Suppose the contrary. Then the student received each of the marks $2$, $3$, $4$, $5$ at least three times. Take three marks of each kind; the sum of these $12$ marks is $42$. Since each of the remaining five marks is at least $2$ and at most $5$, the sum of all $17$ marks is at least $42 + 5 \cdot 2 = 52$ and at most $42 + 5 \cdot 5 = 67$. But none of the numbers from $52$ to $67$ is divisible by $17$, since $3 \cdot 17 = 51$ and $4 \cdot 17 = 68$. Therefore, the arithmetic mean of all the marks is not an integer. Contradiction.