2017 AMC 10B Problem 25

Subtract $90$ from every score: the shifted scores $a_1, \dots, a_7$ are distinct integers from $1$ to $10$, the averages are still integers, and $a_7 = 5$. Let $S_k = a_1 + \dots + a_k$; the condition is $k \mid S_k$ for every $k$.

Bound $S_7$: seven distinct values from $1$ to $10$ sum to at least $1 + \dots + 7 = 28$ and at most $4 + \dots + 10 = 49$. Being a multiple of $7$, $S_7 \in \{28, 35, 42, 49\}$.

Then $S_6 = S_7 - 5 \in \{23, 30, 37, 44\}$ must be a multiple of $6$, so $S_6 = 30$ and $S_7 = 35$.

Next, $S_5 = 30 - a_6$ must be a multiple of $5$, so $a_6$ is a multiple of $5$. It cannot be $5$ (already used by $a_7$), hence $a_6 = 10$, i.e. the sixth score is $100$.

Existence check: the first five shifted scores must be distinct, avoid $5$ and $10$, sum to $20$, and satisfy $2 \mid S_2$, $3 \mid S_3$, $4 \mid S_4$. The order $1, 3, 2, 6, 8$ works ($S_2 = 4$, $S_3 = 6$, $S_4 = 12$, $S_5 = 20$), so the scores $91, 93, 92, 96, 98, 100, 95$ satisfy every condition.

The answer is $\boxed{\textbf{(E)}\ 100}$.

"Average is an integer" is a divisibility condition on partial sums, and shifting by $90$ makes the numbers small enough to bound. Working backwards from the final total, the conditions $7 \mid S_7$, $6 \mid S_6$, $5 \mid S_5$ each eliminate candidates, and distinctness supplies the bounds that make the case list finite. The statistics wording is a disguise for modular arithmetic.

Skip the bound: since $S_7 = S_6 + 5$ with $6 \mid S_6$ and $7 \mid S_7$, we need $S_6 \equiv 2 \pmod 7$ and $S_6 \equiv 0 \pmod 6$, so $S_6 \equiv 30 \pmod{42}$. Six distinct values from $\{1, \dots, 10\} \setminus \{5\}$ sum to between $21$ and $45$, leaving only $S_6 = 30$. Then $5 \mid 30 - a_6$ forces $a_6 = 10$ as before.

• Forgetting that the scores are distinct, which is what bounds the total to 28 through 49, or ignoring that the fifth partial sum must also be divisible by 5.
• Stopping at "the sixth score is a multiple of 5" and choosing $95$ or a nearby option without noticing $95$ is already the seventh score.