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# Remainders and Functions | AIME I, 1994 | Question 7

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

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

But try the problem first…

Source

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.

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