HMMT-Nov 2010 guts, Problem 10

What is the smallest integer greater than 10 such that the sum of the digits in its base 17 representation is equal to the sum of the digits in its base 10 representation?

Solution: We assume that the answer is at most three digits (in base 10). Then our desired number can be expressed in the form $\overline{a b c}_{10}=\overline{d e f}_{17}$, where $a, b, c$ are digits in base 10 , and $d, e, f$ are digits in base 17 . These variables then satisfy the equations \[ \begin{aligned} 100 a+10 b+c & =289 d+17 e+f \\ a+b+c & =d+e+f \end{aligned} \] Subtracting the second equation from the first, we obtain $99 a+9 b=288 d+16 e$, or $9(11 a+b)=$ $16(18 d+e)$. From this equation, we find that $11 a+b$ must be divisible by 16 , and $18 d+e$ must be divisible by 9 . To minimize $\overline{a b c}$, we find the minimal possible value of $a$ : If $a=0$, then the only way for $11 a+b=b$ to be divisible by 16 is to set $b=0$; however, this is disallowed by the problem condition, which stipulates that the number must be greater than 10 . If we try $a=1$, then we find that the only possible value of $b$ which lets $11 a+b=b+11$ be divisible by 16 is $b=5$. Plugging these in and simplifying, we find that we must have $18 d+e=9$. The only possible solution to this is $d=0, e=9$. Now to satisfy $a+b+c=d+e+f$, we must have $1+5+c=0+9+f$, or $c=f+3$. The minimal possible solution to this is $c=3, f=0$. So our answer is $\overline{a b c}=153$, which is also equal to $090_{17}$. $\fbox{153}$.

Answer: 153