NCERT Solutions Ganita Prakash (Part 2) Chapter 6 .6 Decoding Divisibility Tricks — In-text Questions

Book page 1456 Updated on2026-09-05

Q1.

If we choose other 2-digit numbers, and follow the steps, will there always be no remainder?

Answer

Yes — there is never a remainder, for any starting two-digit number.

Let the number be ab, that is 10a + b. Reversed it is ba = 10b + a.
If b > a, the difference is
(10b + a) – (10a + b)
= 10b – b – 10a + a
= 9b – 9a
= 9(b – a)

Since 9(b – a) is 9 times a whole number, dividing it by 9 always leaves remainder 0.

Why it happens: the two numbers are built from the same two digits, just swapped between the tens place and the units place. Each digit therefore gains 10 of itself in one number and loses 1 of itself in the other — a change of 9 of itself. Subtracting leaves nine of each digit, and any combination of nines is a multiple of 9. Notice how little we assumed: only that a and b are digits. That is why the trick can never fail.

Check it yourself: 47 → 74. 74 – 47 = 27 = 9 × 3, and b – a = 7 – 4 = 3 ✓.

Q2.

Can you work out what happens if a > b?

Answer

Exactly the same thing — you just subtract the other way round.

If a > b then ab > ba, so the difference is
(10a + b) – (10b + a)
= 10a – a + b – 10b
= 9a – 9b
= 9(a – b)

So in both cases the difference is 9 × (the difference of the two digits), and dividing by 9 leaves no remainder.

Number	Reversed	Difference	÷ 9
72 (a > b)	27	45	5 = 7 – 2
38 (b > a)	83	45	5 = 8 – 3
91 (a > b)	19	72	8 = 9 – 1

Why it happens: the whole trick depends only on how far apart the two digits are, not on which one is bigger. Writing the difference as 9 × |a – b| covers both cases at once. If a = b the difference is 0 — still a multiple of 9 — which is exactly why the trick asks for a number “of different digits”, so that the answer is a genuine digit from 1 to 8.

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