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NCERT Solutions Ganita Prakash (Part 1) Chapter 5 .1 Is This a Multiple Of? — In-text Questions

Book page 1125 Updated on2026-09-05

Q1.
“Can I write every natural number as a sum of consecutive numbers?”
Answer

No. The powers of 2 — 1, 2, 4, 8, 16, 32, … — cannot be written as a sum of two or more consecutive natural numbers. Every other natural number can.

Write a run of k consecutive numbers starting at a:

a + (a + 1) + … + (a + k – 1)
= ka + (1 + 2 + … + (k – 1))
= ka + k(k – 1)/2
So 2N = k(2a + k – 1)
Why it happens: look at the two factors k and (2a + k – 1). If k is even, then 2a + k – 1 is odd; if k is odd, then 2a + k – 1 is even. So one of them is always odd, and for k ≥ 2 both are bigger than 1. Hence 2N — and therefore N — must have an odd factor greater than 1. A power of 2 has no such factor, so it can never be written this way.

Check the smallest cases:

NOdd factor > 1?Sum of consecutive numbers
631 + 2 + 3
773 + 4
8noneimpossible
93, 94 + 5 and 2 + 3 + 4
16noneimpossible
Q2.
“Which numbers can I write as the sum of consecutive numbers in more than one way?”
Answer

Numbers that have three or more odd factors. In fact the number of ways (using two or more consecutive natural numbers) is exactly one less than the number of odd factors.

15 → odd factors 1, 3, 5, 15 (four of them) → 4 – 1 = 3 ways
15 = 7 + 8 = 4 + 5 + 6 = 1 + 2 + 3 + 4 + 5 ✓ (exactly what Anshu found)
NumberOdd factorsNumber of ways
71, 71 — only 3 + 4
91, 3, 92 — 4 + 5, 2 + 3 + 4
101, 51 — 1 + 2 + 3 + 4
121, 31 — 3 + 4 + 5
451, 3, 5, 9, 15, 455
Why it happens: from Q1, every way of writing N corresponds to a factorisation 2N = k × (2a + k – 1) in which one factor is odd. Each odd factor of N (other than the one that gives k = 1) produces exactly one such run.
Q3.
“Ohh, I know all odd numbers can be written as a sum of two consecutive numbers. Can we write all even numbers as a sum of consecutive numbers?”
Answer

Anshu's first claim is easy to justify, and the answer to his question is no — the even numbers 2, 4, 8, 16, 32, … (the powers of 2) cannot be written this way.

Any odd number = 2n + 1
2n + 1 = n + (n + 1) — two consecutive numbers
e.g. 35 = 17 + 18, 101 = 50 + 51

An even number can never be a sum of two consecutive numbers, because n + (n + 1) is always odd. So it needs a run of three or more:

6 = 1 + 2 + 3   10 = 1 + 2 + 3 + 4
12 = 3 + 4 + 5   14 = 2 + 3 + 4 + 5
18 = 5 + 6 + 7   20 = 2 + 3 + 4 + 5 + 6
Check it yourself: try hard to write 8, 16 or 32 as a sum of consecutive numbers. You will not succeed — they have no odd factor bigger than 1.
Q4.
“Can I write 0 as a sum of consecutive numbers? Maybe I should use negative numbers.”
Answer

Yes, once negative numbers are allowed. Take any run that is symmetric about 0 — the positive and negative parts cancel.

0 = (–1) + 0 + 1
0 = (–2) + (–1) + 0 + 1 + 2
0 = (–3) + (–2) + (–1) + 0 + 1 + 2 + 3
Why it happens: in the run from –m to m, every negative number is paired with its opposite, and 0 sits in the middle. Each pair adds to 0, so the whole sum is 0. Notice that such a run always has an odd number of terms (2m + 1 of them).
Try This: with negatives allowed, every number becomes a sum of consecutive numbers — even 8. For example 8 = (–7) + (–6) + … + 6 + 7 + 8, since everything from –7 to 7 cancels.
Q5.
Take any 4 consecutive numbers. For example, 3, 4, 5, and 6. Place ‘+’ and ‘–’ signs in between the numbers. How many different possibilities exist? Write all of them.
Answer

Eight possibilities. There are 3 gaps between the four numbers and each gap takes either sign, so the count is 2 × 2 × 2 = 2³ = 8.

ExpressionValue
3 + 4 + 5 + 618
3 + 4 + 5 – 66
3 + 4 – 5 + 68
3 + 4 – 5 – 6– 4
3 – 4 + 5 + 610
3 – 4 + 5 – 6– 2
3 – 4 – 5 + 60
3 – 4 – 5 – 6– 12

The tree in the book lists them in exactly this order: branch on the sign before 4, then before 5, then before 6.

Did you notice? Every one of the eight values is an even number.
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