Difference between revisions of "2012 AIME II Problems/Problem 1"
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Dividing by <math>4</math> gives us <math>5m + 3n = 503</math>. Solving for <math>n</math> gives <math>n \equiv 1 \pmod 5</math>. The solutions are the numbers <math>n | Dividing by <math>4</math> gives us <math>5m + 3n = 503</math>. Solving for <math>n</math> gives <math>n \equiv 1 \pmod 5</math>. The solutions are the numbers <math>n | ||
= 1, 6, 11, ... , 166</math>. There are <math>\boxed{034}</math> solutions. | = 1, 6, 11, ... , 166</math>. There are <math>\boxed{034}</math> solutions. | ||
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+ | ==Solution 3== | ||
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+ | Because the x-intercept of the equation is <math>\frac{2012}{20}</math>, and the y-intercept is <math>\frac{2012}{12}</math>, the slope is <math>\frac{\frac{-2012}{12}}{\frac{2012}{20}} = \frac{-5}{3}</math>. Now, notice the first obvious solution: (100,1). From it, we derive all the other solutions by applying the slope in reverse, i.e: <math>(100,1), (97,6), (94,11)...</math> Because the solutions are only positive, we can generate only 33 more solutions, so in total we have <math>33+1=\boxed{034}</math> solutions. | ||
== See Also == | == See Also == |
Revision as of 12:22, 26 May 2018
Problem 1
Find the number of ordered pairs of positive integer solutions to the equation .
Solution
Solution 1
Solving for gives us so in order for to be an integer, we must have The smallest possible value of is obviously and the greatest is so the total number of solutions is
Solution 2
Dividing by gives us . Solving for gives . The solutions are the numbers . There are solutions.
Solution 3
Because the x-intercept of the equation is , and the y-intercept is , the slope is . Now, notice the first obvious solution: (100,1). From it, we derive all the other solutions by applying the slope in reverse, i.e: Because the solutions are only positive, we can generate only 33 more solutions, so in total we have solutions.
See Also
2012 AIME II (Problems • Answer Key • Resources) | ||
Preceded by First Problem |
Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.