2015 AMC 8 Problem 22
Attempt Problem 22 of the 2015 AMC 8 below, then check your answer against the video solution and professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2015 AMC 8 solutions, or check the answer key.
All problems are used with official legal permission of the Mathematical Association of America (MAA).
22.
On June a group of students is standing in rows, with students in each row. On June the same group is standing with all of the students in one long row. On June the same group is standing with just one student in each row. On June the same group is standing with students in each row. This process continues through June with a different number of students per row each day. However, on June they cannot find a new way of organizing the students. What is the smallest possible number of students in the group?
Answer: C
Small hint:
The different row sizes are the divisors of the group size.
Big hint:
You need the smallest multiple of both and with exactly divisors.
Video solution:
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Written solution:
The possible numbers of students per row are exactly the positive divisors of the total number of students. Since June through June give different arrangements and June gives no new one, the total number of students must have exactly positive divisors.
The number must be divisible by both and hence by This number has only divisors.
The smallest multiple of with divisors is which has divisors.
Thus, C is the correct answer.
Problem 22 in other years
- 1985 AMC 8
- 1986 AMC 8
- 1987 AMC 8
- 1988 AMC 8
- 1989 AMC 8
- 1990 AMC 8
- 1991 AMC 8
- 1992 AMC 8
- 1993 AMC 8
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