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

Question

There was a number theory question that I have to do for homework.

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

I know $b=0$ because $10^7\big|34!$ only. But how can I find other variables?

Remark:

No electronic calculation device can be used for calculation. (I need to write the full solution.)

Can someone help me?

Answer 1

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 $\overline{abcd}=2003$.

Answer 2

Sum of digits is divisible by $9$, which gives $$a+c+d\equiv5(\mod9).$$ $$2-9+5-2+3-2+7-9+9-c+d-9+6-0+4-1+4-0+8-4+7-6+1-8+6-0+9-6+4-3+5-a\equiv0(\mod11),$$ which gives $$a+c-d\equiv-1(\mod11).$$ From here we can get $a+c=2$ and $d=3$.

Now, since $a\in\{2,6\}$, we obtain $a=2$, $c=0$ and we are done!