Question 392999
We have


{{{(100a + 10b + c)/(a + b + c) = 26}}} --> {{{100a + 10b + c = 26a + 26b + 26c}}} (a, b, c are digits). Subtracting {{{a+b+c}}} from both sides,


{{{99a + 9b = 25a + 25b + 25c}}}


This implies 9 divides a+b+c and 25 divides 11a+b. Therefore, the only possible values of a+b+c are 9, 18, 27 (since 36 is too high).


Suppose a+b+c = 9. Then, {{{a+b = 9-c}}}, and {{{11a + b = 9 - c + 10a}}} (by substituting) which is divisible by 25. Since 9 - c + 10a is congruent to 9 - c (mod 10) = 5 or 0, the only possibility for c is 4 (since 9 is impossible). This leaves {{{a+b = 5}}} (note that 11a + b is divisible by 25). The easiest way is simply guess and check to find any solutions; we see that a = 2, b = 3 works since 11(2) + 3 is divisible by 25.


Therefore the smallest number is 234.