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2000 AMC 10 Problems/Problem 14

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Problem

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, and 91. What was the last score Mrs. Walter entered?

\mathrm{(A)}\ 71 \qquad\mathrm{(B)}\ 76 \qquad\mathrm{(C)}\ 80 \qquad\mathrm{(D)}\ 82 \qquad\mathrm{(E)}\ 91

Solution

71, 76, 80, 82, 91

The sum of the first 2 scores must be even, so we must choose 2 evens or the 2 odds to be the first two scores.

Let us look at the numbers in mod 3.

2,1,2,1,1

If we choose the two odds, the next number must be a multiple of 3, of which there is none.

Similarly, if we choose 76,80 or 80,82, the next number must be a multiple of 3, of which there is none.

So we choose 76,82 first.

The next number must be 1 in mod 3, of which only 91 remains.

The sum of the first three scores is 249. This is equivalent to 1 in mod 4.

Thus, we need to choose one number that is 3 in mod 4. 71 is the only one that works.

Thus, 80 is the last score entered.

\boxed{\text{C}}

See Also

2000 AMC 10 (ProblemsResources)
Preceded by
Problem 13
Followed by
Problem 15
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
Looking for a challenging geometry text? Preparing for MATHCOUNTS or the AMC exams? Check out Art of Problem Solving's Introduction to Geometry by Richard Rusczyk.
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