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

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== Problem ==
 
== Problem ==
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The [[function]] <math>f</math>, defined on the set of ordered pairs of positive [[integer]]s, satisfies the following properties:
+
The function <math>f</math>, defined on the set of ordered pairs of positive integers, satisfies the following properties:
 +
<cmath> f(x, x) = x,\; f(x, y) = f(y, x), {\rm \ and\ } (x+y)f(x, y) = yf(x, x+y). </cmath>
 +
Calculate <math>f(14,52)</math>.
  
−
<cmath>
+
== Solution 1 (Algebra) ==
−
\begin{eqnarray*} f(x,x) & = & x, \\
+
Let <math>z = x+y</math>. By the substitution <math>z=x+y,</math> we rewrite the third property in terms of <math>x</math> and <math>z,</math> then solve for <math>f(x,z):</math>
−
f(x,y) & = & f(y,x), \quad \text{and} \\
+
<cmath>\begin{align*}
−
(x + y) f(x,y) & = & yf(x,x + y). \end{eqnarray*}
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zf(x,z-x) &= (z-x)f(x,z) \\
−
</cmath>
+
f(x,z) &= \frac{z}{z-x} \cdot f(x,z-x).
−
 
+
\end{align*}</cmath>
−
Calculate <math>f(14,52)</math>.
+
Using the properties of <math>f,</math> we have
 +
<cmath>\begin{align*}
 +
f(14,52) &= \frac{52}{38} \cdot f(14,38) \\
 +
&= \frac{52}{38} \cdot \frac{38}{24} \cdot f(14,24) \\
 +
&= \frac{52}{38} \cdot \frac{38}{24} \cdot \frac{24}{10} \cdot f(14,10)\\
 +
&= \frac{52}{38} \cdot \frac{38}{24} \cdot \frac{24}{10} \cdot f(10,14)\\
 +
&= \frac{52}{38} \cdot \frac{38}{24} \cdot \frac{24}{10} \cdot \frac{14}{4} \cdot f(10,4)\\
 +
&= \frac{52}{38} \cdot \frac{38}{24} \cdot \frac{24}{10} \cdot \frac{14}{4} \cdot f(4,10)\\
 +
&= \frac{52}{38} \cdot \frac{38}{24} \cdot \frac{24}{10} \cdot \frac{14}{4} \cdot \frac{10}{6} \cdot f(4,6)\\
 +
&= \frac{52}{38} \cdot \frac{38}{24} \cdot \frac{24}{10} \cdot \frac{14}{4} \cdot \frac{10}{6} \cdot \frac{6}{2} \cdot f(4,2)\\
 +
&= \frac{52}{38} \cdot \frac{38}{24} \cdot \frac{24}{10} \cdot \frac{14}{4} \cdot \frac{10}{6} \cdot \frac{6}{2} \cdot f(2,4)\\
 +
&= \frac{52}{38} \cdot \frac{38}{24} \cdot \frac{24}{10} \cdot \frac{14}{4} \cdot \frac{10}{6} \cdot \frac{6}{2} \cdot \frac{4}{2} \cdot f(2,2)\\
 +
&= \frac{52}{38} \cdot \frac{38}{24} \cdot \frac{24}{10} \cdot \frac{14}{4} \cdot \frac{10}{6} \cdot \frac{6}{2} \cdot \frac{4}{2} \cdot 2\\
 +
&=\boxed{364}.
 +
\end{align*}</cmath>
 +
~MRENTHUSIASM (credit given to AoPS)
  
−
== Solution 1 ==
+
== 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.  
 
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,
 
Repeating the process several times,
−
<cmath>
+
<cmath>\begin{align*}
−
\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}\times \frac {38}{24}f(14,14 + 24) = \frac {52}{24}f(14,24) \\
+
& = \frac{52}{38}\cdot f(14,38) \\
−
& = & \frac {52}{10}f(10,14) \\
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& = \frac{52}{38}\cdot \frac{38}{24}\cdot f(14,14 + 24) \\
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& = & \frac {52}{10}\times \frac {14}{4}f(10,4) = \frac {91}{5}f(4,10) \\
+
& = \frac{52}{24}\cdot f(14,24) \\
−
& = & \frac {91}{3}f(4,6) \\
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& = \frac{52}{10}\cdot f(10,14) \\
−
& = & 91f(2,4) \\
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& = \frac{52}{10}\cdot \frac{14}{4}\cdot f(10,4) \\
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& = & 91\times 2 f(2,2) = 364. \end{eqnarray*}
+
& = \frac{91}{5}\cdot f(4,10) \\
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</cmath>
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& = \frac{91}{3}\cdot f(4,6) \\
−
 
+
& = 91\cdot f(2,4) \\
−
==Solution 2==
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& = 91\cdot 2 \cdot f(2,2) \\
−
Notice that <math>f(x,y) = \text{lcm}(x,y)</math> satisfies all three properties:
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& = \boxed{364}.
−
 
+
\end{align*}</cmath>
−
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:  
+
==Solution 3 (Number Theory)==
 +
Notice that <math>f(x,y) = \mathrm{lcm}(x,y)</math> satisfies all three properties:  
  
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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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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.  
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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.  
  
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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>.  
+
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 ==
 
== See also ==

Latest revision as of 12:24, 12 October 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 (Algebra)

Let $z = x+y$. By the substitution $z=x+y,$ we rewrite the third property in terms of $x$ and $z,$ then solve for $f(x,z):$ \begin{align*} zf(x,z-x) &= (z-x)f(x,z) \\ f(x,z) &= \frac{z}{z-x} \cdot f(x,z-x). \end{align*} Using the properties of $f,$ we have \begin{align*} f(14,52) &= \frac{52}{38} \cdot f(14,38) \\ &= \frac{52}{38} \cdot \frac{38}{24} \cdot f(14,24) \\ &= \frac{52}{38} \cdot \frac{38}{24} \cdot \frac{24}{10} \cdot f(14,10)\\ &= \frac{52}{38} \cdot \frac{38}{24} \cdot \frac{24}{10} \cdot f(10,14)\\ &= \frac{52}{38} \cdot \frac{38}{24} \cdot \frac{24}{10} \cdot \frac{14}{4} \cdot f(10,4)\\ &= \frac{52}{38} \cdot \frac{38}{24} \cdot \frac{24}{10} \cdot \frac{14}{4} \cdot f(4,10)\\ &= \frac{52}{38} \cdot \frac{38}{24} \cdot \frac{24}{10} \cdot \frac{14}{4} \cdot \frac{10}{6} \cdot f(4,6)\\ &= \frac{52}{38} \cdot \frac{38}{24} \cdot \frac{24}{10} \cdot \frac{14}{4} \cdot \frac{10}{6} \cdot \frac{6}{2} \cdot f(4,2)\\ &= \frac{52}{38} \cdot \frac{38}{24} \cdot \frac{24}{10} \cdot \frac{14}{4} \cdot \frac{10}{6} \cdot \frac{6}{2} \cdot f(2,4)\\ &= \frac{52}{38} \cdot \frac{38}{24} \cdot \frac{24}{10} \cdot \frac{14}{4} \cdot \frac{10}{6} \cdot \frac{6}{2} \cdot \frac{4}{2} \cdot f(2,2)\\ &= \frac{52}{38} \cdot \frac{38}{24} \cdot \frac{24}{10} \cdot \frac{14}{4} \cdot \frac{10}{6} \cdot \frac{6}{2} \cdot \frac{4}{2} \cdot 2\\ &=\boxed{364}. \end{align*} ~MRENTHUSIASM (credit given to AoPS)

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{align*} 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{align*}

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*}

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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