Number Theory Positive Divisor Problems

I'm in number theory and I've been assigned these problems for homework. I've searched throughout the relevant section of the book but I can't seem to find anything that relates to solving these problems.

1. Describe $n\in \mathbb{N}$ when the number of positive divisors of $n$ is 105.

2. Which positive integers $n<100$ have the greatest number of positive divisors?

Any help at all is appreciated, thanks.

Edit: Solved #1:

Note that if $n=\prod^{k}_{i=1}p_i^{c_i}$, then $\prod^{k}_{i=1}(1+c_i)$ is the number of positive divisors of $n$, since any $p_i$ can occur up to $c_i$ times, or not at all.

If $n$ has 105 positive divisors, then $105=\prod^k_{i=1}(1+c_i)$.

Note that $105=3\cdot 5\cdot 7$.

So, $105=(1+2)(1+4)(1+6)\implies n=p_1^2\cdot p_2^4\cdot p_3^6$.

You should have seen a formula for the number of positive divisors of a number. If $n= \prod_{i=1}^k p_i^{v_i}$ with distinct primes $p_i$ and positive integers $v_i$ then the number of divisors is $\prod_{i=1}^k (1+ v_i)$.
For the first question note that $105 = 3 \times 5 \times 7$. This restricts the $k$ and $v_i$ that can appear in the prime factorization of such a number. Be careful though it is not true that $k = 3$, it is only true that $k \le 3$.
• You are welcome. For 1 note that there are other possibilities, as I hinted at with the $k \le 3$ not necessarily $k=3$. For example $p^{104}$ also has $105$ divisors. So does $p^{14}q^6$ and a few others. For 2, you could say since $2 \cdot 3 \cdot 5 \cdot 7 >100$ you have $k\le 3$. Now try which combination of exponents for $2$, $3$, $5$ gives you most divisors (note again that one of them might not appear). Then check if you can get the same with other primes, like $7$ instead of $5$.