$34!=295232799cd96041408476186096435ab000000$, find $a$ ,$b$, $c$, and $d$

Knowing $b=0$ and $\frac{34!}{10^7}$ being a multiple of $8$, the last three non-zero digits $35a$ must be divisible by $8$, so $a=2$ for sure. Now set up linear equations modulo $9$ and $11$ (this is just a digit sum and alternating digit sum, and I learned this from studying the Trachtenberg system a long time ago) and we get $$d+c\equiv3\bmod9$$ $$d-c\equiv3\bmod11$$ since $9,11\mid34!$ By trial and error we see that $d=3$ and $c=0$, so the solution is $\overline{abcd}=2003$.

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Factorial

Elementary Number Theory

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