Solving for mod indirectly

Question

How many positive integers $n$ exist such that $\frac{680}{n}$ is an integer?

So, this is quite obvious,

$680 \equiv 0 \pmod{n}$

How should I solve for $n$? There will be multiple $n$?

Answer 1

$$680=10\cdot68=2^3\cdot17\cdot5$$

So, the number of positive divisors $=(3+1)(1+1)(1+1)$

Answer 2

There as many solutions as divisors of $680$. Decompose $680$ into its prime factors: $$680=2^3\cdot 5^1 \cdot 17^1.$$ Hence there are $4\cdot2\cdot 2=16$ solutions.

Answer 3

Hint $\ $ By the FTA (Fundamental Theorem of Arithmetic - existence and uniqueness of prime factorizations), if $\,p,q,r\,$ are primes then the positive divisors of $\, p^i q^j r^k $ are precisely the naturals of the form $\,p^{\bar i}q^{\bar j} q^{\bar k}\,$ where $\,\bar i< i,\ \bar j< j,\ \bar k < k.\ $ Thus the set of divisors is in bijection with the set of sequences of naturals $(i,j,k)$ satisfying said inequalities. Such sequences are easy to count.

Remark $\ $ The is a protoypical example of how FTA reduces many divisibility problems on integers to simpler problems on (exponent) vectors.