Problem 13 · 2018 AMC 8
Hard
Number Theory
divisibilitycasework
Laila took five math tests, each worth a maximum of 100 points. Laila's score on each test was an integer between 0 and 100, inclusive. Laila received the same score on the first four tests, and she received a higher score on the last test. Her average score on the five tests was 82. How many values are possible for Laila's score on the last test?
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Answer: A — 4 values.
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Hint 1 of 2
The average is 82, but the last test is higher than the other four — so the four equal tests must drag below 82 and the last test makes up the whole shortfall. Picture starting everyone at 82 and then shifting points onto the last test.
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Hint 2 of 2
The technique is a modular constraint plus a range: write the total as 4f + l = 410, get bounds on l, then notice the four identical scores force a step-size on l.
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Approach: modular constraint + range
- Average 82 over 5 tests means the five scores total 5×82 = 410. Let f be the repeated first-four score and l the last, so 4f + l = 410 with f < l ≤ 100.
- Lower bound on l: if all five were 82 the total is exactly 410, but the last must be strictly higher than the others, so l > 82 and f < 82.
- Why l jumps by 4: each point you move off the last test has to be split equally among the four equal tests, but a whole point on each of the four costs 4 points from the last. Concretely, to keep f a whole number, lowering all four from 82 by 1 (to 81) frees 4 points that pile onto the last test — so l climbs 86, 90, 94, 98 in steps of 4.
- (Same fact in modular language: 4f is a multiple of 4 and 410 leaves remainder 2 mod 4, so l ≡ 2 mod 4.) In the window 82 < l ≤ 100 that gives l ∈ {86, 90, 94, 98}.
- 4 values. (Check the top: l = 98 pairs with f = 78 — valid; the next step 102 exceeds 100, so we stop.)
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