Carley made treat bags. Each bag contained exactly 1 chocolate, 1 mint, and 1 caramel. The chocolates came in boxes of 50. The mints came in boxes of 40 . The caramels came in boxes of 25 . Carley made no incomplete treat bags and there were no unused chocolates, mints or caramels. What is the minimum total number of boxes that Carley could have bought?
(A) 19
(B) 17
(C) 44
(D) 25
(E) 9
Suppose that Carley buys $x$ boxes of chocolates, $y$ boxes of mints, and $z$ boxes of caramels. In total, Carley will then have $50 x$ chocolates (since there are 50 chocolates in a box of chocolates), $40 y$ mints (since there are 40 mints in a box of mints), and $25 z$ caramels (since there are 25 caramels in a box of caramels).

Since the contents of each bag was the same and Carley made no incomplete treat bags and there were no left-over candies, then it must be the case that $50 x=40 y=25 z$.

We want to find the minimum possible positive value of $x+y+z$ given this condition.

Dividing by the common factor of 5 , the equation $50 x=40 y=25 z$ becomes $10 x=8 y=5 z$.

Since $10 x$ is a multiple of 10 and $8 y$ is a multiple 8 and $10 x=8 y$, we look for the smallest multiple of 10 which is also a multiple of 8 .

Since 10, 20 and 30 are not multiples of 8 , and 40 is a multiple of 8 , then the smallest possible value of $10 x$ appears to be 40 .

In this case, $x=4, y=5$ and $z=8$ gives $10 x=8 y=5 z=40$, and these are the smallest positive integers that create this equality.

Since $x, y$ and $z$ are each the smallest possible, then their sum $x+y+z$ is also the smallest possible.

Thus, the minimum number of boxes that Carley could have bought is $4+5+8=17$.

ANSWER: (B)

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