Difference between revisions of "2009 AIME II Problems/Problem 9"
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=== Solution 1 === | === Solution 1 === | ||
− | + | It is actually reasonably easy to compute <math>m</math> and <math>n</math> exactly. | |
First, note that if <math>4x+3y+2z=2009</math>, then <math>y</math> must be odd. Let <math>y=2y'-1</math>. We get <math>4x + 6y' - 3 + 2z = 2009</math>, which simplifies to <math>2x + 3y' + z = 1006</math>. For any pair of positive integers <math>(x,y')</math> such that <math>2x + 3y' < 1006</math> we have exactly one <math>z</math> such that the equality holds. Hence we need to count the pairs <math>(x,y')</math>. | First, note that if <math>4x+3y+2z=2009</math>, then <math>y</math> must be odd. Let <math>y=2y'-1</math>. We get <math>4x + 6y' - 3 + 2z = 2009</math>, which simplifies to <math>2x + 3y' + z = 1006</math>. For any pair of positive integers <math>(x,y')</math> such that <math>2x + 3y' < 1006</math> we have exactly one <math>z</math> such that the equality holds. Hence we need to count the pairs <math>(x,y')</math>. |
Revision as of 00:10, 8 May 2017
Problem
Let be the number of solutions in positive integers to the equation , and let be the number of solutions in positive integers to the equation . Find the remainder when is divided by .
Solution
Solution 1
It is actually reasonably easy to compute and exactly.
First, note that if , then must be odd. Let . We get , which simplifies to . For any pair of positive integers such that we have exactly one such that the equality holds. Hence we need to count the pairs .
For a fixed , can be at most . Hence the number of solutions is
Similarly, we can compute that , hence .
Solution 2
We can avoid computing and , instead we will compute directly.
Note that if and only if . Hence there is an almost 1-to-1 correspondence between the positive integer solutions of the two equations. The only exceptions are the solutions of the first equation in which at least one of the variables is equal to . The value is the number of such solutions.
If , we get the equation . The variable must be odd, and it must be between and , inclusive. For each such there is exactly one valid . Hence in this case there are valid solutions.
If , we get the equation , or equivalently . The variable must be between and , inclusive, and for each such there is exactly one valid . Hence in this case there are valid solutions.
If , we get the equation . The variable must be odd, thus let . We get , or equivalently, . Again, we see that must be odd, thus let . We get , which simplifies to . Now, we see that must be between and , inclusive, and for each such we have exactly one valid . Hence in this case there are valid solutions.
Finally, we must note that there are two special solutions: one with , and one with . We counted each of them twice, hence we have to subtract two from the total.
Therefore , and the answer is .
Solution 3
In this solution we will perform a similar operation as in Solution 2, but only on : if and only if . There is a one-to-one correspondence between the solutions of these two equations. Let and require to be positive as well. Then the second equation becomes . Notice that there are several "extra" solutions in the first equation that cannot be included in the second equation (since that would make non-positive). The value is therefore the number of "extra" solutions.
Since , in order for to be non-positive . However, equation (1) requires y to be odd, so we have two cases to consider: and . This results in the two equations and .
simplifies to . There is exactly one valid for each ; must be between and (inclusive) to obtain positive integer solutions. Therefore, there are solutions in this case.
simplifies to . There is exactly one valid for each ; must be between and (inclusive) to obtain positive integer solutions. Therefore, there are solutions in this case.
Thus, .
See Also
2009 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
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.