1955 AMC 12 Problem 35

Attempt Problem 35 of the 1955 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 1955 AMC 12 solutions, or check the answer key.

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35.

Three boys agree to divide a bag of marbles in the following manner. The first boy takes one more than half the marbles. The second takes a third of the number remaining. The third boy finds that he is left with twice as many marbles as the second boy. The original number of marbles:

is none of the following

cannot be determined from the given data

is 2020 or 2626

is 1414 or 3232

is 88 or 3838

Answer: B
Concepts:fractions of a remainderinteger constraintsunderdetermined problem
Difficulty rating: 1490
Small Hint:

Let the original number be nn and express the remainder after the first boy

Big Hint:

Translate the last two boys’ share condition into an equation and see whether it determines nn uniquely

Solution:

The first boy takes n2+1,\frac{n}{2}+1, leaving n21.\frac{n}{2}-1. The second takes one third of that remainder, and the third receives the other two thirds, automatically twice the second boy’s share. Thus the share condition imposes no unique value of n.n. It only requires n21\frac{n}{2}-1 to be a nonnegative multiple of 3,3, so many values such as 8,14,20,26,8,14,20,26,\ldots work.

Therefore the original number cannot be determined, and the correct answer is B.

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