Difference between revisions of "2012 AIME I Problems/Problem 11"
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− | <math>\textbf{Proof:}</math> Denote the total amounts of each specific transformation in the frog's jump sequence to be <math>a,</math> <math>b,</math> <math>c,</math> and <math>d</math> respectively. Then <math>x=7a+2b-5c-10d</math> | + | <math>\textbf{Proof:}</math> Denote the total amounts of each specific transformation in the frog's jump sequence to be <math>a,</math> <math>b,</math> <math>c,</math> and <math>d</math> respectively. Then |
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+ | <math>x=7a+2b-5c-10d</math>, | ||
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+ | <math>y=2a+7b-10c-5d</math>, | ||
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+ | <math>x+y = 9(a+b)-15(c+d) = 3n</math>, and | ||
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+ | <math>x-y = 5(a-b)+5(c-d) = 5m</math> | ||
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+ | together must have integral solutions. But | ||
+ | |||
+ | <math>3(a+b)-5(c+d) = n</math> implies | ||
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+ | <math>(c+d) \equiv n \mod 3</math> and thus | ||
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+ | <math>(a+b) = \lfloor{n/3}\rfloor + 2(c+d).</math> | ||
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+ | Similarly, <math>(a-b)+(c-d) = m</math> implies that <math>(a-b)</math> and <math>(c-d)</math> have the same parity. Now in order for an integral solution to exist, there must always be a way to ensure that the pairs <math>(a+b)</math> and <math>(a-b)</math> and <math>(c+d)</math> and <math>(c-d)</math> have identical parities. The parity of <math>(a+b)</math> is completely dependent on <math>n,</math> so the parities of <math>(a-b)</math> and <math>(c-d)</math> must be chosen to match this value. But the parity of <math>(c+d)</math> can then be adjusted by adding or subtracting <math>3</math> until it is identical to the parity of <math>(c-d)</math> as chosen before, so we conclude that it is always possible to find an integer solution for <math>(a,b,c,d)</math> and thus any point that satisfies <math>x+y = 3n</math> and <math>x-y = 5m</math> can be reached by the frog. | ||
<math>\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}</math> | <math>\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}</math> | ||
Revision as of 09:07, 24 September 2017
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 such that there exists 2 integers and that satisfy and is reachable.
Denote the total amounts of each specific transformation in the frog's jump sequence to be and respectively. Then
,
,
, and
together must have integral solutions. But
implies
and thus
Similarly, implies that and have the same parity. Now in order for an integral solution to exist, there must always be a way to ensure that the pairs and and 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 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.