TITLE: Find the missing digits in the expansion of $34!$
QUESTION [3 upvotes]: 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.

REPLY [8 votes]: $34! = 295232799cd96041408476186096435ab000000$
$\left \lfloor \dfrac{34}{5} \right \rfloor = 6$
$\left \lfloor \dfrac{6}{5} \right \rfloor = 1$
So there are $6+1 = 7$ zeros at the end of $34!$. Hence $$\color{red}{b = 0}$$

THEOREM: Compute the following 

$N = 5q_1 + R_1$
$q_1 = 5q_2 + R_2$
$q_2 = 5q_3 + R_3$
...
$q_{n-1} = 5q_n + R_n$

where $0 \le R_i < 5$ for all $i$ and $0 \le q_n < 5$.
Then the first non zero digit in $N!$ is 
$U(N!)
= 2^P \times Q! \times R_1! \times R_2! \times R_3! \dots \times R_n! \pmod{10}$
Where    

$P = q_1 + q_2 + \cdots + q_n$
$Q = q_n$

We compute
\begin{align}
    34 &= 5(6) + 4 \\
     6 &= 5(1) + 1 \\
     1 &= 5(0) + 1 \\
\end{align}
$P = 6 + 1 = 7$
$Q = 0$
\begin{align}
    U(34!)
        &= 2^7 \times 0! \times 4! \times 1! \times 1! \pmod{10} \\
        &=   8 \times 0! \times  4 \times 1! \times 1! \pmod{10} \\ 
        &=   2
\end{align}
So $$\color{red}{a = 2}$$

$34! = 2\; 95\; 23\; 27\; 99\; \color{red}{cd}\; 96\; 04\; 14\; 08\; 47\; 61\; 86\; 09\; 64\; 35\; 20\; 00\; 00\; 00$
Clearly $99 \mid 34!$ So, when we cast out $99's$, we should get $0$. Pairing off the numbers in $34!$ from right to left, skipping $cd$, and adding modulo $99$, we get
$ 2 + 95 + 23 + 27 + 99 + 96 + 04 + 14 + 08 + 
 47 + 61 + 86 + 09 + 64 + 35 + 20 + 00 + 00 + 00 \pmod{99} = 96$
So $cd = 99 - 96 = 03$
Hence
$$ \color{red}{c = 0} $$
$$ \color{red}{d = 3} $$
{"set_name": "stack_exchange", "score": 3, "question_id": 1803146}