TITLE: Find the three digit number whose sum of 5 Permutations is 3194
QUESTION [5 upvotes]: Let $N={abc}$ be Three Digit Number such that $$abc+bca+bac+cab+cba=3194$$ Find the Number
My Try: I added both sides the left over number $acb$ both sides Then we get
$$222(a+b+c)-3194=b+10c+100a$$
Help needed from here

REPLY [5 votes]: The equation at the end of your post can be rewritten as
$$122a+221b+212c=3194$$
This in turn can be written as
$$122(a+b+c)+90(b+c)+9b=3194$$
From this we may conclude that $3194-122(a+b+c)$ must be a multiple of $9$.
This occurs when $a+b+c\equiv 7\pmod{9}$.
$a+b+c=7$ is too small.  But if $a+b+c=16$ (the next possible value), we are left with $$90(b+c)+9b=1242$$
To get the ones digit right, we must have $b=8$.  Then $c=5$.  And from $a+b+c=16$, we get $a=3$.
You can further verify that there are no solutions if $a+b+c=25$ (or more than 25).