2013 AMC 12A Problems/Problem 16
Let pile have
rocks, and so on.
The mean weight of and
together is
, so the total weight of
and
is
To get the total weight of and
, we need to add the total weight of
and subtract the total weight of
And then dividing by the number of rocks and
together, to get the mean of
and
,
Simplifying,
Now, to get rid of the in the numerator, we use two definitions of the total weight of
and
Substituting back in,
Note that , and the maximal value of this factor occurs when
Also note that must cancel to give an integer value, and the only fraction that satisfies both these conditions is
Plugging in, we get
Solution 2
Suppose there are rocks in the three piles, and that the mean of pile C is
, and that the mean of the combination of
and
is
. We are going to maximize
, subject to the following conditions:
which can be rearranged as:
Let us test is possible. If so, it is already the answer. If not, there will be some contradiction. So the third equation becomes
So ,
,
, therefore,
, which gives us a consistent solution. Therefore
is the answer.
(Note: To further illustrate the idea, let us look at and see what happens. We then get
, which is a contradiction!)