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NCERT Solutions Ganita Prakash (Part 1) Chapter 5 Always, Sometimes, or Never / What Remains? — In-text Questions

Book page 121 Updated on2026-09-05

Q1.
Examine each of the following statements, and determine whether it is ‘Always true’, ‘Sometimes true’, ‘Never true’. 6. If a number is divisible by both 9 and 4, it must be divisible by 36.
Answer

Always true.

9 = 3², 4 = 2² — they share no prime
LCM (9, 4) = 3² × 2² = 36

If the number is divisible by 9, its prime factorisation contains 3 × 3. If it is also divisible by 4, the factorisation contains 2 × 2. Since these are different primes, both blocks sit in the factorisation together, so 3 × 3 × 2 × 2 = 36 divides the number.

Examples: 36, 72, 108, 180, 900 — each divisible by 9, by 4, and by 36
Tip: the general rule is if A is divisible by k and also by m, then A is divisible by LCM (k, m).
Q2.
7. If a number is divisible by both 6 and 4, it must be divisible by 24.
Answer

Sometimes true. The counterexample is small: 12.

12 ÷ 6 = 2 ✓   12 ÷ 4 = 3 ✓   12 ÷ 24 ✗
Why it happens: 6 = 2 × 3 and 4 = 2 × 2 share a factor 2. The rule promises only LCM (6, 4) = 2² × 3 = 12, not 24. So every number divisible by 6 and 4 is a multiple of 12 — and multiples of 12 are alternately 12, 24, 36, 48, …, only half of which are multiples of 24.

So the statement holds for 24, 48, 72 but fails for 12, 36, 60. Compare it with statement 6, where 9 and 4 had no common factor and the product was the LCM.

Q3.
8. When you add an odd number to an even number we get a multiple of 6.
Answer

Never true. The sum of an odd and an even number is odd, while every multiple of 6 is even.

Suppose (2n) + (2m + 1) = 6j
2n + 2m = 6j – 1
2(n + m) = 6j – 1

The left side is even; the right side is 1 less than an even number, so it is odd. An even number can never equal an odd number, so no values of n, m, j exist.

Did you know? Anshu's doubt in the margin — “Can I write an even and an odd number as 2n and 2n+1 instead?” — is worth answering: no, not here. Writing 2n and 2n + 1 forces the two numbers to be consecutive. Using 2n and 2m + 1 keeps them independent, which is what the statement needs.
Q4.
Find a number that has a remainder of 3 when divided by 5. Write more such numbers.
Answer

3, 8, 13, 18, 23, 28, 33, 38, … — each one is 3 more than a multiple of 5.

k012345
5k + 33813182328
Check: 18 = 5 × 3 + 3   28 = 5 × 5 + 3   103 = 5 × 20 + 3

The gap between consecutive numbers in the list is always 5, because moving to the next multiple of 5 keeps the leftover 3 unchanged.

Q5.
Which algebraic expression(s) capture all such numbers? (i) 3k + 5 (ii) 3k – 5 (iii) 3k/5 (iv) 5k + 3 (v) 5k – 2 (vi) 5k – 3
Answer

(iv) 5k + 3 and (v) 5k – 2. They look different but describe exactly the same set of numbers.

5k – 2 = 5k – 5 + 3 = 5(k – 1) + 3
OptionFirst few valuesRemainder on ÷ 5
(i) 3k + 55, 8, 11, 14, 17varies — 0, 3, 1, 4, 2
(ii) 3k – 5– 2, 1, 4, 7, 10varies
(iii) 3k/50.6, 1.2, 1.8, …not a whole number
(iv) 5k + 33, 8, 13, 18, 233 — always ✓
(v) 5k – 23, 8, 13, 18, 23 (k ≥ 1)3 — always ✓
(vi) 5k – 32, 7, 12, 17, 222 — always ✗
Why it happens: to leave remainder 3 on division by 5, a number must be a multiple of 5 with 3 added. Only expressions of the form 5k + c, where c itself leaves remainder 3 on division by 5, can do it. Here c = 3 in (iv) and c = – 2 in (v), and – 2 = – 5 + 3.
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