Problem:
The function f has the property that, for each real number x,
f(x)+f(x−1)=x2
If f(19)=94, what is the remainder when f(94) is divided by 1000?
Solution:
Using f(x)=x2−f(x−1) repeatedly, we have
f(94)​=942−f(93)=942−932+f(92)=942−932+922−f(91)⋮=942−932+922−⋯+202−f(19)=(94+93)(94−93)+(92+91)(92−91)+⋯+(22+21)(22−21)+202−94=(94+93+92+⋯+21)+306=294+21​⋅74+306=4561​
Thus, when f(94) is divided by 1000, the remainder is 561​.
The problems on this page are the property of the MAA's American Mathematics Competitions