AIME 2015 Problem 3

We know that the answer is less than $1000$ due to constraints on the $AIME$ , and so we can say that $100a+10b+c=17k$ for some integers $a,b,c,k$ . We also know that $a+b+c=17$ and $a,b,c \leq 9$ .

We can then substitute $17-a-b$ for $c$ and get $100a+10b+17-a-b=17k$ .

Working in $\mod 17$ we get that $99a+9b=17k\\9b-3a=17k\\3(3b-a)=17k$

We now know that $k$ is a multiple of $3$ , so let $k=3p$

$3b-a=17p$

For $p=1$ , we get:

$a=1,b=6,c=10\\a=4,b=7,c=6$

We can reject the first one since $c \leq 9$

If $p \geq 2$ , then $b \geq 10$ , so $\boxed{476}$ is our smallest possibility

I dunno tat answer shud b less than 1000...

Wat abt using a c++ program for this? (I very well know tat there are many other prog. Language available .. But c++ is my first prog. Language im learning)

nice problem