November 2, 2009
Without
performing multiplications, find the digits $a$ and $b$ in
23! = 2585201ab38884976640000.
Solution.
The sum of the given digits is 86, while the sum of all the
digits must be a multiple of 9 (since 9 divides $23!$). Therefore
$a+b=4$ or $a+b=13$. Also, the alternating sum of the given digits is
10, while the alternating sum of all the digits must be 11 (since 11
divides $23!$). Hence $b-a=1$. The only integer solution to these
equations is $a=6$ and $b=7$.