Dividing Digit sum

Number Theory Level 4

FInd the sum of all $n$ such that $n^{2}-10n-135$ is the digital sum of $n$

Note- Digital sum denotes sum of digits. Like for 279 it is equal to 2+7+9. Here $n$ ranges over all integers.

The answer is 18.

2 solutions

Isaac Jiménez
Aug 22, 2014

First, let say $n$ is a one-digit number. So ${ n }^{ 2 }-10n-135=n\rightarrow n(n-11)=135$ which is not possible.

Now, let say $n$ is a two digit number. So, ${ n }^{ 2 }-10n-135\le 18$ , which is $n(n-10)\le 153$ so $10\le n\le 18$ . $n$ only works with $18$ .

If $n$ is a three digit number, then ${ n }^{ 2 }-10n-135\le 27\rightarrow n(n-10)\le 162$ , so $n$ can`t be a three digit number, neither greater.

So, $n=\boxed { 18 }$ .

I think my solution is very short, so I will explain some steps here, for those who don´t understand the solution:

In the second praragraph ${ n }^{ 2 }-10n-135\le 18$ because in this case I consider $n$ as a two digit number, and the TWO digit number with the maximum sum is $99$ with its sum equals $18$ .

In the third paragraph, $n$ is a three digit number, so as $999$ is the number with the greatest sum $27$ then ${ n }^{ 2 }-10n-135\le 27$ .

Continue posting great problems @Krishna Ar

Isaac Jiménez - 6 years, 9 months ago

You too post equally great solutions to my problems @Isaac Jiménez . Really awesome! In fact, I too had the same solution! :D

Krishna Ar - 6 years, 9 months ago

Brian Moehring
Aug 4, 2018

The digital sum of a non-negative number $n$ is at least $0$ and at most $n$ . Then solving $0\leq n^2-10n-135\leq n \iff n=18$

Then checking $(18)^2-10(18)-135 = 9 = 1+8$ shows the only solution is $\boxed{18}.$

Dividing Digit sum

The answer is 18.

2 solutions

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