Difference between revisions of "1988 AIME Problems/Problem 4"
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Since <math>|x_i| < 1</math> then | 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> | + | <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\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>. |
Latest revision as of 16:11, 19 December 2018
Problem
Suppose that for . Suppose further that What is the smallest possible value of ?
Solution
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 .
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 |
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