Difference between revisions of "1988 AIME Problems/Problem 8"

m (See also)
(solution, revised version of 4everwise's)
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
 +
The [[function]] <math>f</math>, defined on the set of ordered pairs of positive [[integer]]s, satisfies the following properties:
 +
<cmath>
 +
\begin{eqnarray*} f(x,x) & = & x, \\
 +
f(x,y) & = & f(y,x), \quad \text{and} \\
 +
(x + y) f(x,y) & = & yf(x,x + y). \end{eqnarray*}
 +
</cmath>
 +
Calculate <math>f(14,52)</math>.
  
 
== Solution ==
 
== Solution ==
 +
Since all of the function's properties contain a recursive definition except for the first one, we know that <math>f(x,x) = x</math> in order to obtain an integer answer. So, we have to transform <math>f(14,52)</math> 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
 +
 +
<cmath>f(14,52) = \frac {38}{38}f(14,14 + 38) = \frac {52}{38}f(14,38)</cmath>
 +
 +
Repeating the process several times,
 +
<cmath>
 +
\begin{eqnarray*}f(14,52) & = & \frac {38}{38}f(14,14 + 38) = \frac {52}{38}f(14,38) \\
 +
& = & \frac {52}{38}\times \frac {24}{24}f(14,14 + 24) = \frac {52}{24}f(14,24) \\
 +
& = & \frac {52}{10}f(10,14) \\
 +
& = & \frac {52}{10}\times \frac {14}{4}f(10,4) = \frac {92}{5}f(4,10) \\
 +
& = & \frac {91}{3}f(4,6) \\
 +
& = & 91f(2,4) \\
 +
& = & 91\times 2 f(2,2) = 364. \end{eqnarray*}
 +
</cmath>
  
 
== See also ==
 
== See also ==
* [[1988 AIME Problems]]
+
{{AIME box|year=1988|num-b=7|num-a=9}}
  
{{AIME box|year=1988|num-b=7|num-a=9}}
+
[[Category:Intermediate Algebra Problems]]

Revision as of 15:41, 28 September 2007

Problem

The function $f$, defined on the set of ordered pairs of positive integers, satisfies the following properties: \begin{eqnarray*} f(x,x) & = & x, \\ f(x,y) & = & f(y,x), \quad \text{and} \\ (x + y) f(x,y) & = & yf(x,x + y). \end{eqnarray*} Calculate $f(14,52)$.

Solution

Since all of the function's properties contain a recursive definition except for the first one, we know that $f(x,x) = x$ in order to obtain an integer answer. So, we have to transform $f(14,52)$ 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

\[f(14,52) = \frac {38}{38}f(14,14 + 38) = \frac {52}{38}f(14,38)\]

Repeating the process several times, \begin{eqnarray*}f(14,52) & = & \frac {38}{38}f(14,14 + 38) = \frac {52}{38}f(14,38) \\ & = & \frac {52}{38}\times \frac {24}{24}f(14,14 + 24) = \frac {52}{24}f(14,24) \\ & = & \frac {52}{10}f(10,14) \\ & = & \frac {52}{10}\times \frac {14}{4}f(10,4) = \frac {92}{5}f(4,10) \\ & = & \frac {91}{3}f(4,6) \\ & = & 91f(2,4) \\ & = & 91\times 2 f(2,2) = 364. \end{eqnarray*}

See also

1988 AIME (ProblemsAnswer KeyResources)
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