1994 AIME Problems/Problem 3

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?

$f(94)=94^2-f(93)=94^2-93^2+f(92)=94^2-93^2+92^2-f(91)=\ldots$

$f(94) = (94^2-93^2) + (92^2-91^2) +\ldots+ (22^2-21^2)+ 20^2-f(19) = 94+93+\ldots+21+400-94$

$f(94) = 4561$

So, the remainder is 561.

1994 AIME (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
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