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1988 AIME Problems/Problem 8

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Problem

The function , 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 .

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)

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 (ProblemsResources)
Preceded by
Problem 7
Followed by
Problem 9
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
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