2004 AIME I Problems/Problem 2

Problem

Set $A$ consists of $m$ consecutive integers whose sum is $2m$ , and set $B$ consists of $2m$ consecutive integers whose sum is $m.$ The absolute value of the difference between the greatest element of $A$ and the greatest element of $B$ is $99$ . Find $m.$

Simple Solution

Look at the problem... consecutive integers. Now, since set $A$ has the properties of $m$ integers that sum to $2m$ , it's obvious that the middle integer is just $2$ (the average that "balances out" everything), and the largest is $2 + \frac{m-1}{2}$ . That's because there are $m-1$ to go after taking $2$ , and they are "evenly balanced" on either side.

From there, we see that set $B$ 's average is $0.5$ . How can that happen? Only if the middle TWO values are $0$ and $1$ . Since set $B$ has $2m$ integers, its maximum must be $99$ larger than the maximum of set $A$ unless $m=1$ , which is impossible.

From there, the largest element of set $B$ is $1 + (m-1) = m$ .

Solving, we get $m - 2 - \frac{m-1}{2} = 99$ $m-\frac{m}{2}+\frac{1}{2}=101$ $\frac{m}{2}=100\frac{1}{2}.$ There we go- $m$ is equal to none other than $\boxed{201}$ .

Solution 1

Let us give the elements of our sets names: $A = \{x, x + 1, x + 2, \ldots, x + m - 1\}$ and $B = \{y, y + 1, \ldots, y + 2m - 1\}$ . So we are given that $2m = x + (x + 1) + \ldots + (x + m - 1) = mx + (1 + 2 + \ldots + (m - 1)) = mx + \frac{m(m -1)}2,$ so $2 = x + \frac{m - 1}2$ and $x + (m - 1) = \frac{m + 3}2$ . Also, $m = y + (y + 1) + \ldots + (y + 2m - 1) = 2my + \frac{2m(2m - 1)}2,$ so $1 = 2y + (2m - 1)$ so $2m = 2(y + 2m - 1)$ and $m = y + 2m - 1$ .

Then by the given, $99 = |(x + m - 1) - (y + 2m - 1)| = \left|\frac{m + 3}2 - m\right| = \left|\frac{m - 3}2\right|$ . $m$ is a positive integer so we must have $99 = \frac{m - 3}2$ and so $m = \boxed{201}$ .

Solution 2

The thing about this problem is, you have some "choices" that you can make freely when you get to a certain point, and these choices won't affect the accuracy of the solution, but will make things a lot easier for us.

First, we note that for set $A$

$\frac{m(f + l)}{2} = 2m$

Where $f$ and $l$ represent the first and last terms of $A$ . This comes from the sum of an arithmetic sequence.

Solving for $f+l$ , we find the sum of the two terms is $4$ .

Doing the same for set B, and setting up the equation with $b$ and $e$ being the first and last terms of set $B$ ,

$m(b+e) = m$

and so $b+e = 1$ .

Now we know, assume that both sequences are increasing sequences, for the sake of simplicity. Based on the fact that set $A$ has half the number of elements as set $B$ , and the difference between the greatest terms of the two two sequences is $99$ (forget about absolute value, it's insignificant here since we can just assume both sets end with positive last terms), you can set up an equation where $x$ is the last term of set A:

$2(x-(-x+4)+1) = 1+(x+99)-(-x-99+1)$

Note how i basically just counted the number of terms in each sequence here. It's made a lot simpler because we just assumed that the first term is negative and last is positive for each set, it has absolutely no effect on the end result! This is a great strategy that can help significantly simplify problems. Also note how exactly i used the fact that the first and last terms of each sequence sum to $4$ and $1$ respectively (add $x$ and $(-x+4)$ to see what i mean).

Solving this equation we find $x = 102$ . We know the first and last terms have to sum to $4$ so we find the first term of the sequence is $-98$ . Now, the solution is in clear sight, we just find the number of integers between $-98$ and $102$ , inclusive, and it is $201$ .

Note how this method is not very algebra heavy. It seems like a lot by the amount of text but really the first two steps are quite simple.

2004 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 1	Followed by Problem 3
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

Page

Toolbox

Search

2004 AIME I Problems/Problem 2

Contents

Problem

Simple Solution

Solution 1

Solution 2

See also