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Difference between revisions of "2002 AMC 12P Problems/Problem 20"

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== Solution ==
 
== Solution ==
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If <math>\log_{b} 729 = n</math>, then <math>b^n = 729</math>. Since <math>729 = 3^6</math>, <math>b</math> must be <math>3</math> to some [[factor]] of 6. Thus, there are four (3, 9, 27, 729) possible values of <math>b \Longrightarrow \boxed{\mathrm{E}}</math>.
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When <math>x = 2</math>, then we get <math> f(2) + 2f(1001) = 6</math>; we can also substitute <math>x</math> as <math>1001</math>, then we will get <math>f(1001) + 2f(2) =3003</math>. Solve this system of equations, then we get <math>f(2)= 2000</math> <math>\Longrightarrow \boxed{\mathrm{B}}</math>.
  
 
== See also ==
 
== See also ==
 
{{AMC12 box|year=2002|ab=P|num-b=19|num-a=21}}
 
{{AMC12 box|year=2002|ab=P|num-b=19|num-a=21}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 15:55, 17 January 2024

Problem

Let $f$ be a real-valued function such that

\[f(x) + 2f(\frac{2002}{x}) = 3x\]

for all $x>0.$ Find $f(2).$

$\text{(A) }1000 \qquad \text{(B) }2000 \qquad \text{(C) }3000 \qquad \text{(D) }4000 \qquad \text{(E) }6000$

Solution

When $x = 2$, then we get $f(2) + 2f(1001) = 6$; we can also substitute $x$ as $1001$, then we will get $f(1001) + 2f(2) =3003$. Solve this system of equations, then we get $f(2)= 2000$ $\Longrightarrow \boxed{\mathrm{B}}$.

See also

2002 AMC 12P (Problems • Answer Key • Resources)
Preceded by
Problem 19 		Followed by
Problem 21
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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