Difference between revisions of "1988 AIME Problems/Problem 8"
(solution, revised version of 4everwise's) |
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\begin{eqnarray*}f(14,52) & = & \frac {38}{38}f(14,14 + 38) = \frac {52}{38}f(14,38) \\ | \begin{eqnarray*}f(14,52) & = & \frac {38}{38}f(14,14 + 38) = \frac {52}{38}f(14,38) \\ | ||
| − | & = & \frac {52}{38}\times \frac { | + | & = & \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 {92}{5}f(4,10) \\ | & = & \frac {52}{10}\times \frac {14}{4}f(10,4) = \frac {92}{5}f(4,10) \\ | ||
Revision as of 17:37, 4 March 2012
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
![\[f(14,52) = \frac {38}{38}f(14,14 + 38) = \frac {52}{38}f(14,38)\]](http://latex.artofproblemsolving.com/7/6/b/76bbfac5877869d14c76ef97573bb8da397804b1.png)
Repeating the process several times,
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 | ||
