# Proving or disproving that if $\Gamma(a)+\Gamma(b)= 121\,645\,106\,635\,852\,800$ both $a$ and $b$ are integers.

I found some formula about special function very complicated, so I am curious how you people solve this by hand.

$$\Gamma(a)+\Gamma(b)= 121\,645\,106\,635\,852\,800$$

but $a$ and $b$ are very small actually, condition: at least one of $a$ and $b$ are integers.

Can we prove or disprove both of them must be integer?

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The equation $$\Gamma(a)+\Gamma(b)=121645106635852800$$ does not have unique solutions in $a,b$. Are you looking for integers? –  anon Apr 19 '12 at 21:09
@anon - Is my version of problem looks harder? –  Victor Apr 19 '12 at 21:13
To try solving by hand one could say that since the number is 18 digits long it probably involves something like 18!. At that point you'd see this is not quite big enough and compute 19!. Then subtract 19! and get 6227020800 which is 10 digits long. Computing 10!, 11!, 12!, 13! one finally finds that your number is $19!+13!=\Gamma(20)+\Gamma(14)$. –  Bill Cook Apr 19 '12 at 21:20
In general, title+tags should be informative & tell the front page reader about the actual question being asked. –  user2468 Apr 19 '12 at 21:43
There are only $2 \ 0$s on the end so $a,b$ cannot both be $>15$, and neither is $<11$. The last digit non-zero digit of $6227020800$ is an $8$, and the digit before that is a $0$, and the last $2$ digits can easily be calculated to eliminate $10!,11!,12!,14!$. –  Angela Richardson Apr 21 '12 at 10:38