Difference between revisions of "2010 AIME I Problems/Problem 5"

Revision as of 17:02, 9 August 2018

Problem

Positive integers $a$ , $b$ , $c$ , and $d$ satisfy $a > b > c > d$ , $a + b + c + d = 2010$ , and $a^2 - b^2 + c^2 - d^2 = 2010$ . Find the number of possible values of $a$ .

Solution

Solution 1

Using the difference of squares, $2010 = (a^2 - b^2) + (c^2 - d^2) = (a + b)(a - b) + (c + d)(c - d) \ge a + b + c + d = 2010$ , where equality must hold so $b = a - 1$ and $d = c - 1$ . Then we see $a = 1004$ is maximal and $a = 504$ is minimal, so the answer is $\boxed{501}$ .

Solution 2

Since $a+b$ must be greater than $1005$ , it follows that the only possible value for $a-b$ is $1$ (otherwise the quantity $a^2 - b^2$ would be greater than $2010$ ). Therefore the only possible ordered pairs for $(a,b)$ are $(504, 503)$ , $(505, 504)$ , ... , $(1004, 1003)$ , so $a$ has $\boxed{501}$ possible values.

@@ Line 1: / Line 1: @@
-==Problem==
 __TOC__
 == Problem ==

2010 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

Page

Toolbox

Search