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

 
m (→‎Solution 2 (Algebra))
 
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== Problem ==
 
== Problem ==
 +
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>.
  
−
== Solution ==
+
== Solution 1 (Algebra) ==
 +
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>
 +
<cmath>\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*}</cmath>
 +
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 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 <cmath>f(14,52) = f(14,14 + 38) = \frac{52}{38}\cdot f(14,38).</cmath>
 +
 
 +
Repeating the process several times,
 +
<cmath>\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*}</cmath>
 +
 
 +
==Solution 3 (Number Theory)==
 +
Notice that <math>f(x,y) = \mathrm{lcm}(x,y)</math> satisfies all three properties:
 +
 
 +
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>.
 +
 
 +
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>.
 +
 
 +
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 ==
−
* [[1988 AIME Problems]]
+
{{AIME box|year=1988|num-b=7|num-a=9}}
 +
 
 +
[[Category:Intermediate Algebra Problems]]
 +
{{MAA Notice}}

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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