Counting the Distinct Factors of the Positive Integer $36^{2}$

Question

How many factors does $36^2$ have?

So, I'll start.

$36$ has these prime factors: \begin{align*} 36=&2 \times 18 \\ = &2 \times 2 \times 9\\ = &2 \times 2 \times 3 \times 3 \end{align*}

\begin{align*} 36^2&=(2\times2\times3\times3)^2\\ &= (2^2 \times 3^2) ^ 2\\ &= 2^4 \times 3^4 \end{align*}

So there are $8$ factors, $4$ factors of $2$ and $4$ factors of $3$.

1) Are there $5$ factors of $2$ and $5$ factors of $3$? Does $2^0$ count as a factor?

2) I think I'm supposed to multiply $4 \times 4$ or $5 \times 5$ and come up with the answer. I've never understood why multiplying makes sense. Can someone help me with that intuition?

Answer 1

So there are 8 factors, 4 factors of 2 and 4 factors of 3.

Those are just the prime factors (with the appropriate multiplicity).

The number itself and $1$ are always factors of a number.

Some other factors of $36^2$: $4, 8, 24, 648$.

Basically you'll need all numbers that divide $1296$ evenly. You can have zero to four factors of $2$, and zero to four factors of $3$.

Can you take it from there?

Answer 2

You factored $36^2$ as $2^43^4$. Think about this, if $x$ is a factor of $36^2$, then $x = 2^n 3^m$ for some $n, m$ between $0$ and $4$. Thus there are $5$ choices for $n$ and $5$ choices for $m$. A total of $5 \cdot 5 = 25$ total factors including $1$ and $36^2$.