Problem 391

Suppose that Carley buys $x$ boxes of chocolates, $y$ boxes of mints, and $z$ boxes of caramels.

In total, Carley will then have $50x$ chocolates (since there are 50 chocolates in a box of chocolates), $40y$ mints (since there are 40 mints in a box of mints), and $25z$ 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 $50x = 40y = 25z$.

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 $50x = 40y = 25z$ becomes $10x = 8y = 5z$.

Since $10x$ is a multiple of 10 and $8y$ is a multiple 8 and $10x=8y$, 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 $10x$ appears to be 40.

In this case, $x=4$, $y=5$ and $z=8$ gives $10x = 8y = 5z = 40$, 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 $4+5+8=17$.