AMC 10 · 2017 · #21

Grade 6 algebra
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 seventh score was 95. Find her sixth score.

Pick an answer.

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

Subtracting 90 makes it a 1 to 10 problem.

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

Bound the seven-test total

All different narrows the total's range.

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

Use the six-test rule to fix the total

The condition after the sixth test pins the total.

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

Use the five-test rule to find the sixth score

Using the five-test condition gives 100.

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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