Question

The number of ‘three-digit numbers’ which are multiples of 9 are
[a] 101
[b] 99
[c] 100
[d] 98

Answer 1

Hint: Use the fact that the smallest three-digit number is 100 and the largest three-digit number is 999. Use the fact that the smallest number that must be added to a number “a” so that it is a multiple of b is the remainder obtained on dividing a by b subtracted from b. Use the fact that the smallest number that must be subtracted from a number “a” so that the result is a multiple of b is the remainder obtained on dividing “a” by b. Use the fact that the number of multiples of c between two multiples a and b of c is given by $\dfrac{b-a}{c}+1$. Hence find the total number of three-digit numbers which are multiples of 3.
Complete step by step solution:
We know that the smallest three-digit number is 100.
We know that the smallest number that must be added to a number a so that it is a multiple of b is the remainder obtained on dividing a by b subtracted from b.
Now, we have
$100=9\times 11+1$
Hence the smallest number that must be added to 100 to make it a multiple of 9 is 9-1 = 8
Hence the smallest three-digit number divisible by 9 is 100+8 = 108
We know that the largest three-digit number is 999.
Since 999 is a multiple of 9, we have the largest three-digit number divisible by 9 is 999.
We know that the number of multiples of c between two multiples a and b of c is given by $\dfrac{b-a}{c}+1$
Hence, we have
The number of multiples of 9 $=\dfrac{999-108}{9}+1=111-12=99+1=100$
Hence option [c] is correct.
Note: Alternative solution:
We know that the number of multiples of x in the numbers from 1-n is $\left[ \dfrac{n}{k} \right]$, where [x] denotes the greatest integer less or equal to x.
Hence, we have
Number of multiple of 9 from 1-999 is $\left[ \dfrac{999}{9} \right]=111$
Number of multiples of 9 from 1-99 is $\left[ \dfrac{99}{9} \right]=11$
Hence the number of multiples of 9 from 100-999 is 111-11 = 100.
Hence option [d] is correct.