1991 AMC 8 Problem 17

Below is the professionally curated solution for Problem 17 of the 1991 AMC 8, 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.

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Concepts:arrangements with restrictions arithmetic sequence floor and ceiling functions

Difficulty rating: 1140

17.

An auditorium with $20$ rows of seats has $10$ 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

$150$

$180$

$200$

$400$

$460$

Solution:

Row $k$ has $9+k$ seats, so it holds $\left\lceil (9+k)/2 \right\rceil$ students. For rows $1$ through $20$ (seats $10$ through $29$ ) the maxima are $5, 6, 6, 7, 7, 8, 8, \ldots, 14, 14, 15.$

These sum to $200.$

Thus, the correct answer is C .

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