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NCERT Solutions Ganita Prakash (Part 1) Chapter 5 A Shortcut for Divisibility by 9 — In-text Questions

Book page 125 Updated on2026-09-05

Q1.
Will this work with bigger numbers?
Answer

Yes — for every place value, however large. That is the whole point of the pattern:

1 = 0 + 1
10 = 9 + 1
100 = 99 + 1
1000 = 999 + 1
10000 = 9999 + 1, and so on

Each place value is 1 more than a multiple of 9, so each digit contributes exactly itself to the remainder. Take 7309:

7 × 1000 + 3 × 100 + 0 × 10 + 9 × 1
= 7 × (999 + 1) + 3 × (99 + 1) + 0 × (9 + 1) + 9 × (0 + 1)
= (7 × 999 + 3 × 99 + 0 × 9 + 9 × 0) + (7 + 3 + 0 + 9)
= (a multiple of 9) + 19

Now shrink 19 the same way: 1 + 9 = 10, then 1 + 0 = 1. So 7309 is 1 more than a multiple of 9, and 7309 ÷ 9 leaves remainder 1.

Check: 7309 = 9 × 812 + 1 = 7308 + 1 ✓

Tip: the rule to remember — a number is divisible by 9 if and only if the sum of its digits is divisible by 9, and adding the digits repeatedly down to one digit gives the remainder (with 9 standing for remainder 0).
Q2.
Look at each of the following statements. Which are correct and why? (i) If a number is divisible by 9, then the sum of its digits is divisible by 9. (ii) If the sum of the digits of a number is divisible by 9, then the number is divisible by 9. (iii) If a number is not divisible by 9, then the sum of its digits is not divisible by 9. (iv) If the sum of the digits of a number is not divisible by 9, then the number is not divisible by 9.
Answer

All four are correct.

Everything follows from one fact established on this page:

N = (a multiple of 9) + (sum of the digits of N)
So N and its digit sum leave the same remainder on division by 9
StatementFormCorrect?
(i) 9 | N ⇒ 9 | digit sumthe 'only if' halfYes
(ii) 9 | digit sum ⇒ 9 | Nthe 'if' half — the converse of (i)Yes
(iii) 9 ∤ N ⇒ 9 ∤ digit sumcontrapositive of (ii)Yes
(iv) 9 ∤ digit sum ⇒ 9 ∤ Ncontrapositive of (i)Yes
Why all four survive: a statement and its contrapositive are always either both true or both false, so (iii) rides on (ii) and (iv) rides on (i). What is not automatic is that (i) and (ii) are both true — a statement and its converse usually part company. Here they do not, because the remainders are equal, not merely related.
Check it yourself: 405 → 9 ✓ and 405 = 9 × 45 ✓ (both (i) and (ii) hold). 8888 → 32 ✗ and 8888 = 9 × 987 + 5 ✗ (both (iii) and (iv) hold).
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