$2$ s in Divisor Product of $36000$

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 $\displaystyle\prod_{i=0}^{5} {(2^i)}^{12} = {(2^{1+2+3+4+5})}^{12} = 2^{180}$ .

Here is a Mathematica code that outputs the Prime factors of this product

FactorInteger[Times @@ Divisors@36000]

which returns

{{2, 180}, {3, 72}, {5, 108}}

so, the answer is 180

First of all note that, $36000= 2^5×3^2×5^3$ . As we know that from number theory,

the product of positive divisors of $n∈N =n^{\frac{τ(n)}{2}}$ where $τ(n)=$ number of positive divisors of $n∈N$ .

$τ(36000) =τ( 2^5×3^2×5^3)$ $= (5+1)×(2+1)×(3+1)$ $= 72$

This imply,

$P_{36000}= 36000^{36}$ $=(2^5×3^2×5^3)^{36}$ $= (2^5)^{36} × (3^2)^{36} ×(5^3)^{36}$

Which clearly imply the highest integer powers of $2$ that divides $P_{36000}$ is $(2^5)^{36}= 2^{180}$ this imply $m=180$