Difference between revisions of "1994 AIME Problems/Problem 3"
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== Solution == | == Solution == | ||
| − | + | <math>f(94)=94^2-f(93)=94^2-93^2+f(92)=94^2-93^2+92^2-f(91)=\ldots</math> | |
| + | <math>f(94) = (94^2-93^2) + (92^2-91^2) +\ldots+ (22^2-21^2)+ 20^2-f(19) = 94+93+\ldots+21+400-94 </math> | ||
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| + | <math>f(94) = 4561</math> | ||
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| + | So, the remainder is 561. | ||
== See also == | == See also == | ||
{{AIME box|year=1994|num-b=2|num-a=4}} | {{AIME box|year=1994|num-b=2|num-a=4}} | ||
Revision as of 05:24, 31 December 2007
Problem
The function 
has the property that, for each real number 
.
If 
what is the remainder when 
is divided by 1000?
Solution
So, the remainder is 561.
See also
| 1994 AIME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 2 |
Followed by Problem 4 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
