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- #1

- Feb 14, 2012

- 3,802

Determine all positive integers $a,\,b$ and $c$ that satisfy equation $(a+b)!=4(b+c)!+18(a+c)!$.

- Thread starter anemone
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- Thread starter
- Admin
- #1

- Feb 14, 2012

- 3,802

Determine all positive integers $a,\,b$ and $c$ that satisfy equation $(a+b)!=4(b+c)!+18(a+c)!$.

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- #2

- Feb 7, 2012

- 2,738

There seems to be no obvious reason why $b\geqslant a$, so we should also look at the possibility $a>b$. Then a similar calculation to the one above leads to an equation like $xy = 4y + 18$, but with the $4$ and $18$ interchanged. However, that did not lead to any new solutions of the factorial equation. So there are only two solutions, namely

$7! = 4\cdot6! + 18\cdot5!$ (when $(a,b,c) = (3,4,2)$)

and

$22! = 4\cdot21! + 18\cdot21!$ (when $(a,b,c) = (11,11,10)$).