# 2013 AMC 12A Problems/Problem 16

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Let pile $A$ have $A$ rocks, and so on.

The mean weight of $A$ and $C$ together is $44$, so the total weight of $A$ and $C$ is $44(A + C)$

To get the total weight of $B$ and $C$, we need to add the total weight of $B$ and subtract the total weight of $A$ $44A + 44C + 50B - 40A = 4A + 44C + 50B$

And then dividing by the number of rocks $B$ and $C$ together, to get the mean of $B$ and $C$, $\frac{4A + 44C + 50B}{B + C}$

Simplifying, $\frac{4A + 44C + 44B + 6B}{B + C}$ $44 + \frac{4A + 6B}{B + C}$

Now, to get rid of the $A$ in the numerator, we use two definitions of the total weight of $A$ and $B$ $40A + 50B = 43A + 43B$ $3A = 7B$ $A = \frac{7}{3}B$

Substituting back in, $44 + \frac{4(\frac{7}{3}B) + 6B}{B + C}$ $44 + \frac{46}{3}*\frac{B}{B + C}$

Note that $\frac{B}{B + C} < 1$, and the maximal value of this factor occurs when $C = 1$

Also note that $\frac{46}{3}$ must cancel to give an integer value, and the only fraction that satisfies these conditions is $\frac{45}{46}$

Plugging in, we get $44 + \frac{46}{3} * \frac{45}{46} = 44 + 15 = 59$