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Difference between revisions of "University of South Carolina High School Math Contest/1993 Exam/Problem 5"

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== Solution ==
 
== Solution ==
 
Notice that <math>f(3)=f(2+1)=f(2)+f(1)+1=f(2)+3</math>. Also, <math>f(2)=f(1+1)=f(1)+f(1)+1=5</math>. Thus, <math>f(3)=3+5=8</math>.
 
Notice that <math>f(3)=f(2+1)=f(2)+f(1)+1=f(2)+3</math>. Also, <math>f(2)=f(1+1)=f(1)+f(1)+1=5</math>. Thus, <math>f(3)=3+5=8</math>.
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In general, <math>f(x + 1) = f(x) + f(1) + 1 = f(x) + 3</math>, so we have a simple [[recursion|recursive]] definition for the [[function]] <math>f</math>.  From here we can see that <math>f(n) = 3n - 1</math> for all [[positive integer]]s <math>n</math>.
  
 
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Latest revision as of 14:55, 29 July 2006

Problem

Suppose that $f$ is a function with the property that for all $x$ and $y$, $f(x + y) = f(x) + f(y) + 1$ and $f(1) = 2.$ What is the value of $f(3)$?

$\mathrm{(A) \ }4 \qquad \mathrm{(B) \ }5 \qquad \mathrm{(C) \ }6 \qquad \mathrm{(D) \ }7 \qquad \mathrm{(E) \ }8$

Solution

Notice that $f(3)=f(2+1)=f(2)+f(1)+1=f(2)+3$. Also, $f(2)=f(1+1)=f(1)+f(1)+1=5$. Thus, $f(3)=3+5=8$.

In general, $f(x + 1) = f(x) + f(1) + 1 = f(x) + 3$, so we have a simple recursive definition for the function $f$. From here we can see that $f(n) = 3n - 1$ for all positive integers $n$.

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