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 $\tfrac12$ . Let $k_1<k_2<\cdots<k_{100}$ be the $100$ smallest possible values for the total number of marbles in the urn. Compute the remainder when $k_1+k_2+k_3+\cdots+k_{100}$ is divided by $1000$

Solution

Let the number of black marbles be $b$ and the number of white marbles be $w$ .

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 $\frac{b}{b+w}*\frac{w}{b+w-1}$ , and the second case by $\frac{w}{b+w}*\frac{b}{b+w-1}$ . Multiplying out, we get $\frac{bw}{(b+w)(b+w-1)}$ for both cases. Adding the cases, we get the equation $2*\frac{bw}{(b+w)(b+w-1)}=\frac{1}{2}$ or $\frac{bw}{(b+w)(b+w-1)}=\frac{1}{4}$ .

Cross multiplying, we get $(b+w)(b+w-1)=4bw$ . We can expand the left side of the equation to $b^2+bw-b+bw+w^2-w=b^2+2bw+w^2-b-w$ . We can then subtract $4bw$ from both sides of the equation to get $b^2-2bw+2^w-b-w=0$ , and rearranging, we get $(b-w)^2=b+w$ .

Note that we want to find the the smallest possible values for the total amount of marbles, or $b+w$ . From the equation, we learn that $b+w$ must be a perfect square, since $b-w$ is an integer. Note that since $b$ is at least $2$ , we can state that $b+w > 2$ , and since $b+w$ must be a perfect square, $b+w>4$ .

We then want to see which values for $b+w$ yield integer solutions for $b$ and $w$ . To do so, we can set $b-w$ equal to an even integer, $2n$ . This gives the equation $(b-w)^2=b+w=4n^2$ . We get $b-w=2n$ , and $b+w=4n^2$ . Solving, we get $b=n+2n^2$ and $w=n-2n^2$ . Therefore, $b+w$ can be all even squares.

The next case we have is when $b-w$ is an odd integer, $2n+1$ . This gives the equation $(b-w)^2=b+w=4n^2+4n+1$ . We get $b-w=2n+1$ , and $b+w=4n^2+4n+1$ . Solving, we get $b=2n^2+3n+1$ and $w=2n^2+n$ . Therefore, both even and odd squares work as solutions for $b+w$ .

We can then see that the first $100$ solutions for $b+w$ will be the first $100$ perfect squares, starting from $2^2=4$ . Therefore, $k_1+k_2+k_3+\cdots+k_{100}=2^2+3^2+4^2+\cdots+101^2$ . Using the formula to find the sum of the squares from $1^2$ to $101^2$ , we get $\frac{(101)(102)(101 * 2 + 1)}{6}$ Evaluating, we get $348551$ . We subtract $1^2$ to get the total sum as $348550$ . Taking the remainder after dividng by 1000, we get $550$ as our answer.

-treetor10145

Page

Toolbox

Search

Mock AIME I 2015 Problems/Problem 5

Problem 5

Solution