2012 AIME I Problems/Problem 11
Problem 11
A frog begins at and makes a sequence of jumps according to the following rule: from the frog jumps to which may be any of the points or There are points with that can be reached by a sequence of such jumps. Find the remainder when is divided by
Solution
First of all, it is easy to see by induction that for any in the frog's jump sequence, will be a multiple of and will be a multiple of The base case obviously satisfies the constraints and if and any of the four transformations will sustain this property:
So we know that any point the frog can reach will satisfy and
Any point in the form and is reachable.
Denote the total amounts of each specific transformation in the frog's jump sequence to be and respectively. Then and and the equations and must be solvable in integers. But implies and thus Similarly implies and have the same parity. Now in order for an integer solution to exist, there must always be a way to ensure and have identical parities and also and have identical parities. The parity of is completely dependent on so the parities of and must be chosen to match this value. But the parity of can then be adjusted by adding or subtracting until it is identical to the parity of as chosen before, so we conclude that it is always possible to find an integer solution for and thus any point that satisfies and can be reached by the frog.
To count the number of such points in the region we first note that any such point will lie on the intersection of one line of the form and another line of the form The intersection of two such lines will yield the point which will be integral if and only if and have the same parity. Now since we find that
So there are possible odd values and possible even values for and possible odd values and possible even values for Every pair of lines described above will yield a valid accessible point for all pairs of and with the same parity, and the number of points is thus
See also
2012 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
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