Question with Natural numbers

m and n are natural numbers. 1/m + 1/n is equal to 0,1 , what is the sum of the values that n can take ?

Do you mean 1/m + 1/n = 1/10 ? (Your use of comma for decimal is ill-advised.)

If so then this can be rearranged to n = 10m / (m - 10)

Obviously you need m > 10 so try any value from 11 upwards to see whether n is an integer.

For example m = 11 leads to n = 110 and 1/110 + 1/11 = 1/10

EDIT. How can Tom say that n = 20 is the only solution when I have already shown another!

Here's two more. 1/12 + 1/60 = 1/10 and 1/14 + 1/35 = 1/10

The sum of n values is over 300.

EDIT. Oh, I see that he's changed his answer following my advice. But why give me a TD for pointing out his error in the first place. That's just childish.

(1/m) + (1/n) = 1/10

=> (n+m)/(m*n) = 1/10

=> 10*(n+m) = m*n

=> 10n + 10m = m*n /-10n

=> (m-10)*n = 10m

first case: m=10
=> 0 = 100 contradiction!

second case: m≠10
=> n = 10m/(m-10)

polynomial divission:

10m:(m-10) = 10 + 100/(m-10)
-(10m-100)
-----------------
100
-100
-------
0

By what numbers 100 is divissible?
By 1;2;4;5;10;20;25;50 and 100.

So m-10 can take the values
1;2;4;5;10;20;25;50 or 100.

=> m can take the values
11;12;14;15;20;30;35;60;110

=> The solution set of this problem are the
ordered pairs

L={(m;n)∈|N: (11;110);(12;60);(14;35);
(15;30);(20;20);(30;15);(35;14);(60;12)…
==================================

The sum of the all possible values of n is
110+60+35+30+20+15+14+12+11 = 307
==============================

M=N=1 OR M=N=2