Problem 2. Determine the smallest natural number that has exactly 2015 divisors.
Solution. If a natural number decomposes into $p_{1}^{a_{1}} \cdot p_{2}^{a_{2}} \cdot \ldots \cdot p_{n}^{a_{n}}$, where $p_{1}, p_{2}, \ldots p_{n}$ are different prime numbers, then the number of its natural divisors is equal to $\left(a_{1}+1\right)\left(a_{2}+1\right) \ldots\left(a_{n}+1\right)$.

Since $2015=1 \cdot 2015=5 \cdot 403=13 \cdot 155=31 \cdot 65=5 \cdot 13 \cdot 31$, it follows that the sought number has one of the forms $a=2^{2014}, b=2^{402} \cdot 3^{4}, c=2^{154} \cdot 3^{12}, d=2^{64} \cdot 3^{30}$ or $e=2^{30} \cdot 3^{12} \cdot 5^{4}$, because to obtain the smallest numbers, the largest exponents must correspond to the smallest prime numbers $4 p$

We have $a>e$ since from $2^{2014}>2^{30} \cdot 3^{12} \cdot 5^{4}$ we get $2^{1984}>3^{12} \cdot 5^{4}$ which is obviously true. We have $b>e$ since from $2^{402} \cdot 3^{4}>2^{30} \cdot 3^{12} \cdot 5^{4}$ we get $2^{372}>3^{8} \cdot 5^{4}$ which is obviously true. We have $c>e$ since from $2^{154} \cdot 3^{12}>2^{30} \cdot 3^{12} \cdot 5^{4}$ we get $2^{124}>5^{4}$ which is obviously true.

We have $d>e$ since from $2^{64} \cdot 3^{30}>2^{30} \cdot 3^{12} \cdot 5^{4}$ we get $2^{34} \cdot 3^{18}>5^{4}$ which is obviously true.

Therefore, the smallest natural number with exactly 2015 divisors is $2^{30} \cdot 3^{12} \cdot 5^{4} \ldots$