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

Revision as of 08:11, 19 June 2021

Problem

The function $f$ , defined on the set of ordered pairs of positive integers, satisfies the following properties: $f(x, x) = x,\; f(x, y) = f(y, x), {\rm \ and\ } (x+y)f(x, y) = yf(x, x+y).$ Calculate $f(14,52)$ .

Solution 1 (Generalized)

Solution 2 (Algebra)

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 $f(14,52) = f(14,14 + 38) = \frac{52}{38}\cdot f(14,38).$

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

Solution 3 (Number Theory)

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

For the first two properties, it is clear that $\mathrm{lcm}(x,x) = x$ and $\mathrm{lcm}(x,y) = \mathrm{lcm}(y,x)$ .

For the third property, using the identities $\gcd(x,y) \cdot \mathrm{lcm}(x,y) = x\cdot y$ and $\gcd(x,x+y) = \gcd(x,y)$ gives $\begin{align*} y \cdot \mathrm{lcm}(x,x+y) &= \dfrac{y \cdot x(x+y)}{\gcd(x,x+y)} \\ &= \dfrac{(x+y) \cdot xy}{\gcd(x,y)} \\ &= (x+y) \cdot \mathrm{lcm}(x,y). \end{align*}$ Hence, $f(x,y) = \mathrm{lcm}(x,y)$ 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, we have $\begin{align*} f(14,52) &= \mathrm{lcm}(14,52) \\ &= \mathrm{lcm}(2 \cdot 7,2^2 \cdot 13) \\ &= 2^2 \cdot 7 \cdot 13 \\ &= \boxed{364}. \end{align*}$

@@ Line 4: / Line 4: @@
 Calculate <math>f(14,52)</math>.
-== Solution 1 ==
+== Solution 1 (Generalized) ==
+== Solution 2 (Algebra)==
 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
+Note that <cmath>f(14,52) = f(14,14 + 38) = \frac{52}{38}\cdot f(14,38).</cmath>
-<cmath>f(14,52) = f(14,14 + 38) = \frac {52}{38}f(14,38)</cmath>
 Repeating the process several times,
-<cmath>
+<cmath>\begin{eqnarray*}
-\begin{eqnarray*}f(14,52) & = & f(14,14 + 38) = \frac {52}{38}f(14,38) \\
+f(14,52) & = & f(14,14 + 38) = \frac{52}{38}\cdot f(14,38) \\
-& = & \frac {52}{38}\times \frac {38}{24}f(14,14 + 24) = \frac {52}{24}f(14,24) \\
+& = & \frac{52}{38}\cdot \frac{38}{24}\cdot f(14,14 + 24) \\
-& = & \frac {52}{10}f(10,14) \\
+& = & \frac{52}{24}\cdot f(14,24) \\
-& = & \frac {52}{10}\times \frac {14}{4}f(10,4) = \frac {91}{5}f(4,10) \\
+& = & \frac{52}{10}\cdot f(10,14) \\
-& = & \frac {91}{3}f(4,6) \\
+& = & \frac{52}{10}\cdot \frac{14}{4}\cdot f(10,4) \\
-& = & 91f(2,4) \\
+& = & \frac{91}{5}\cdot f(4,10) \\
-& = & 91\times 2 f(2,2) = 364. \end{eqnarray*}
+& = & \frac{91}{3}\cdot f(4,6) \\
+& = & 91\cdot f(2,4) \\
+& = & 91\cdot 2 \cdot f(2,2) \\
+& = & \boxed{364}.
+\end{eqnarray*}
 </cmath>
-==Solution 2==
+==Solution 3 (Number Theory)==
-Notice that <math>f(x,y) = \text{lcm}(x,y)</math> satisfies all three properties:
+Notice that <math>f(x,y) = \mathrm{lcm}(x,y)</math> satisfies all three properties:
-Clearly, <math>\text{lcm}(x,x) = x</math> and <math>\text{lcm}(x,y) = \text{lcm}(y,x)</math>.
-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:
-<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>.
+For the first two properties, it is clear that <math>\mathrm{lcm}(x,x) = x</math> and <math>\mathrm{lcm}(x,y) = \mathrm{lcm}(y,x)</math>.
-Hence, <math>f(x,y) = \text{lcm}(x,y)</math> is a solution to the functional equation.
+For the third property, using the identities <math>\gcd(x,y) \cdot \mathrm{lcm}(x,y) = x\cdot y</math> and <math>\gcd(x,x+y) = \gcd(x,y)</math> gives
+<cmath>\begin{align*}
+y \cdot \mathrm{lcm}(x,x+y) &= \dfrac{y \cdot x(x+y)}{\gcd(x,x+y)} \\
+&= \dfrac{(x+y) \cdot xy}{\gcd(x,y)} \\
+&= (x+y) \cdot \mathrm{lcm}(x,y).
+\end{align*}</cmath>
+Hence, <math>f(x,y) = \mathrm{lcm}(x,y)</math> 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 <math>f(14,52)</math>.
+Since this is an AIME problem, there is exactly one correct answer, and thus, exactly one possible value of <math>f(14,52)</math>.
-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>.
+Therefore, we have
+<cmath>\begin{align*}
+f(14,52) &= \mathrm{lcm}(14,52) \\
+&= \mathrm{lcm}(2 \cdot 7,2^2 \cdot 13) \\
+&= 2^2 \cdot 7 \cdot 13 \\
+&= \boxed{364}.
+\end{align*}</cmath>
 == 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
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