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

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& = & 91\times 2 f(2,2) = 364. \end{eqnarray*}
 
& = & 91\times 2 f(2,2) = 364. \end{eqnarray*}
 
</cmath>
 
</cmath>
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==Solution 2==
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Notice that <math>f(x,y) = \text{lcm}(x,y)</math> satisfies all three properties:
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Clearly, <math>\text{lcm}(x,x) = x</math> and <math>\text{lcm}(x,y) = \text{lcm}(y,x)</math>.
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Using the identities <math>\text{gcd}(x,y) \cdot \text{lcm}(x,y) = xy</math> and <math>\text{gcd}(x,x+y) = \text{gcd}(x,y)</math>, we have:
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<math>y \cdot \text{lcm}(x,x+y) </math> <math>= \dfrac{y \cdot x(x+y)}{\text{gcd}(x,x+y)} </math> <math>= \dfrac{(x+y) \cdot xy}{\text{gcd}(x,y)} </math> <math>= (x+y) \cdot \text{lcm}(x,y)</math>.
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Hence, <math>f(x,y) = \text{lcm}(x,y)</math> is a solution to the functional equation.
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Since this is an [[AIME]] problem, there is exactly one correct answer, and thus, exactly one possible value of <math>f(14,52)</math>.
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Therefore, <math>f(14,52) = \text{lcm}(14,52) = \text{lcm}(2 \cdot 7,2^2 \cdot 13) = 2^2 \cdot 7 \cdot 13 = \boxed{364}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 17:23, 21 September 2012

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 {38}{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*}

Solution 2

Notice that $f(x,y) = \text{lcm}(x,y)$ satisfies all three properties:

Clearly, $\text{lcm}(x,x) = x$ and $\text{lcm}(x,y) = \text{lcm}(y,x)$.

Using the identities $\text{gcd}(x,y) \cdot \text{lcm}(x,y) = xy$ and $\text{gcd}(x,x+y) = \text{gcd}(x,y)$, we have:

$y \cdot \text{lcm}(x,x+y)$ $= \dfrac{y \cdot x(x+y)}{\text{gcd}(x,x+y)}$ $= \dfrac{(x+y) \cdot xy}{\text{gcd}(x,y)}$ $= (x+y) \cdot \text{lcm}(x,y)$.

Hence, $f(x,y) = \text{lcm}(x,y)$ 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 $f(14,52)$.

Therefore, $f(14,52) = \text{lcm}(14,52) = \text{lcm}(2 \cdot 7,2^2 \cdot 13) = 2^2 \cdot 7 \cdot 13 = \boxed{364}$.

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
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