Difference between revisions of "2010 AIME I Problems/Problem 5"
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Positive integers <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math> satisfy <math>a > b > c > d</math>, <math>a + b + c + d = 2010</math>, and <math>a^2 - b^2 + c^2 - d^2 = 2010</math>. Find the number of possible values of <math>a</math>. | Positive integers <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math> satisfy <math>a > b > c > d</math>, <math>a + b + c + d = 2010</math>, and <math>a^2 - b^2 + c^2 - d^2 = 2010</math>. Find the number of possible values of <math>a</math>. | ||
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== Solution == | == Solution == | ||
− | Using the [[difference of squares]], <math>2010 = (a^2 - b^2) + (c^2 - d^2) = (a + b)(a - b) + (c + d)(c - d) | + | === 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>. | ||
− | == Solution 2 == | + | === Solution 2 === |
Since <math>a+b</math> must be greater than <math>1005</math>, it follows that the only possible value for <math>a-b</math> is <math>1</math> (otherwise the quantity <math>a^2 - b^2</math> would be greater than <math>2010</math>). Therefore the only possible ordered pairs for <math>(a,b)</math> are <math>(504, 503)</math>, <math>(505, 504)</math>, ... , <math>(1004, 1003)</math>, so <math>a</math> has <math>\boxed{501}</math> possible values. | Since <math>a+b</math> must be greater than <math>1005</math>, it follows that the only possible value for <math>a-b</math> is <math>1</math> (otherwise the quantity <math>a^2 - b^2</math> would be greater than <math>2010</math>). Therefore the only possible ordered pairs for <math>(a,b)</math> are <math>(504, 503)</math>, <math>(505, 504)</math>, ... , <math>(1004, 1003)</math>, so <math>a</math> has <math>\boxed{501}</math> possible values. | ||
Revision as of 11:24, 31 March 2010
Problem
Positive integers , , , and satisfy , , and . Find the number of possible values of .
Solution
Solution 1
Using the difference of squares, , where equality must hold so and . Then we see is maximal and is minimal, so the answer is .
Solution 2
Since must be greater than , it follows that the only possible value for is (otherwise the quantity would be greater than ). Therefore the only possible ordered pairs for are , , ... , , so has possible values.
See also
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 |