AMC 10 · 2017 · #25

Grade 6 arithmetic
Archetype Small-Domain Constrained Search └ Sub-archetype Factor / Divisibility Constraint
divisibility-rulesmean-median-mode-rangemodular-arithmetic bound-inequality-then-enumerateeasier-related-problem ↑ Prerequisites: divisibility-rules
📏 Long solution 💡 3 insights
Problem
Isabella took 7 math tests and got 7 different whole-number scores, each from 91 to 100. After every test, the average of all her scores so far came out to a whole number. Her 7th score was 95. Find her 6th score.

Pick an answer.

(A)
92
(B)
94
(C)
96
(D)
98
(E)
100

AMC 10 2017 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Eliminate Possibilities

Tool #9 (Easier Related Problem): subtracting 90 from every score turns the ugly 91-100 range into the clean set 1-10 without changing whether any average is a whole number, because shifting all scores by 90 just shifts every average by 90. Tool #14 (Extreme Principle): the smallest and largest possible totals pin the seven-test sum into a short list of multiples of 7. Tool #3 (Eliminate Possibilities): the six-test and five-test divisibility conditions knock out every candidate but one, forcing the sixth score.

1STEP 1

Shift to the numbers 1 to 10

Subtract 90 from each score: seven distinct integers in 1-10, the 7th becomes 5, and each running sum is a multiple of its test count.

score-90 ∈ {1,2,…,10}, a₇ = 5, k ∣ (a₁+…+a_k)
2STEP 2

Bound the seven-test total

The seven distinct shifted scores sum between 1+…+7=28 and 4+…+10=49, and being a multiple of 7 leaves only 28, 35, 42, and 49.

28 ≤ S₇ ≤ 49, 7 ∣ S₇ → S₇ ∈ {28,35,42,49}
3STEP 3

Use the six-test rule to fix the total

Removing the 7th score (5) leaves a first-six sum that must be a multiple of 6; of 23, 30, 37, 44 only 30 qualifies, so S₆=30.

6 ∣ (S₇-5): 23, 30, 37, 44 → S₇=35, S₆=30
4STEP 4

Use the five-test rule to find the sixth score

Since 30 is a multiple of 5, a₆ must be too; the only free multiple of 5 is 10 (5 is taken), so the real sixth score is 10+90=100 (E).

5 ∣ (30-a₆) → 5 ∣ a₆ → a₆=10 → 10+90=100 (E)
Answer
100
A full order actually exists, so the answer is consistent. Shifted scores 1,3,2,6,8,10,5 give running sums 1,4,6,12,20,30,35, which divide evenly by 1,2,3,4,5,6,7. Adding 90 back, the real scores 91,93,92,96,98,100,95 are seven distinct values in 91-100, with the sixth equal to 100 and the seventh equal to 95 as required. Among the answer choices, 100 is the only multiple of 5, which also fits the rule that the sixth score had to be a multiple of 5.
💡Key takeaway

Whole-number averages mean each running sum is a clean multiple, so the last two steps force the sixth score to be a multiple of 5 — and 100 is the only choice that fits, choice (E).

  • Shift to the numbers 1 to 10
  • Bound the seven-test total
  • Use the six-test rule to fix the total
  • Use the five-test rule to find the sixth score

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