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.

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