Big Factorials And Rules With Their Digits This is the issue I have. 
Given that $34! = 295,232,799,**cd**9,604,140,847,618,609,643,5**ab**,000,000$ determine the digits $a$, $b$, $c$, $d$. 
As far as I've gotten is $34!$ contains these multiples of $5$. $5$, $10$, $15$, $20$, $25$, $30$. So $34!$ is divisible by $5^7$ ($25$ has two powers of $5$), and since there's at least that power of $2$, the last $7$ digits are $0$, so b is $0$.

Yes, $b=0$ for the reason you stated ($34!$ is divisible by $10^7$ but not $10^8$ so $b=0$ but $a\neq0$). That's step $1$.
Step $2$. Now say you divided $34!$ by the $10^7$ so you take away seven products of $2$ and seven product of $5$ (speaking in terms of prime factors). What you're left with has at least three products of $2$ (since every other integer up to $34$ is even so there's at least seventeen products of $2$). If a number is divisible by $8$ then it's last three digits is also divisible by $8$ since $1000$ (and any multiple of it) is divisible by $8$. Therefore $35a$ is divisible by $8$.
$35a = 300 + 50 + a = (4 + 2 + a) ($mod $ 8)$. $4+2+a$ can only equal $8$ (not $16$ since $a$ would have to be $10$), there $a=2$.
Step $3$. Now, $34!$ is also divisible by $9$ (which is relatively prime to $10$) so it's digits must add up to a multiple of $9$. The sum of the digits (including $a$ and $b$) comes to $141+c+d$. $141+c+d = (6+c+d)($mod $9)$. $6+c+d=9$ or $18$.
So $c+d=3$ or $12$. Let's call this equation $(1)$.
Step $4$. Now, $34!$ is also divisible by $11$ (which is relatively prime with $10$ and $9$) so (according to a proof I'll address in a second) the sum of the odd-positioned numbers subtracted from the evenly-positioned numbers is a multiple of $11$. This gives the equation $80 - 61 + d - c = (8 + d - c)($mod $11) = 0$ $ ($mod $11)$. So $8+d-c=0$ or $11$ .
So $d-c=-8$ or $3$. Let's call this equation $(2)$.
$(1)+(2)$ gives: $2d=$ $...$
This must be the sum of two evens or two odds for $d$ to be an integer.
So the two potential pairs give the following:
$(c,d)=(10,2)$ or $(0,3)$.
Finally, $(a,b,c,d)=(2,0,0,3)$.
The proof for step $4$ is as follows. Basically, $10=(-1)($mod $11)$ and $100=10^2=1$ $($mod $11)$. Every odd power of ten gives $(-1)($mod $11)$ and every even power gives $1$ $($mod $11)$. So the sum of the evenly-positioned digits (are multiplied by an odd power of $10$ and therefore are essentially multiplied by $-1$. Equally the sum of the oddly-positioned digits (are multiplied by an even power of $10$ and therefore are essentially multiplied by $1$.
