Difference between revisions of "2012 AIME II Problems/Problem 1"
(Added one more solution.) |
Tempaccount (talk | contribs) (Adding problem section) |
||
Line 1: | Line 1: | ||
+ | |||
+ | ==Problem== | ||
== Problem 1 == | == Problem 1 == | ||
Find the number of ordered pairs of positive integer solutions <math>(m, n)</math> to the equation <math>20m + 12n = 2012</math>. | Find the number of ordered pairs of positive integer solutions <math>(m, n)</math> to the equation <math>20m + 12n = 2012</math>. |
Revision as of 15:43, 9 August 2018
Problem
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.