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

Book page 124 Updated on2026-09-05

Q1.
Can we say that any number made up of only the digits ‘0’ and ‘9’, in any order, will always be divisible by 9?
Answer

Yes. If every digit is 0 or 9, then in the expanded form every single term is a multiple of 9.

99009 = 9 × 10000 + 9 × 1000 + 0 × 100 + 0 × 10 + 9 × 1
= 9 × (10000 + 1000 + 1)
= 9 × 11001 = 99009 ✓

The same works for 90, 909, 9900, 90909, 999000 — take 9 out as a common factor and what remains is a whole number.

Tip: this shortcut finds some multiples of 9 but not all of them — 18, 27, 36 and 405 are multiples of 9 with no 9 in sight. Unlike 2, 5 and 10, the units digit alone tells you nothing about 9: both 99 and 109 end in 9, but only 99 is a multiple of 9.
Q2.
Is 10 divisible by 9? If not, what is the remainder? Check the divisibility of other multiples of 10 (10, 20, 30, ...) by 9.
Answer

No — 10 leaves a remainder of 1, because 10 = 9 + 1.

Multiple of 10102030405060708090
Remainder ÷ 9123456780

So for any multiple of 10 the remainder is the number of tens.

Why it happens: 10 = 9 + 1, so b tens is 10b = 9b + b. The 9b part is a whole number of 9s and disappears; the b is what is left over. (At b = 9 the leftover 9 forms one more complete group, which is why 90 leaves 0.)
Q3.
Similarly, look at the remainder when the multiples of 100 (100, 200, 300, … ) are divided by 9. What do you notice?
Answer

The remainder is the number of hundreds.

Multiple of 100100200300400500700900
Remainder ÷ 91234570
100 = 99 + 1, and 99 is a multiple of 9
So 100c = 99c + c → remainder c

Check 700: 700 = 693 + 7, and 693 = 9 × 77 ✓ so the remainder is 7.

Did you notice? The tens behaved this way because 10 = 9 + 1; the hundreds behave this way because 100 = 99 + 1. The same trick is about to work for every place value.
Q4.
Using this observation, find the remainder when 427 is divided by 9.
Answer

The remainder is 4.

427 has 4 hundreds → contributes remainder 4
427 has 2 tens → contributes remainder 2
427 has 7 units → contributes remainder 7
Total leftover = 4 + 2 + 7 = 13
13 makes one more group of 9, leaving 4

Check: 427 = 9 × 47 + 4 = 423 + 4 ✓

Why it happens: 400 = 396 + 4 and 20 = 18 + 2, where 396 and 18 are multiples of 9. So 427 = (396 + 18) + (4 + 2 + 7) — a big multiple of 9 plus the digit sum. Only the digit sum can leave a remainder.
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