Second Round of the 73rd Czech and Slovak Mathematical Olympiad (January 16th, 2024)

Denote the numbers $x_1,\dots ,x_{74}$. The assumption $4\le x_i\le 10$ guarantees that $(x_i-4)(10-x_i)\ge 0$ holds for all $i$. Expanding this to $x_i^2\le 14x_i-40$ and summing over $i$ gives us an upper bound $$x_1^2+x_2^2+\cdots +x_{74}^2\le 14(x_1+\cdots +x_{74})-40\cdot 74=14\cdot 356-40\cdot 74=2024.$$ Therefore, it is now sufficient to show that there exists a set of 74 real numbers that attains this bound. Our approach above shows that the bound is attained if and only if $x_i\in \{4,10\}$ for all $i$, so it suffices to check that such a 74-tuple with sum 356 exists. Suppose that the tuple contains $a$ fours and $b$ tens, then the non-negative integers $a,b$ must satisfy the system of equations: $$\begin{array}{rl}a+b& =74,\\ 4a+10b& =356.\end{array}$$ It is easy to see that $(a,b)=(64,10)$ is the only solution, so a set consisting of 64 fours and 10 tens proves that the upper bound is tight.