Difference between revisions of "1994 AIME Problems/Problem 3"
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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. | 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> | + | 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 02:47, 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, where is the th triangular number.
Using this, as well as using the fact that the value of directly determines the value of and we conclude that for all odd and for all even where is a constant real number.
Since and we see that It follows that so the answer 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.