2017 AMC 10B Problem 25

"The average after $k$ tests is a whole number" means "the total of the first $k$ scores is a multiple of $k$." So this is really a problem about totals.

1. Make the numbers small. Subtract $90$ from every score. Now the scores are seven different whole numbers from $1$ to $10$, and the seventh one is $95 - 90 = 5$. Subtracting $90$ from each of $k$ scores lowers the total by $90k$, a multiple of $k$, so the averages are still whole numbers.
2. The total after 7 tests. Seven different numbers from $1$ to $10$ add to at least $1 + 2 + \cdots + 7 = 28$ and at most $4 + 5 + \cdots + 10 = 49$. It must be a multiple of $7$: so it is $28$, $35$, $42$ or $49$.
3. The total after 6 tests. Take away the seventh score, $5$: the total after six tests is $23$, $30$, $37$ or $44$. It must be a multiple of $6$, and only $30$ is. So the six-test total is $30$ (and the seven-test total is $35$).
4. The total after 5 tests. It is $30$ minus the sixth score, and it must be a multiple of $5$. So the sixth score is a multiple of $5$: either $5$ or $10$. The $5$ is already used by the seventh test, and the scores are all different, so the sixth score is $10$.
5. Back to real scores. $10 + 90 = 100$.
6. Check it can really happen. The first five scores must be different, avoid $5$ and $10$, and add to $20$, with each running total working. The order $1, 3, 2, 6, 8$ does it: running totals $1, 4, 6, 12, 20$ are multiples of $1, 2, 3, 4, 5$. So the real scores $91, 93, 92, 96, 98, 100, 95$ satisfy everything.

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

"The average is a whole number" is a divisibility fact about running totals. Shifting every score by the same amount keeps those facts but makes the numbers tiny, and "all different" gives the limits on the total. Then work backwards from the last test: each earlier total is pinned down by one more divisibility rule.

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 $\boxed{\textbf{(E)}\ 100}$.