NCERT Solutions Ganita Prakash (Part 1) Chapter 5 A Shortcut for Divisibility by 9 — Figure it Out

Book page 126 Updated on2026-09-05

Q1.

Find, without dividing, whether the following numbers are divisible by 9. (i) 123 (ii) 405 (iii) 8888 (iv) 93547 (v) 358095

Answer

Only (ii) 405 is divisible by 9.

Number	Sum of digits	Reduced	Divisible by 9?
123	1 + 2 + 3 = 6	6	No (remainder 6)
405	4 + 0 + 5 = 9	9	Yes
8888	8 + 8 + 8 + 8 = 32	3 + 2 = 5	No (remainder 5)
93547	9 + 3 + 5 + 4 + 7 = 28	2 + 8 = 10 → 1	No (remainder 1)
358095	3 + 5 + 8 + 0 + 9 + 5 = 30	3 + 0 = 3	No (remainder 3)

Tip: while adding, throw away any 9s and any pairs that make 9. For 358095 the 9 goes, 3 + 5 + 8 + 5 = 21, then 2 + 1 = 3 — same answer, less work.

Q2.

Find the smallest multiple of 9 with no odd digits.

Answer

The smallest one is 288.

Allowed digits: 0, 2, 4, 6, 8 — all even
A sum of even digits is even
So the digit sum must be an even multiple of 9 → 18, 36, …

Two digits? The largest possible digit sum is 8 + 8 = 16, which is less than 18. Impossible.

Three digits? Keep the hundreds digit as small as possible, then the tens:

Hundreds digit 2 → the other two even digits must add to 16
Only 8 + 8 = 16 works
→ 288

Check: 2 + 8 + 8 = 18, a multiple of 9, and 288 = 9 × 32 ✓. Every digit is even ✓. Nothing smaller can work, because any number below 288 with all even digits has at most three digits and either starts with 2 (forcing 8 and 8) or is a two-digit number.

Q3.

Find the multiple of 9 that is closest to the number 6000.

Answer

The closest multiple of 9 is 6003.

6 + 0 + 0 + 0 = 6, so 6000 is 6 more than a multiple of 9
Multiple of 9 just below: 6000 – 6 = 5994 (= 9 × 666)
Multiple of 9 just above: 5994 + 9 = 6003 (= 9 × 667)

6000 – 5994 = 6 6003 – 6000 = 3

6003 is only 3 away, so it is the closer of the two.

Tip: the digit sum told us the remainder is 6 without any division. Whenever the remainder is more than 4½, the multiple above is the nearer one.

Q4.

How many multiples of 9 are there between the numbers 4300 and 4400?

Answer

There are 11 of them.

4 + 3 + 0 + 0 = 7, so 4300 is 7 more than a multiple of 9
First multiple of 9 above 4300: 4300 + (9 – 7) = 4302 = 9 × 478
Last multiple of 9 below 4400: 4 + 4 + 0 + 0 = 8, so 4400 – 8 = 4392 = 9 × 488

Count = 488 – 478 + 1 = 11

The eleven numbers are 4302, 4311, 4320, 4329, 4338, 4347, 4356, 4365, 4374, 4383, 4392.

Why 11 and not 100 ÷ 9 ≈ 11.1: in a stretch of 100 consecutive numbers there are either 11 or 12 multiples of 9, depending on where the stretch starts. Counting the first and last one exactly is the safe way.

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