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

Revision as of 17:34, 13 March 2011

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 3: / Line 3: @@
 __TOC__
-===Solution===
+==Solution==
 === Solution 1 ===
 Using the [[difference of squares]], <math>2010 = (a^2 - b^2) + (c^2 - d^2) = (a + b)(a - b) + (c + d)(c - d) \ge a + b + c + d = 2010</math>, where equality must hold so <math>b = a - 1</math> and <math>d = c - 1</math>. Then we see <math>a = 1004</math> is maximal and <math>a = 504</math> is minimal, so the answer is <math>\boxed{501}</math>.

2010 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
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