Difference between revisions of "1988 AIME Problems/Problem 4"
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So <math>n \ge 20</math>. We now just need to find an example where <math>n = 20</math>: suppose <math>x_{2k-1} = \frac{19}{20}</math> and <math>x_{2k} = -\frac{19}{20}</math>; then on the left hand side we have <math>\left|\frac{19}{20}\right| + \left|-\frac{19}{20}\right| + \dots + \left|-\frac{19}{20}\right| = 20\left(\frac{19}{20}\right) = 19</math>. On the right hand side, we have <math>19 + \left|\frac{19}{20} - \frac{19}{20} + \dots - \frac{19}{20}\right| = 19 + 0 = 19</math>, and so the equation can hold for <math>n = \boxed{020}</math>. | So <math>n \ge 20</math>. We now just need to find an example where <math>n = 20</math>: suppose <math>x_{2k-1} = \frac{19}{20}</math> and <math>x_{2k} = -\frac{19}{20}</math>; then on the left hand side we have <math>\left|\frac{19}{20}\right| + \left|-\frac{19}{20}\right| + \dots + \left|-\frac{19}{20}\right| = 20\left(\frac{19}{20}\right) = 19</math>. On the right hand side, we have <math>19 + \left|\frac{19}{20} - \frac{19}{20} + \dots - \frac{19}{20}\right| = 19 + 0 = 19</math>, and so the equation can hold for <math>n = \boxed{020}</math>. | ||
| − | ===Solution 2=== | + | ===Solution 2 (???dubious)=== |
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>. | 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>. | ||
Revision as of 00:23, 6 June 2014
Problem
Suppose that 
for 
. Suppose further that
What is the smallest possible value of 
?
Solution
Solution 1
Since then
![\[|x_1| + |x_2| + \dots + |x_n| = 19 + |x_1 + x_2 + \dots + x_n| < n\]](http://latex.artofproblemsolving.com/d/c/7/dc78223b6120412e17e5d89e5fe59d41f5d758cc.png)
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 (???dubious)
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. 
