NCERT Solutions for Class 7th Maths Chapter 3 In-text Questions — Least, but not Last!

Book page 56 Updated on2026-09-19

Q1.

Do you remember the ‘Idli-Vada’ game from Grade 6 (see chapter ‘Prime Time’)? Two numbers are chosen and whenever players come to their multiples, ‘idli’ or ‘vada’ should be called out depending on whose multiple the number is. If the number happens to be a common multiple, then ‘idli-vada’ should be called out. In each problem below, the two numbers corresponding to ‘idli’ and ‘vada’ are given. Find the first number for which ‘idli-vada’ will be called out: (a) 4 and 6 (b) 7 and 11 (c) 14 and 30 (d) 15 and 55

Answer

'Idli-vada' is called out at the first number that is a multiple of both — the LCM.

	Numbers	Prime factorisations	First 'idli-vada'
(a)	4 and 6	4 = 2 × 2, 6 = 2 × 3	12
(b)	7 and 11	7, 11 (both prime)	77
(c)	14 and 30	14 = 2 × 7, 30 = 2 × 3 × 5	210
(d)	15 and 55	15 = 3 × 5, 55 = 5 × 11	165

(a) LCM = 2 × 2 × 3 = 12 (multiples of 4: 4, 8, 12…; of 6: 6, 12…)
(b) LCM = 7 × 11 = 77
(c) LCM = 2 × 3 × 5 × 7 = 210
(d) LCM = 3 × 5 × 11 = 165

Tip: for (c) and (d), listing multiples takes a long time — 210 is the 15th multiple of 14. Prime factorisation gets there in one line.

Q2.

Is the answer always the LCM of the two numbers? Explain.

Answer

Yes, always.

Why it happens: 'idli' is called at every multiple of the first number and 'vada' at every multiple of the second. 'Idli-vada' is called only when a number is on both lists, that is, at a common multiple. Counting starts at 1 and goes up, so the first such number reached is the smallest common multiple — the LCM, by definition.

(a) LCM(4, 6) = 12 ✓ (8 is only 'idli', 6 is only 'vada')
(b) LCM(7, 11) = 77 ✓
(c) LCM(14, 30) = 210 ✓
(d) LCM(15, 55) = 165 ✓

Check it yourself: every later 'idli-vada' is a multiple of the first one — 12, 24, 36, … for (a). The common multiples of two numbers are exactly the multiples of their LCM.

Q3.

How do we find the LCM of two numbers using their prime factors?

Answer

Collect every prime that appears in either number, and take each one the greatest number of times it occurs in either factorisation.

Step 1 — write both prime factorisations
Step 2 — list every prime that appears anywhere
Step 3 — for each prime, take the maximum number of occurrences
Step 4 — multiply

Example: 12 = 2 × 2 × 3 and 18 = 2 × 3 × 3.

2s: 12 has two, 18 has one → take two
3s: 12 has one, 18 has two → take two
LCM = 2 × 2 × 3 × 3 = 36

Why it happens: a multiple of 12 must contain 2 × 2 × 3 inside it, and a multiple of 18 must contain 2 × 3 × 3. To hold both, the number needs at least two 2s and at least two 3s. Taking exactly that many — and no spare primes — gives the smallest such number. Notice the contrast with the HCF: there we took the minimum count, here the maximum.

Was this helpful?