50th Mathematical Olympiad in Ukraine, Fourth Round (March 23, 2010)

Let $n$ be a natural number that satisfies the condition of the problem. We first calculate the sum of all given numbers: $$3n^2-31n+100=2n^2-30n+n(n-1)+100$$ which is even. This implies that at least one of them is even, in other words, it equals to $2$. Solving the following three equations $$n^2-10n+23=2,n^2-9n+31=2,n^2-12n+46=2,$$ we get $n=3$, or $n=7$. Simple check shows that these values of $n$ meet all the requirements. Indeed,

for $n=3$ we have: $n^2-10n+23=2$, $n^2-9n+31=13$ and $n^2-12n+46=19$.

for $n=7$ we have: $n^2-10n+23=2$, $n^2-9n+31=17$ and $n^2-12n+46=11$.