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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

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1 Answer 1

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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).

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