Difference between revisions of "1988 AIME Problems/Problem 4"
(→Solution) |
(→Solution 2) |
||
Line 18: | Line 18: | ||
===Solution 2=== | ===Solution 2=== | ||
− | Let <math>|x_1 + x_2 + \dots + x_n| = 0</math> and <math>|x_1| + |x_2| + \dots + |x_n| = 19</math> | + | Let <math>|x_1 + x_2 + \dots + x_n| = 0</math> and <math>|x_1| + |x_2| + \dots + |x_n| = 19</math>. Then the smallest value of <math>n = \boxed {20}</math> because <math>|x_i| < 1</math>, and therefore <math>n > 19</math>. |
== See also == | == See also == |
Revision as of 21:20, 15 February 2014
Contents
[hide]Problem
Suppose that for . Suppose further that What is the smallest possible value of ?
Solution
Solution 1
Since then
So . We now just need to find an example where : suppose and ; then on the left hand side we have . On the right hand side, we have , and so the equation can hold for .
Solution 2
Let and . Then the smallest value of because , and therefore .
See also
1988 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
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.