1958 AMC 12 Problem 14

Attempt Problem 14 of the 1958 AMC 12 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 1958 AMC 12 solutions, or check the answer key.

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

14.

At a dance party a group of boys and girls exchange dances as follows: one boy dances with 55 girls, a second boy dances with 66 girls, and so on, the last boy dancing with all the girls. If bb represents the number of boys and gg the number of girls, then:
b=gb=g
b=g5b=\dfrac g5
b=g−4b=g-4
b=g−5b=g-5
It is impossible to determine a relation between bb and gg without knowing b+gb+g
Answer: C
Concepts:arithmetic sequencelinear equation
Small hint:
The numbers of partners form the sequence 5,6,7,…,g5,6,7,\ldots,g
Big hint:
There are bb terms, so express the last term using the arithmetic-sequence formula
Solution:
The last of the bb terms in the sequence 5,6,7,…5,6,7,\ldots is 5+(b−1)=b+4. 5+(b-1)=b+4. This last number equals the total number gg of girls. Hence g=b+4,g=b+4, or b=g−4.b=g-4.
Thus, the correct answer is C.
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