2004 AIME I Problems/Problem 2

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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 $m.$

Solution

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 $m = \boxed{201}$ .

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

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2004 AIME I Problems/Problem 2

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