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NCERT Solutions for Class 8th Maths Chapter 5 .1 Is This a Multiple Of? — In-text Questions

Book page 1135 Updated on2026-09-19

Q1.
Evaluate each expression and write the result next to it. Do you notice anything interesting?
Answer

The eight values for 3, 4, 5, 6 are 18, 6, 8, – 4, 10, – 2, 0, – 12 — and the interesting thing is that every one of them is even.

ExpressionValueExpressionValue
3 + 4 + 5 + 6183 – 4 + 5 + 610
3 + 4 + 5 – 663 – 4 + 5 – 6– 2
3 + 4 – 5 + 683 – 4 – 5 + 60
3 + 4 – 5 – 6– 43 – 4 – 5 – 6– 12

A second thing worth noticing: the eight values are 12 apart at the extremes and fall in steps of 2 — no odd number ever appears, and no value is repeated.

Q2.
Now, take four other consecutive numbers. Place the ‘+’ and ‘–’ signs as you have done before. Find out the results of each expression. What do you observe?
Answer

Taking 5, 6, 7, 8:

ExpressionValueExpressionValue
5 + 6 + 7 + 8265 – 6 + 7 + 814
5 + 6 + 7 – 8105 – 6 + 7 – 8– 2
5 + 6 – 7 + 8125 – 6 – 7 + 80
5 + 6 – 7 – 8– 45 – 6 – 7 – 8– 16

Observation: again all eight values are even. And three of them — 0, – 2 and – 4 — are exactly the same as before.

Q3.
Repeat this for one more set of 4 consecutive numbers. Share your findings.
Answer

Taking 7, 8, 9, 10:

ExpressionValueExpressionValue
7 + 8 + 9 + 10347 – 8 + 9 + 1018
7 + 8 + 9 – 10147 – 8 + 9 – 10– 2
7 + 8 – 9 + 10167 – 8 – 9 + 100
7 + 8 – 9 – 10– 47 – 8 – 9 – 10– 20

The three constant answers appear once more. Writing the numbers as n, n + 1, n + 2, n + 3 shows why they must:

n – (n + 1) – (n + 2) + (n + 3) = 0
n – (n + 1) + (n + 2) – (n + 3) = – 2
n + (n + 1) – (n + 2) – (n + 3) = – 4
Why it happens: in each of these three, two of the n's are added and two are subtracted, so all the n's cancel. What is left is a fixed number that does not depend on which four consecutive numbers you started with.
Q4.
Do these patterns occur no matter which 4 consecutive numbers are chosen? Is there a way to find out through reasoning? Hint: Use algebra and describe the 8 expressions in a general form.
+ − + − + − + − + − + − + − 3 4 4 5 5 5 5 6 6 6 6 6 6 6 6 3 + 4 + 5 + 6 3 + 4 + 5 − 6
The diagram on page 113: the eight ways of putting a ‘+’ or a ‘−’ sign before 4, before 5 and before 6, starting from 3. The first two branches are written out in full.
Answer

Yes — and algebra settles it in one step, without testing more sets.

The general expression is
n ± (n + 1) ± (n + 2) ± (n + 3)
= n(1 ± 1 ± 1 ± 1) + (0 ± 1 ± 2 ± 3)

Look at the two brackets separately.

  • 1 ± 1 ± 1 ± 1 is one of 4, 2, 0, – 2 — always even. So the first part is an even multiple of n.
  • 0 ± 1 ± 2 ± 3 starts from 1 + 2 + 3 = 6, which is even, and changing any sign alters it by twice that number — again an even change. So the second part is always even too.
even + even = even
Why it happens: the value can only change by an even amount, so it can never cross from even to odd. That is why the answer is even for every choice of four consecutive numbers — the three fixed values 0, – 2, – 4 turn up every single time, and no one in the class will ever report an odd result.
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