Difference between revisions of "1994 AIME Problems/Problem 3"
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If <math>f(19)=94,\,</math> what is the remainder when <math>f(94)\,</math> is divided by <math>1000</math>? | If <math>f(19)=94,\,</math> what is the remainder when <math>f(94)\,</math> is divided by <math>1000</math>? | ||
| − | == Solution == | + | == Solution 1 == |
<cmath>\begin{align*}f(94)&=94^2-f(93)=94^2-93^2+f(92)=94^2-93^2+92^2-f(91)=\cdots \\ | <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 \\ | &= (94^2-93^2) + (92^2-91^2) +\cdots+ (22^2-21^2)+ 20^2-f(19) \\ &= 94+93+\cdots+21+400-94 \\ | ||
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So, the remainder is <math>\boxed{561}</math>. | So, the remainder is <math>\boxed{561}</math>. | ||
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| + | == Solution 2 == | ||
| + | Those familiar with triangular numbers and some of their properties will quickly recognize the equation given in the problem. It is well-known (and easy to show) that the sum of two consecutive triangular numbers is a perfect square; that is, | ||
| + | <math></math>T_{n-1} + T_n = n^2,<math> | ||
| + | where </math>T_n = 1+2+...+n = \frac{n(n+1)}{2}<math> is the </math>n$th triangular number. | ||
== See also == | == See also == | ||
Revision as of 01:42, 6 July 2016
Contents
Problem
The function 
has the property that, for each real number 
.
If 
what is the remainder when 
is divided by 
?
Solution 1
So, the remainder is 
.
Solution 2
Those familiar with triangular numbers and some of their properties will quickly recognize the equation given in the problem. It is well-known (and easy to show) that the sum of two consecutive triangular numbers is a perfect square; that is,
$$ (Error compiling LaTeX. Unknown error_msg)T_{n-1} + T_n = n^2,
T_n = 1+2+...+n = \frac{n(n+1)}{2}
n$th triangular number.
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
