Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1994 based on Remainders and Functions.
The function f has the property that, for each real number x, \(f(x)+f(x-1)=x^{2}\) if f(19)=94, find the remainder when f(94) is divided by 1000.
Integers
Remainder
Functions
Answer: is 561.
AIME I, 1994, Question 7
Elementary Number Theory by David Burton
f(94)=\(94^{2}-f(93)=94^{2}-93^{2}+f(92)\)
=\(94^{2}-93^{2}+92^{2}-f(91)\)
=\((94^{2}-93^{2})+(92^{2}-91^{2})\)
\(+....+(22^{2}-21^{2})+20^{2}-f(19)\)
=94+93+.....+21+400-94
=4561
\(\Rightarrow\) remainder =561.
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