Difference between revisions of "2010 AIME I Problems/Problem 5"
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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>. | 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>. | ||
Revision as of 17:33, 13 March 2011
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 | ||
