NCERT Solutions for Class 8 Maths Ganita Prakash Chapter 5 — Figure it Out (Page 126)#
Every place value is one more than a multiple of 9: $10 = 9 + 1$, $100 = 99 + 1$, $1000 = 999 + 1$, … So a number leaves the same remainder on division by 9 as the sum of its digits. In particular, a number is divisible by 9 exactly when the sum of its digits is divisible by 9.
1. Find, without dividing, whether the following numbers are divisible by 9. (i) 123 (ii) 405 (iii) 8888 (iv) 93547 (v) 358095 #
Solution
We add the digits of each number:
| Number | Sum of digits | Divisible by 9? |
|---|---|---|
| (i) 123 | $1 + 2 + 3 = 6$ | No |
| (ii) 405 | $4 + 0 + 5 = 9$ | Yes |
| (iii) 8888 | $8 + 8 + 8 + 8 = 32$ | No |
| (iv) 93547 | $9 + 3 + 5 + 4 + 7 = 28$ | No |
| (v) 358095 | $3 + 5 + 8 + 0 + 9 + 5 = 30$ | No |
∴ Only 405 is divisible by 9 ($405 = 9 \times 45$).
2. Find the smallest multiple of 9 with no odd digits. #
Solution
The digits can only be 0, 2, 4, 6 and 8, and the digit sum must be a multiple of 9.
- Even digits always add up to an even number, so the digit sum cannot be 9. The smallest possible digit sum is therefore 18.
- One or two even digits add up to at most $8 + 8 = 16$, so we need at least three digits.
- Three even digits adding to 18: the only choices are $\{2, 8, 8\}$, $\{4, 6, 8\}$ and $\{6, 6, 6\}$. The smallest number we can make is with the smallest first digit, 2: 288.
Check: $288 = 9 \times 32$.
∴ The smallest multiple of 9 with no odd digits is 288.
3. Find the multiple of 9 that is closest to the number 6000. #
Solution
The digit sum of 6000 is 6, so 6000 is 6 more than a multiple of 9:
$6000 - 6 = 5994$ (a multiple of 9, 6 below 6000)
$5994 + 9 = 6003$ (the next multiple of 9, 3 above 6000)
∴ The multiple of 9 closest to 6000 is 6003.
4. How many multiples of 9 are there between the numbers 4300 and 4400? #
Solution
First multiple after 4300. The digit sum of 4300 is 7, so $4300 = 9k + 7$. The next multiple of 9 is 2 more:
First multiple $= 4300 + 2 = 4302$
Last multiple before 4400. The digit sum of 4400 is 8, so:
Last multiple $= 4400 - 8 = 4392$
The multiples go up in steps of 9:
Number of multiples $= \dfrac{4392 - 4302}{9} + 1$
Number of multiples $= 10 + 1$
Number of multiples $= 11$
∴ There are 11 multiples of 9 between 4300 and 4400: 4302, 4311, …, 4392.