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

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== Problem ==
 
== Problem ==
−
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>
+
<cmath> f(x, x) = x,\; f(x, y) = f(y, x), {\rm \ and\ } (x+y)f(x, y) = yf(x, x+y). </cmath>
−
\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*}
 
−
</cmath>
 
 
Calculate <math>f(14,52)</math>.
 
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.  
 
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>
 +
 
 +
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:
  
−
<cmath>f(14,52) = \frac {38}{38}f(14,14 + 38) = \frac {52}{38}f(14,38)</cmath>
+
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>.
  
−
Repeating the process several times,
+
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>
+
<cmath>\begin{align*}
−
\begin{eqnarray*}f(14,52) & = & \frac {38}{38}f(14,14 + 38) = \frac {52}{38}f(14,38) \\
+
y \cdot \mathrm{lcm}(x,x+y) &= \dfrac{y \cdot x(x+y)}{\gcd(x,x+y)} \\
−
& = & \frac {52}{38}\times \frac {38}{24}f(14,14 + 24) = \frac {52}{24}f(14,24) \\
+
&= \dfrac{(x+y) \cdot xy}{\gcd(x,y)} \\
−
& = & \frac {52}{10}f(10,14) \\
+
&= (x+y) \cdot \mathrm{lcm}(x,y).
−
& = & \frac {52}{10}\times \frac {14}{4}f(10,4) = \frac {92}{5}f(4,10) \\
+
\end{align*}</cmath>
−
& = & \frac {91}{3}f(4,6) \\
+
Hence, <math>f(x,y) = \mathrm{lcm}(x,y)</math> is a solution to the functional equation.
−
& = & 91f(2,4) \\
+
 
−
& = & 91\times 2 f(2,2) = 364. \end{eqnarray*}
+
Since this is an AIME problem, there is exactly one correct answer, and thus, exactly one possible value of <math>f(14,52)</math>.
−
</cmath>
+
 
 +
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>
 +
 
 +
-------------------
 +
 
 +
Proof that <math>f(x,y)=\mathrm{lcm}(x,y)</math> is the only function instead of just using the fact that this is an AIME problem (Proof credited to AMSP's 101 Algebra Problems):
 +
 
 +
FTSOC, suppose that there is another function <math>g(x,y)</math> also satisfying the given conditions. Let <math>S</math> be the set of all pairs of positive integers <math>(x,y)</math> such that <math>f(x,y) \ne g(x,y)</math>, and let <math>(m,n)</math> be such a pair with minimal sum <math>m+n</math>. It is clear that <math>m \ne n</math>, otherwise <math>f(m,n) = f(m,m) = m = g(m,m) = g(m,n)</math>. By symmetry, we can also assume that <math>n>m</math>. Note that <cmath>nf(m,n-m)=[m+(n-m)]f(m,n-m)</cmath><cmath>=(n-m)f(m,m+(n-m))=(n-m)f(m,n)</cmath>
 +
or <cmath>f(m,n-m=\dfrac{n-m}{n} \cdot f(m,n).</cmath>
 +
Likewise, <cmath>g(m,n-m) = \dfrac{n-m}{n} \cdot g(m,n).</cmath>
 +
Since <math>f(m,n) \ne g(m,n)</math>, <math>f(m,n-m) \ne g(m,n-m)</math>. Thus <math>(m,n-m) \in S</math>. But <math>(m,n-m)</math> has a smaller sum then <math>(m,n)</math>, a contradiction. Therefore, <math>f(x,y) = \mathrm{lcm}(x,y)</math> is the only solution.
 +
 
 +
~[[Yiyj1]]
  
 
== See also ==
 
== See also ==
Line 30: Line 85:
  
 
[[Category:Intermediate Algebra Problems]]
 
[[Category:Intermediate Algebra Problems]]
 +
{{MAA Notice}}

Latest revision as of 16:51, 28 March 2025

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*}

Proof that $f(x,y)=\mathrm{lcm}(x,y)$ is the only function instead of just using the fact that this is an AIME problem (Proof credited to AMSP's 101 Algebra Problems):

FTSOC, suppose that there is another function $g(x,y)$ also satisfying the given conditions. Let $S$ be the set of all pairs of positive integers $(x,y)$ such that $f(x,y) \ne g(x,y)$, and let $(m,n)$ be such a pair with minimal sum $m+n$. It is clear that $m \ne n$, otherwise $f(m,n) = f(m,m) = m = g(m,m) = g(m,n)$. By symmetry, we can also assume that $n>m$. Note that \[nf(m,n-m)=[m+(n-m)]f(m,n-m)\]\[=(n-m)f(m,m+(n-m))=(n-m)f(m,n)\] or \[f(m,n-m=\dfrac{n-m}{n} \cdot f(m,n).\] Likewise, \[g(m,n-m) = \dfrac{n-m}{n} \cdot g(m,n).\] Since $f(m,n) \ne g(m,n)$, $f(m,n-m) \ne g(m,n-m)$. Thus $(m,n-m) \in S$. But $(m,n-m)$ has a smaller sum then $(m,n)$, a contradiction. Therefore, $f(x,y) = \mathrm{lcm}(x,y)$ is the only solution.

~Yiyj1

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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