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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

Check the Answer


Answer: is 561.

AIME I, 1994, Question 7

Elementary Number Theory by David Burton

Try with Hints


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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