2017 AMC 10B Problem 25

P21-25~8 minPrint

Modular Arithmetic Statistics & Data Bound the quantity above/below or estimate to pin it down Start from the end state / desired conclusion and reverse

Last year Isabella took $7$ math tests and received $7$ different scores, each an integer between $91$ and $100$ , inclusive. After each test she noticed that the average of her test scores was an integer. Her score on the seventh test was $95$ . What was her score on the sixth test?

Show solution without answering

Key insight

Subtract 90 from each score. The 7-test total is a multiple of 7 from 28 to 49, and minus 5 it must be a multiple of 6: only 35.

Solution

"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.

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.
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 $49$ .
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 $35$ ).
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 $10$ .
Back to real scores. $10 + 90 = 100$ .
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 $\boxed{\textbf{(E)}\ 100}$ .

Why this works

"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.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

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

Review in the AoPS books

Introduction to Number Theory, Chapter 12: Introduction to Modular Arithmetic
Prealgebra, Chapter 13: Data and Statistics
Prealgebra, Chapter 15: Problem-Solving Strategies (§15.4)
Introduction to Number Theory, Chapter 10: Units Digits