1994 AIME Problems/Problem 3

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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 $\boxed{561}$ .

1994 AIME (Problems • Resources)
Preceded by Problem 2	Followed by Problem 4
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

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1994 AIME Problems/Problem 3

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Problem

Solution

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