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

Oct 31, 2017

$31$

#### Explanation:

Suppose that the number is $a + 10 b + 100 c + 1000 d + 10000 e + \ldots$ 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, $a + 10 b + 100 c + 1000 d + 10000 e + \ldots + 5 = 9 \left(a + b + c + d + e + \ldots\right)$

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

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 $8 a$.

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

As most of the terms evaluate to zero, we have $b + 5 = 8 a$ 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 $a + 10 b = 31$.