Question: Problem: Given that $34!=295232799cd96041408476186096435ab000000$. Find $a, b, c, d$.
$a, b, c, d$ are single digits.

I am able to find $a$ and $b$ but cant find $c, d$.
I did the prime factorisation of $34!$ using De-Polignac's formula.
I got $\frac{34!}{10^7}=2^{25}\times3^{15}\times7^4\times11^3\times13^2\times17^2\times19\times23\times29\times31$

So, i got $b=0$ and $a=2$ since $2^{25}\times3^{15}\times7^4\times11^3\times13^2\times17^2\times19\times23\times29\times31 \equiv 2 \mod 10$.

So, please help me find $c$ and $d$

Answer: 
HINT: The number is a multiple of $9$ and a multiple of $11$.
You know that $34!$ is divisible by $9$, so, because the sum of all other digits is $141$, you know that $c+d=3$ or $c+d=12$.

Now do the same for divisibility by $11$ (remember the alternating sums criterion?).