Let $P_n$ be the product of all positive divisors of a positive integer $n$ . For example, $P_1 = 1$ and $P_4 = 1 \times 2 \times 4 = 8.$

If $2^m$ is the highest integer power of $2$ that divides $P_{36000}$ , what is $m$ ?

The answer is 180.

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Note that $36000 = 2^5 \cdot 3^2 \cdot 5^3$ . Also, note that a divisor of $36000$ is of the form $2^a \cdot 3^b \cdot 5^c (a, b, c \in \mathbb{N}_0)$ , where $a \leq 5$ , $b \leq 2$ and $c \leq 3$ . Then, for a fixed $a$ , $2^a$ appears $3 \cdot 4\ = 12$ times among the divisors of $36000$ .

Knowing that, as $a$ takes values from $0$ to $5$ , the highest integer power of $2$ that divides $P_{36000}$ is $\displaystyle\prod_{i=0}^{5} {(2^i)}^{12} = {(2^{1+2+3+4+5})}^{12} = 2^{180}$ .