Mock AIME I 2015 Problems/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
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.