2007 AMC 12A Problems/Problem 24
Contents
Problem
For each integer , let be the number of solutions to the equation on the interval . What is ?
Solution 1
By looking at various graphs, we obtain that, for most of the graphs
Notice that the solutions are basically reflections across . However, when , the middle apex of the sine curve touches the sine curve at the top only one time (instead of two reflected points), so we get here .
Solution 2
So if and only if or .
The first occurs whenever , or for some nonnegative integer . Since , . So there are solutions in this case.
The second occurs whenever , or for some nonnegative integer . Here so that there are solutions here.
However, we overcount intersections. These occur whenever
which is equivalent to dividing . If is even, then is odd, so this never happens. If , then there won't be intersections either, since a multiple of 8 can't divide a number which is not even a multiple of 4.
This leaves . In this case, the divisibility becomes dividing . Since and are relatively prime (subtracting twice the second number from the first gives 1), must divide . Since , . Then there is only one intersection, namely when .
Therefore we find is equal to , unless , in which case it is one less, or . The problem may then be finished as in Solution 1.
See also
2007 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 23 |
Followed by Problem 25 |
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