Highest Possible Number for $x$

$\begin{aligned} y & = \frac {100!}{2^x} & \small \color{#3D99F6} \text{Factorize all }k! \\ & = \frac {2^{q_2}3^{q_5}5^{q_5}\cdots 97^{q_{97}}}{2^x} & \small \color{#3D99F6} \text{where }q_j \text{ is the power of prime factor }p_j \end{aligned}$

For $y$ to be an integer $2^x \mid 2^{q_2}3^{q_5}5^{q_5}\cdots 97^{q_{97}}$ . 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,

$\begin{aligned} p_2 & = \sum_{k=1}^\infty \left \lfloor \frac {100}{2^k} \right \rfloor \\ & = \left \lfloor \frac {100}2 \right \rfloor + \left \lfloor \frac {100}4\right \rfloor + \left \lfloor \frac {100}8 \right \rfloor + \left \lfloor \frac {100}{16} \right \rfloor + \left \lfloor \frac {100}{32} \right \rfloor + \left \lfloor \frac {100}{64} \right \rfloor \\ & = 50 + 25 + 12 + 6 + 3 + 1 \\ & = \boxed{97} \end{aligned}$

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

It's the number of powers of 2 that are less than 100 + the number of powers of 4 that are less than 100 + ... + the number of powers of 64 that are less than 100.