10.1. Find all natural numbers $n$ such that $\frac{1}{n}=\frac{1}{p}+\frac{1}{q}+\frac{1}{p q}$ for some primes $p$ and $q$
Answer. $n=1$.

Solution. Bringing the expression in the condition to a common denominator, we get: $n(p+q+1)=pq$. From the simplicity of $p$ and $q$, it follows that the divisors of the right-hand side can only be the numbers $1, p, q$, and $pq$, one of which must equal $p+q+1$. Since $1, p$, and $q$ are less than $p+q+1$, we obtain $pq=p+q+1$ and $n=1$. Rewriting the last equation as $(p-1)(q-1)=2$, we get $p=2, q=3$ or $p=3, q=2$.

Grading Criteria. Guessed $n=1$ and $p=2, q=3$ or $p=3, q=2$: 1 point. Proved that $n=1$ is the only number that can satisfy the condition: 5 points. Shown that it satisfies the condition for $p=2, q=3$ or $p=3, q=2$: 2 points. (It is sufficient to simply provide the example $p=2, q=3$ or $p=3, q=2$)