58th Ukrainian National Mathematical Olympiad

Let $A=p_1^{k_1}{p}_2^{k_2}\dots p_n^{k_n}$ be a canonical prime factorization of $A$, where $p_i$ are different prime factors, and $k_i$ are positive integers, $i=1,n$. Knowing the total number of positive integer factors of $A$ we can conclude that $$(k_1+1)(k_2+1)\dots (k_n+1)=2018.$$

Since the left-hand side comprises factors each of which is greater or equal to 2, we have only two cases:

1) $k_1+1=2018$. 2) $(k_1+1)(k_2+1)=2\cdot 1009$.

From the first case, $A$ has a canonical prime factorization of the form: $A=p_1^{2017}$, which contradicts the fact that $A\ne 2018$, since in that case $A$ can't be divisible by both 2 and the prime number 1009.

Hence, $(k_1+1)(k_2+1)=2\cdot 1009$, which yields, without loss of generality, $k_1=1$, $k_2=1008$. Therefore, $A=p_1{p}_2^{1008}$, moreover from divisibility by 2 and 1009 we obtain $A=2\cdot 1009^{1008}$ or $A=1009\cdot 2^{1008}$. However, none of these numbers is divisible by $2018^2$, Q.E.D.