Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1994 AIME Problems/Problem 3"

(→‎Solution 2)
(→‎Solution 2)
Line 15: Line 15:
 
<cmath>T_{n-1} + T_n = n^2,</cmath>
 
<cmath>T_{n-1} + T_n = n^2,</cmath>
 
where <math>T_n = 1+2+...+n = \frac{n(n+1)}{2}</math> is the <math>n</math>th triangular number.
 
where <math>T_n = 1+2+...+n = \frac{n(n+1)}{2}</math> is the <math>n</math>th triangular number.
 +
 +
Using this, as well as using the fact that the value of <math>f(x)</math> directly determines the value of <math>f(x+1)</math> and <math>f(x-1),</math> we conclude that <math>f(n) = T_n + K</math> for all odd <math>n</math> and <math>f(n) = T_n - K</math> for all even <math>n,</math> where <math>K</math> is a constant real number.
 +
 +
Since <math>f(19) = 94</math> and <math>T_19 = 190,</math> we see that <math>K = -96.</math> It follows that <math>f(94) = T_94 - (-96) = \frac{94\cdot 95}{2} + 96 = 4561,</math> so the answer is <math>561.</math>
  
 
== See also ==
 
== See also ==

Revision as of 01:47, 6 July 2016

Problem

The function $f_{}^{}$ has the property that, for each real number $x,\,$

$f(x)+f(x-1) = x^2\,$

.

If $f(19)=94,\,$ what is the remainder when $f(94)\,$ is divided by $1000$?

Solution 1

\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*}

So, the remainder is $\boxed{561}$.

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, \[T_{n-1} + T_n = n^2,\] where $T_n = 1+2+...+n = \frac{n(n+1)}{2}$ is the $n$th triangular number.

Using this, as well as using the fact that the value of $f(x)$ directly determines the value of $f(x+1)$ and $f(x-1),$ we conclude that $f(n) = T_n + K$ for all odd $n$ and $f(n) = T_n - K$ for all even $n,$ where $K$ is a constant real number.

Since $f(19) = 94$ and $T_19 = 190,$ we see that $K = -96.$ It follows that $f(94) = T_94 - (-96) = \frac{94\cdot 95}{2} + 96 = 4561,$ so the answer 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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1994_AIME_Problems/Problem_3&oldid=79191"