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

 
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== Problem ==
 
== Problem ==
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The function <math>f_{}^{}</math> has the property that, for each real number <math>x,\,</math>
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<center><math>f(x)+f(x-1) = x^2.\,</math></center>
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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>?
  
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== Solution ==
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== Solution 1 ==
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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>
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So, the remainder is <math>\boxed{561}</math>.
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== Solution 2 ==
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Those familiar with triangular numbers and some of their properties will quickly recognize the equation given in the problem. It is well-known (and easy to show) that the sum of two consecutive triangular numbers is a perfect square; that is,
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<cmath>T_{n-1} + T_n = n^2,</cmath>
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where <math>T_n = 1+2+...+n = \frac{n(n+1)}{2}</math> is the <math>n</math>th triangular number.
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Using this, as well as using the fact that the value of <math>f(x)</math> directly determines the value of <math>f(x+1)</math> and <math>f(x-1),</math> we conclude that <math>f(n) = T_n + K</math> for all odd <math>n</math> and <math>f(n) = T_n - K</math> for all even <math>n,</math> where <math>K</math> is a constant real number.
 +
 
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Since <math>f(19) = 94</math> and <math>T_{19} = 190,</math> we see that <math>K = -96.</math> It follows that <math>f(94) = T_{94} - (-96) = \frac{94\cdot 95}{2} + 96 = 4561,</math> so the answer is <math>\boxed{561}</math>.
  
 
== See also ==
 
== See also ==
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* [[1994 AIME Problems]]
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{{AIME box|year=1994|num-b=2|num-a=4}}
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[[Category:Intermediate Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 10:12, 3 July 2023

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 1

\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 $\boxed{561}$.

Solution 2

Those familiar with triangular numbers and some of their properties will quickly recognize the equation given in the problem. It is well-known (and easy to show) that the sum of two consecutive triangular numbers is a perfect square; that is, \[T_{n-1} + T_n = n^2,\] where $T_n = 1+2+...+n = \frac{n(n+1)}{2}$ is the $n$th triangular number.

Using this, as well as using the fact that the value of $f(x)$ directly determines the value of $f(x+1)$ and $f(x-1),$ we conclude that $f(n) = T_n + K$ for all odd $n$ and $f(n) = T_n - K$ for all even $n,$ where $K$ is a constant real number.

Since $f(19) = 94$ and $T_{19} = 190,$ we see that $K = -96.$ It follows that $f(94) = T_{94} - (-96) = \frac{94\cdot 95}{2} + 96 = 4561,$ so the answer is $\boxed{561}$.

See also

1994 AIME (Problems • Answer Key • Resources)
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

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