Factorial of 34 is really Huge!

Though its looks very hard, it's quite easy and interesting. It is completely based on divisibility rules.

First, find out the number of trailing zeros in $34!$ . There are actually $7$ trailing zeroes in $34!$ . So, $\color{#3D99F6}{B= \boxed{0}}$ .

Since Highest power of $2$ in $34!$ is $32$ . It implies that $2^{32-7}$ divides the number obtained after removing the $7$ zeroes including $B$ . It implies that $2^{4} | 435A$ ( divisibilty rule for 16 ). It implies that $\color{#3D99F6}{A=\boxed{2}}$ (Because $2$ is the only number which satisfies this.)

Since, $9|34!$ . It implies sum of the digits of $34!$ is divisible by $9$ .

So, $9|A+B+C+D+139$ , it implies that $\color{#20A900}{9|C+D+6}$ .

Also, $11|34!$ . It implies that difference of the sums of even and odd digits is divisible by 11.

So, $11|21+D-A-C$ , It implies that $\color{#20A900}{11|19+D-C}$ .

But, $0 < C+D+6 \leq 24$ and $0 < 19+D-C \leq 28$ .

So figuring out the possibilities

$\begin{cases} C+D+6 = 9 (or) \\ C+D+6 = 18 \end{cases}$ $\begin{cases} 19+D-C = 11 (or)\\ 19+D - C = 22 \end{cases}$

Solving the equations by taking the different combination of equations we get $(C,D) = \left\{(10,2),(0,3)\right\}$ as integral solutions as $C$ , $D$ are digits. But observe that $(C,D) = (10, 0)$ is not a valid solution because $C \leq 9$ . So, $\color{#3D99F6}{C=\boxed{0}}$ and $\color{#3D99F6}{D=\boxed{3}}$ .

Therefore, $\color{#D61F06}{A+B+C+D = \boxed{5}}$ .

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To know how to find the highest power of prime in a factorial, click here

To know how to find the number of trailing, it is actually finding the highest powers of $2$ and $5$ in that factorial.