combinatorics - What is the highest power of $18$ contained in $frac{50!}{25!(50-25)!}$?

What is the highest power of $18$ contained in $\frac{50!}{25!(50-25)!}$?

How will I be able to find the answer to such questions? Is there any special technique to find the answer to such problems? Thank you.

Answer

$18 = 2\cdot 3^2$. We can find the power of a small prime in a large factorial by successive division to find base divisibility, then divisibility by squares, etc. So the multiplicity of powers of $2$ in $50!$, $v_2(50!),$ is $$ v_2(50!) = \left\lfloor\frac{50}{2}\right\rfloor + \left\lfloor\frac{50}{4}\right\rfloor + \left\lfloor\frac{50}{8}\right\rfloor + \cdots = 25+12+6+3+1 = 47$$

and similarly $v_2(25!)=22$, $v_3(50!)=16+5+1 = 22$ and $v_3(25!)=8+2 =10$, so

$$v_2\left(\frac{50!}{25!25!}\right) = 47-2\cdot22=3 \\ v_3\left(\frac{50!}{25!25!}\right) = 22-2\cdot 10=2 $$

and only $2$ available powers of $3$ means that $v_{18}\left(\frac{50!}{25!25!}\right)=1$ - the highest power of $18$ dividing the given expression is $18^1=18$.