Difference between revisions of "1994 AIME Problems/Problem 3"
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== Solution 2 == | == 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, | 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, | ||
− | < | + | <cmath>T_{n-1} + T_n = n^2,</cmath> |
− | where < | + | where <math>T_n = 1+2+...+n = \frac{n(n+1)}{2}</math> is the <math>n</math>th triangular number. |
== See also == | == See also == |
Revision as of 02:43, 6 July 2016
Contents
Problem
The function has the property that, for each real number
![$f(x)+f(x-1) = x^2\,$](http://latex.artofproblemsolving.com/6/7/8/67827ba82e465559d945580a479bca7bcce2fdd9.png)
.
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,
where
is the
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.