Difference between revisions of "1994 AIME Problems/Problem 3"
m (minor fmt) |
|||
| Line 2: | Line 2: | ||
The function <math>f_{}^{}</math> has the property that, for each real number <math>x,\,</math> | The function <math>f_{}^{}</math> has the property that, for each real number <math>x,\,</math> | ||
<center><math>f(x)+f(x-1) = x^2\,</math></center>. | <center><math>f(x)+f(x-1) = x^2\,</math></center>. | ||
| − | If <math>f(19)=94,\,</math> what is the remainder when <math>f(94)\,</math> is divided by 1000? | + | If <math>f(19)=94,\,</math> what is the remainder when <math>f(94)\,</math> is divided by <math>1000</math>? |
== Solution == | == Solution == | ||
| − | < | + | <cmath>\begin{align*}f(94)&=94^2-f(93)=94^2-93^2+f(92)=94^2-93^2+92^2-f(91)=\cdots \\ |
| + | &= (94^2-93^2) + (92^2-91^2) +\cdots+ (22^2-21^2)+ 20^2-f(19) \\ &= 94+93+\cdots+21+400-94 \\ | ||
| + | &= 4561 \end{align*}</cmath> | ||
| − | <math> | + | So, the remainder is <math>\boxed{561}</math>. |
| − | |||
| − | |||
| − | |||
| − | |||
== See also == | == See also == | ||
{{AIME box|year=1994|num-b=2|num-a=4}} | {{AIME box|year=1994|num-b=2|num-a=4}} | ||
| + | |||
| + | [[Category:Intermediate Algebra Problems]] | ||
Revision as of 22:16, 8 October 2008
Problem
The function 
has the property that, for each real number 
.
If 
what is the remainder when 
is divided by 
?
Solution
So, the remainder is 
.
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 | ||
