1991 AMC 8 Problem 17

Attempt Problem 17 of the 1991 AMC 8 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 1991 AMC 8 solutions, or check the answer key.

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

17.

An auditorium with 2020 rows of seats has 1010 seats in the first row. Each successive row has one more seat than the previous row. If students taking an exam are permitted to sit in any row, but not next to another student in that row, then the maximum number of students that can be seated for an exam is
150150
180180
200200
400400
460460
Answer: C
Concepts:arrangements with restrictionsarithmetic sequencefloor and ceiling functions
Small hint:
In a row of nn seats, the most students with no two adjacent is ⌈n2⌉\left\lceil \frac{n}{2} \right\rceil
Big hint:
Row kk has 9+k9+k seats; add ⌈9+k2⌉\left\lceil \frac{9+k}{2} \right\rceil over the 2020 rows
Solution:
Row kk has 9+k9+k seats, so it holds ⌈9+k2⌉\left\lceil \frac{9+k}{2} \right\rceil students. For rows 11 through 2020 (seats 1010 through 2929) the maxima are 5,5, 6,6, 6,6, 7,7, 7,7, 8,8, 8,8, …,\ldots, 14,14, 14,14, 15.15.
These sum to 200.200.
Thus, the correct answer is C .
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