C1. How many three-digit multiples of 9 consist only of odd digits?
Solution
We use the fact that 'an integer is a multiple of 9 when the sum of its digits is a multiple of 9 , and not otherwise'.
Consider a three-digit integer with the required properties. Each digit is between 0 and 9 , and none of them is zero, so the sum of the digits is between 1 and 27 . Since we want the integer to be a multiple of 9 , the sum of the digits is therefore 9,18 or 27 .
However, it is not possible to write 18 as a sum of three odd numbers, and the only way of making the sum of the digits equal to 27 is with 999 , which is thus one possible integer. But the remaining question is 'how can we make the sum of the digits equal to 9 ?'
If one of the digits is 1 , then we can make the remaining 8 in two ways:
$$
1+7
$$
giving the three integers 117,171 and 711 ;
$$
3+5
$$
giving the six integers $135,153,315,351,513$ and 531.
If we do not use a 1 , then the only possible integer is 333 .
Hence there are eleven three-digit multiples of 9 consisting only of odd digits.