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

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

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Answer 1 · 2017-10-31T13:32:37

$31$

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 $a+10b=31$ .

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

1 Answer

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