Highest Possible Number for $x$ (Part 3)

$\begin{aligned} y & = \frac {\prod_{k=1}^{12}k!}{2^x} & \small \color{#3D99F6} \text{Factorize all }k! \\ & = \frac {2^{q_2}3^{q_5}5^{q_5}7^{q_7}11^{q_{11}}}{2^x} & \small \color{#3D99F6} \text{where }q_j \text{ is the power of prime factor }p_j \end{aligned}$

For $y$ to be an integer $2^x \mid 2^{q_2}3^{q_5}5^{q_5}7^{q_7}11^{q_{11}}$ . Then $\max (x) = q_2$ .

We note that the formula to find the power of prime factor $r_p$ in $n!$ is given by $r_p(n) = \displaystyle \sum_{k=1}^\infty \left \lfloor \dfrac n{p^k} \right \rfloor$ . Therefore,

$\begin{aligned} p_2 & = \sum_{n=1}^{12} r_2(n) = \sum_{n=1}^{12} \sum_{k=1}^\infty \left \lfloor \frac n{2^k} \right \rfloor \\ & = \sum_{n=1}^{12} \left \lfloor \frac n2 \right \rfloor + {\color{#3D99F6}\sum_{n=1}^{12} \left \lfloor \frac n4 \right \rfloor} + \color{#D61F06}\sum_{n=1}^{12} \left \lfloor \frac n8 \right \rfloor \\ & = \sum_{n=2}^{12} \left \lfloor \frac n2 \right \rfloor + {\color{#3D99F6}\sum_{n=4}^{12} \left \lfloor \frac n4 \right \rfloor} + \color{#D61F06}\sum_{n=8}^{12} \left \lfloor \frac n8 \right \rfloor \\ & = 2(1+2+3+4+5)+6 + {\color{#3D99F6}4(1+2)+3} + \color{#D61F06} 5 \\ & = 36 + {\color{#3D99F6}15} + \color{#D61F06} 5 \\ & = \boxed{56} \end{aligned}$

---

Notation: $\lfloor \cdot \rfloor$ denotes the floor function .

The product written out looks like this:

$1!\times2!\times3!\times4!\times5!\times6!\times7!\times8!\times9!\times10!\times11!\times12!$

The factorials all written out look like this:

$1\times2\times3\times4\times5\times6\times7\times8\times9\times10\times11\times12\times$

$1\times2\times3\times4\times5\times6\times7\times8\times9\times10\times11\times$

$1\times2\times3\times4\times5\times6\times7\times8\times9\times10\times$

$1\times2\times3\times4\times5\times6\times7\times8\times9\times$

$1\times2\times3\times4\times5\times6\times7\times8\times$

$1\times2\times3\times4\times5\times6\times7\times$

$1\times2\times3\times4\times5\times6\times$

$1\times2\times3\times4\times5\times$

$1\times2\times3\times4\times$

$1\times2\times3\times$

$1\times2\times$

$1$

We want to divide this by as many 2s as possible to get the highest value for x, so we need to know exactly how many 2s there are in here.

Each 2 is divisible by 2 once.

Each 4 is divisible by 2 twice.

Each 6 is divisible by 2 once.

Each 8 is divisible by 2 three times.

Each 10 is divisible by 2 once.

Each 12 is divisible by 2 twice.

This gives us the total number of 2s in the prime factorisation of this huge number:

$(\#\,of\,2s)+2(\#\,of\,4s)+(\#\,of\,6s)+3(\#\,of\,8s)+(\#\,of\,10s)+2(\#\,of\,12s)$

$11+2(9)+7+3(5)+3+2(1)$

$=56$

So the highest possible integer value of x is 56.