So I have these parts to this problem:
If $n \in \mathbb N$, then let $\tau(n)$ be the number of positive divisors of $n$. For example, $\tau(6) = 4$.
(a) If $n= p_1^{a_1}p_2^{a_2}\ldots p_k^{a_k}$ where $p_i$ are distinct primes and $a_i\in \mathbb N$, then prove that the number of positive divisors of $\tau(n)=(a_1+1)(a_2+1)\ldots(a_k+1)$.
(b) Find the number of positive divisors of $1,\!633,\!500$.
(c) Prove or disprove that $\tau(nm) = \tau(n)\tau(m)$ for all $m$, $n\in \mathbb N$.
I'm not sure how to prove part a. For part b, I know I can just do a series of divisions to find that answer, so I got $2^2\times 5^3\times 3^3\times11^2$ for the divisors, so I think $\tau(1,\!633,\!500)$ would be 4 in this case. For part c, I think I can do a counterexample of 4 and 6, where $\tau(4)=3$, and $\tau(6)=4$, but $\tau(24)$ is not equal to 12, which is $\tau(4)\tau(6)$, if I did my math right.
So, can anyone help me on part a and tell me if I am doing b and c wrong?
Best Answer
In (a) you can write each positive divisor uniquely in the form
$$p_1^{\ell_1}p_2^{\ell_2}\ldots p_k^{\ell_k}\;,$$
where $\ell_i\in\{0,1,\ldots,a_i\}$ for $i=1,\ldots,k$. Choosing a positive divisor amounts to choosing the exponents $\ell_1,\ldots,\ell_k$; in how many ways can that be done?
Your answer to (b) is way off, partly because you did not in fact apply the formula, and partly because you didn’t finish decomposing $1,633,500$ as a product of powers of distinct primes.
Your answer to (c) is fine.