Difference between revisions of "1994 AIME Problems/Problem 3"

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The function <math>f_{}^{}</math> has the property that, for each real number <math>x,\,</math>
 
The function <math>f_{}^{}</math> has the property that, for each real number <math>x,\,</math>
 
<center><math>f(x)+f(x-1) = x^2\,</math></center>.
 
<center><math>f(x)+f(x-1) = x^2\,</math></center>.
If <math>f(19)=94,\,</math> what is the remainder when <math>f(94)\,</math> is divided by 1000?
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If <math>f(19)=94,\,</math> what is the remainder when <math>f(94)\,</math> is divided by <math>1000</math>?
  
 
== Solution ==
 
== Solution ==
<math>f(94)=94^2-f(93)=94^2-93^2+f(92)=94^2-93^2+92^2-f(91)=\ldots</math>
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<cmath>\begin{align*}f(94)&=94^2-f(93)=94^2-93^2+f(92)=94^2-93^2+92^2-f(91)=\cdots \\
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&= (94^2-93^2) + (92^2-91^2) +\cdots+ (22^2-21^2)+ 20^2-f(19) \\ &= 94+93+\cdots+21+400-94  \\
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&= 4561 \end{align*}</cmath>
  
<math>f(94) = (94^2-93^2) + (92^2-91^2) +\ldots+ (22^2-21^2)+ 20^2-f(19) = 94+93+\ldots+21+400-94 </math>
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So, the remainder is <math>\boxed{561}</math>.
  
<math>f(94) = 4561</math>
 
 
 
So, the remainder is 561.
 
 
== See also ==
 
== See also ==
 
{{AIME box|year=1994|num-b=2|num-a=4}}
 
{{AIME box|year=1994|num-b=2|num-a=4}}
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[[Category:Intermediate Algebra Problems]]

Revision as of 21:16, 8 October 2008

Problem

The function $f_{}^{}$ has the property that, for each real number $x,\,$

$f(x)+f(x-1) = x^2\,$

.

If $f(19)=94,\,$ what is the remainder when $f(94)\,$ is divided by $1000$?

Solution

\begin{align*}f(94)&=94^2-f(93)=94^2-93^2+f(92)=94^2-93^2+92^2-f(91)=\cdots \\ &= (94^2-93^2) + (92^2-91^2) +\cdots+ (22^2-21^2)+ 20^2-f(19) \\ &= 94+93+\cdots+21+400-94  \\ &= 4561 \end{align*}

So, the remainder is $\boxed{561}$.

See also

1994 AIME (ProblemsAnswer KeyResources)
Preceded by
Problem 2
Followed by
Problem 4
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions