How many Positive Integers divide $72^{100}$?

Question

So I began by finding the prime factorization of $72$, which is: $$ 72 = 2^{3} \times 3^{2}$$ then exponentiating by $100$: $$ 72^{100} = 2^{300} \times 3^{200}$$ and so now the question is asking to count the positive integers (natural numbers) which divide this prime factorization.

So basically is this question asking me to compute the natural numbers which divide $2$ and $3$? wouldn't that just be $1,2$ and $3$? Or am I misinterpreting this (could very well be the case). Thanks for any assistance!

Answer 1

By your analysis, any number of the form $2^i3^j$ with $0 \le i \le 300$ and $0 \le j \le 200$ will divide your number. So, you have $301*201=60501$ possible numbers like that.

Answer 2

Numbers that also divide $2^{300}\cdot 3^{200}$ are things like $2^{50}, 2^{51}, 3^{20},3^{2}, 3^{37},2^{52}\cdot 3^{167},\ldots$ (think about it with $\dfrac{x^{a}}{x^{b}}=x^{a-b}$). So, you need to figure out how many combinations of powers of $2$ and powers of $3$ can be made. There are $301$ such powers of $2$ and $201$ such powers of $3$ giving $201\cdot301=\boxed{60501}$. (If you need to see why I multiplied these bnumbers, try reading a bit here : https://www.mathsisfun.com/combinatorics/combinations-permutations.html)