Question

Number is 5 less than 9 times the sum of the digits. How do you find the number?

Answer 1

`31`

Explanation:

Suppose that the number is `a+10b+100c+1000d+10000e+ldots` where `a,b,c,d,e,ldots` are positive integers less than `10`.

The sum of its digits is `a+b+c+d+e+ldots`

Then, according to the problem statement, `a+10b+100c+1000d+10000e+ldots+5=9(a+b+c+d+e+ldots)`

Simplify to get `b+91c+991d+9991e+ldots+5=8a`.

Recall that all variables are integers between `0` and `9`. Then, `c,d,e,ldots` must be `0`, else it is impossible for the left-hand side to add up to `8a`.

This is because the maximum value `8a` can be is `8*9=72`, while the minimum value of `91c,991d,9991e,ldots` where `c,d,e,ldots≠0` is `91,991,9991,ldots`

As most of the terms evaluate to zero, we have `b+5=8a` left.

Since the maximum possible value for `b+5` is `9+5=14`, it must be the case that `a<2`.

So only `a=1` and `b=3` work. Thus, the sole possible answer is `a+10b=31`.