<s>[INST] Find the missing digits in the expansion of $34!$ If $34!=\overline{295232799cd96041408476186096435ab000000}$ then find the value of $a,b,c$ and $d.$
My Attempt: I can find that $b=0$ because it has seven five integers.
$\lfloor{\frac{34}{5}}\rfloor+\lfloor{\frac{34}{25}}\rfloor=6+1=7$
Also I think $a$ can be found using divisibility rule of $8$ that means $\overline{35a}$ is divisible by $8$ which gives us $a=2$. But I'm stuck in finding $c$ and $d$. Using the divisibility rule of $9$ I get an equation. And using divisibility rule of $11$ I get another equation but when I solve them I get two values for $c$ and $d$.
Note: calculator is not allowed.
 [/INST] This is a partial solution...
Use all the divisibility rules.
But first, you can notice that $32!=2^31\times5^7\times\cdots$, so there are $7$ zeros at the end of the number. So $b=0$, and $a\ne 0$.
$34!$ is divisible by $9$, so: 
$$4+a+c+d=0\pmod 9.$$
It's divisible by $7$, so: 
$$000-000+5a0-643+609-618+847-140+604-cd9+799+327-952+2=0\pmod 7,$$
so 
$$835+5a0-cd9=0\pmod 7.$$
It's divisible by $11$, so: 
$$2-2+5-2+9-7+2-3+9-9+7-9+d-c+4-0+6-0+4-1+7-4+1-7+4-8+8-1+6-9+0-6+3-4+6-0+a-5=0\pmod {11},$$
so
$$6+a-c+d=0\pmod {11}.$$
It's also divisible by $13$, so you can look at the alternate sum of three digits from the left.
And so on...
Try to look at this article about divisibility rules.
I hope this helps!
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