2017 AMC 12B Problem 21

Attempt Problem 21 of the 2017 AMC 12B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2017 AMC 12B solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

21.

Last year Isabella took 77 math tests and received 77 different scores, each an integer between 9191 and 100,100, inclusive. After each test she noticed that the average of her test scores was an integer. Her score on the seventh test was 95.95. What was her score on the sixth test?
9292
9494
9696
9898
100100
Answer: E
Concepts:divisibilitymeanbounding to limit cases
Small hint:
The sum SS of all 77 scores is a multiple of 77 and lies between 658658 and 679679
Big hint:
Also S−95S - 95 is a multiple of 6;6; then the first 55 scores sum to a multiple of 5,5, forcing the sixth score’s residue
Solution:
Let SS be the sum of all seven scores. Then SS is a multiple of 77 with 658≤S≤679,658 \le S \le 679, so S∈{658,665,672,679}.S \in \{658, 665, 672, 679\}. Since the average after six tests is an integer, S−95S - 95 is a multiple of 6,6, which forces S=665.S = 665. Then the first six scores sum to 570,570, a multiple of 5;5; the average after five tests is an integer, so the first five scores also sum to a multiple of 5,5, making the sixth score a multiple of 5.5. Since all scores differ and the seventh is 95,95, the sixth must be 100.100.
Thus, the correct answer is E.
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