Romanian Mathematical Olympiad

Consider the set $A = \{2a + 5b \mid a, b = 1, 2, \dots, 100\}$ and notice that $1 \notin A$ and $3 \notin A$.

The largest even number from $A$ is equal to $700$ and it is obtained for $a = b = 100$. The number $698$ is obtained for $a = 99, b = 100$. The largest odd number from $A$ is equal to $695$ and it is obtained for $b = 99, a = 100$, implying that $697 \notin A$ and $699 \notin A$.

We claim that all the integers between $4$ and $695$ belong to $A$.
Let $y \le 500$ and let $r$ be the remainder left by $y$ upon division by $5$. Write $y = 5c + r$ with $0 \le c \le 100$ and $0 \le r \le 4$. If $r$ is even, then $y = 5c + 2k$, where $r = 2k$. If $r$ is odd, then $y \ge 5$, so $c \ge 1$ and $y = 5(c - 1) + 2(k + 3)$, where $r = 2k + 1$.

For $y > 500$, write $y = 500 + z$ with $z \ge 200$. If $z$ is even, then $y = 5 \cdot 100 + 2k$, where $z = 2k$. If $z$ is odd, then $y \le 695$, so $z \le 195$ and $y = 5 \cdot 99 + 2(k + 3)$, where $z = 2k + 1$. A quick inspection of all the above cases shows that the claim holds.

$$S = \frac{1}{10}((1 + 2 + \dots + 700) - (1 + 3 + 697 + 699) = 350 \cdot 697) = 35 \cdot 697.$$