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Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1994 based on Remainders and Functions.

## Remainders and Functions – AIME I, 1994

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.

- is 107
- is 561
- is 840
- cannot be determined from the given information

**Key Concepts**

Integers

Remainder

Functions

## Check the Answer

But try the problem first…

Answer: is 561.

Source

Suggested Reading

AIME I, 1994, Question 7

Elementary Number Theory by David Burton

## Try with Hints

First hint

f(94)=\(94^{2}-f(93)=94^{2}-93^{2}+f(92)\)

=\(94^{2}-93^{2}+92^{2}-f(91)\)

Second Hint

=\((94^{2}-93^{2})+(92^{2}-91^{2})\)

\(+….+(22^{2}-21^{2})+20^{2}-f(19)\)

Final Step

=94+93+…..+21+400-94

=4561

\(\Rightarrow\) remainder =561.

## Other useful links

- https://www.cheenta.com/rational-number-and-integer-prmo-2019-question-9/
- https://www.youtube.com/watch?v=lBPFR9xequA

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