NCERT Solutions – Number Play
Exercise 5.21. Check divisibility by 9 (using sum of digits rule):
Rule: A number is divisible by 9 if the sum of its digits is divisible by 9.
- (i) 123 → 1+2+3 = 6 → Not divisible by 9
- (ii) 405 → 4+0+5 = 9 → Divisible by 9
- (iii) 8888 → 8+8+8+8 = 32 → Not divisible by 9
- (iv) 93547 → 9+3+5+4+7 = 28 → Not divisible by 9
- (v) 358095 → 3+5+8+0+9+5 = 30 → Not divisible by 9
2. Smallest multiple of 9 with no odd digits:
We need digits only from {0, 2, 4, 6, 8}
Try smallest numbers:
18 → has odd digit ❌
36 → has odd digit ❌
72 → has odd digit ❌
90 → digits (9,0) ❌
Answer: 0 (smallest multiple of 9 with no odd digits)
3. Multiple of 9 closest to 6000:
6000 ÷ 9 = 666 remainder 6
Closest multiples:
9 × 666 = 5994
9 × 667 = 6003
Difference:
6000 − 5994 = 6
6003 − 6000 = 3
Answer: 6003 (closest)
4. Number of multiples of 9 between 4300 and 4400:
First multiple ≥ 4300:
4302 (9 × 478)
Last multiple ≤ 4400:
4392 (9 × 488)
Count:
= 488 − 478 + 1 = 11
Answer: 11 multiples