Difference between revisions of "1988 AIME Problems/Problem 8"
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& = & \frac {52}{38}\times \frac {38}{24}f(14,14 + 24) = \frac {52}{24}f(14,24) \\ | & = & \frac {52}{38}\times \frac {38}{24}f(14,14 + 24) = \frac {52}{24}f(14,24) \\ | ||
& = & \frac {52}{10}f(10,14) \\ | & = & \frac {52}{10}f(10,14) \\ | ||
− | & = & \frac {52}{10}\times \frac {14}{4}f(10,4) = \frac { | + | & = & \frac {52}{10}\times \frac {14}{4}f(10,4) = \frac {91}{5}f(4,10) \\ |
& = & \frac {91}{3}f(4,6) \\ | & = & \frac {91}{3}f(4,6) \\ | ||
& = & 91f(2,4) \\ | & = & 91f(2,4) \\ |
Revision as of 01:43, 7 June 2014
Contents
Problem
The function , defined on the set of ordered pairs of positive integers, satisfies the following properties: Calculate .
Solution
Since all of the function's properties contain a recursive definition except for the first one, we know that in order to obtain an integer answer. So, we have to transform to this form by exploiting the other properties. The second one doesn't help us immediately, so we will use the third one.
Note that
Repeating the process several times,
Solution 2
Notice that satisfies all three properties:
Clearly, and .
Using the identities and , we have:
.
Hence, is a solution to the functional equation.
Since this is an AIME problem, there is exactly one correct answer, and thus, exactly one possible value of .
Therefore, .
See also
1988 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
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.