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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.

## 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

Integers

Remainder

Functions

## Check the Answer

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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