# Mock AIME I 2015 Problems/Problem 5

## Problem 5

In an urn there are a certain number (at least two) of black marbles and a certain number of white marbles. Steven blindfolds himself and chooses two marbles from the urn at random. Suppose the probability that the two marbles are of opposite color is . Let be the smallest possible values for the total number of marbles in the urn. Compute the remainder when is divided by

## Solution

Let the number of black marbles be and the number of white marbles be .

We have two cases for when the two marbles chosen from the bag and are different colors, Black-White and White-Black. The first case can be represented by , and the second case by . Multiplying out, we get for both cases. Adding the cases, we get the equation or .

Cross multiplying, we get . We can expand the left side of the equation to . We can then subtract from both sides of the equation to get , and rearranging, we get .

Note that we want to find the the smallest possible values for the total amount of marbles, or . From the equation, we learn that must be a perfect square, since is an integer. Note that since is at least , we can state that , and since must be a perfect square, .

We then want to see which values for yield integer solutions for and . To do so, we can set equal to an even integer, . This gives the equation . We get , and . Solving, we get and . Therefore, can be all even squares.

The next case we have is when is an odd integer, . This gives the equation . We get , and . Solving, we get and . Therefore, both even and odd squares work as solutions for .

We can then see that the first solutions for will be the first perfect squares, starting from . Therefore, . Using the formula to find the sum of the squares from to , we get
Evaluating, we get . We subtract to get the total sum as . Taking the remainder after dividng by 1000, we get as our answer.

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