2000 AMC 12 Problem 9

Below is the professionally curated solution for Problem 9 of the 2000 AMC 12, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2000 AMC 12 solutions, or check the answer key.

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Concepts:divisibilitymodular arithmeticmean

Difficulty rating: 1580

9.

Mrs. Walter gave an exam in a mathematics class of five students. She entered the scores in random order into a spreadsheet, which recalculated the class average after each score was entered. Mrs. Walter noticed that after each score was entered, the average was always an integer. The scores (listed in ascending order) were 71,76,80,82,71, 76, 80, 82, and 91.91. What was the last score Mrs. Walter entered?

7171

7676

8080

8282

9191

Solution:

The total is 71+76+80+82+91=400,71 + 76 + 80 + 82 + 91 = 400, which is divisible by 5.5. The sum of the first three scores must be divisible by 3.3.

Modulo 3,3, the scores are 2,1,2,1,1.2, 1, 2, 1, 1. The only triple summing to a multiple of 33 is 76+82+91=249,76 + 82 + 91 = 249, so these are the first three (with 9191 third, since the first two, 7676 and 82,82, must have equal parity).

Since 2491(mod4),249 \equiv 1 \pmod 4, the fourth score must be 3(mod4),\equiv 3 \pmod 4, which is 71.71. That leaves 8080 as the last score entered.

Thus, the correct answer is C.

Problem 9 in Other Years