51st Ukrainian National Mathematical Olympiad, 3rd Round

The number of divisors of $n^2=11^{4022}\cdot 2011^{22}$ is: $$N=(4022+1)(22+1)=92529.$$ We can pair the divisors of $n^2$ in a way $d$ and $\frac{n^2}{d}$ where one divisor is greater than $n$, another one is less. The number $n$ does have a pair, therefore the number of divisors that are less than $n$ is $\frac{92529-1}{2}=46264$.

The number $N_1$ of divisors of $n$ that are less than $n$ is: $$N_1=(2011+1)(11+1)-1=24143.$$ So the number we are looking for is $46264-24143=22121$.