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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+ (...

So, the remainder is \boxed{561}.

See also

1994 AIME (ProblemsResources)
Preceded by
Problem 2
Followed by
Problem 4
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Looking for a challenging algebra text? Preparing for MATHCOUNTS or the AMC exams?
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