Difference between revisions of "1988 AIME Problems/Problem 4"
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== Problem == | == Problem == | ||
| + | Suppose that <math>|x_i| < 1</math> for <math>i = 1, 2, \dots, n</math>. Suppose further that | ||
| + | <math> | ||
| + | |x_1| + |x_2| + \dots + |x_n| = 19 + |x_1 + x_2 + \dots + x_n|. | ||
| + | </math> | ||
| + | What is the smallest possible value of <math>n</math>? | ||
== Solution == | == Solution == | ||
| + | Since <math>|x_i| < 1</math> then | ||
| + | |||
| + | <cmath>|x_1| + |x_2| + \dots + |x_n| = 19 + |x_1 + x_2 + \dots + x_n| < n</cmath> | ||
| + | |||
| + | 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(\frac{19}{20}) = 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 = 020</math>. | ||
== See also == | == See also == | ||
| − | + | {{AIME box|year=1988|num-b=3|num-a=5}} | |
| − | + | [[Category:Intermediate Algebra Problems]] | |
Revision as of 17:35, 26 September 2007
Problem
Suppose that 
for 
. Suppose further that
What is the smallest possible value of 
?
Solution
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 
.
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 | ||
