Determine $4$ specific digits in $34!$ 
Find the values of $a,b,c,d\in\mathbb{N}$ such that
$$
34!=295232799cd9604140847618609643ab0000000
$$

My Attempt:
The factorial of $34$ contains a $3$, so the RHS must be divisible by 
$3$. Similarly, it must be divisible by $7$, $11$, $13$, $19$ etc.
But I do not understand how can I calculate $a,b,c,d$ in this equation.
 A: $34!$ has 7 powers of 5, which explains the last 7 0's. 
Since $34!$ has $17+8+4+2+1 = 32$ powers of 2, $34! / 10^{7}$ has $32-7=25$ powers of 2. Doing a divisibility by $2^{7} = 128$ on the last 7 digits, we get that $ab = 52$. 
(Note: If we had the last 3 digits missing, we could do a divisibility by $2^{10} = 1024$ on the last 10 digits. This is a useful approach that isn't often mentioned.)
Now, since these are digits, $0 \leq c, d \leq 9$, so $-9 \leq c-d \leq 9$ and $0 \leq c+d \leq 18$.
Use the fact that $34!$ is a multiple of 9, to tell you the value of $c+d$. We get that $c+d = 3$ or $12$. Use the fact that $34!$ is a multiply of 11, to tell you the value of $c-d$. We get that $c-d = -3$ or $8$. Since $2c$ is an even number from 0 to 18, we conclude that $c=0, d=3$.
A: So $34!=(\ldots) + b10^7+a10^8+c10^{27}+d10^{28}$.
Any divisibility rule by $k$ with $k\mid10^{20}-1$ cannot help here because it would allow to increase $a$ at the cost of $c$ and $b$ at the cost of $d$.
Unfortunately, this rules out $9$ and $11$. But $7$ and $13$ combined should be helpful (note that $7\cdot 13=10^2-10+1$).
A: Ah, nice. Have you tried applying the divisibility criteria (in terms of decimal digits) for the various numbers you mention? Did I take it right that you have just covered those criteria?
