. Write the factorisation with powers and add 1 to each power, then multiply.
90 = 2¹ × 3² × 5¹ → 2 × 3 × 2 = 12 factors
105 = 3¹ × 5¹ × 7¹ → 2 × 2 × 2 = 8 factors
132 = 2² × 3¹ × 11¹ → 3 × 2 × 2 = 12 factors
360 = 2³ × 3² × 5¹ → 4 × 3 × 2 = 24 factors
840 = 2³ × 3¹ × 5¹ × 7¹ → 4 × 2 × 2 × 2 = 32 factors
The counts for (d) and (e) match what the book says, so nothing has been missed.
Why it happens: to build a factor of 360 = 2³ × 3² × 5 you decide how many 2s to keep (0, 1, 2 or 3 — four choices), how many 3s (0, 1 or 2 — three choices) and how many 5s (0 or 1 — two choices). Each different set of choices gives a different factor, so there are 4 × 3 × 2 = 24 of them.
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