Difference between revisions of "1988 AIME Problems/Problem 4"

Revision as of 00:22, 6 June 2014

Problem

Suppose that $|x_i| < 1$ for $i = 1, 2, \dots, n$ . Suppose further that $|x_1| + |x_2| + \dots + |x_n| = 19 + |x_1 + x_2 + \dots + x_n|.$ What is the smallest possible value of $n$ ?

Solution

Solution 1

Since $|x_i| < 1$ then

$|x_1| + |x_2| + \dots + |x_n| = 19 + |x_1 + x_2 + \dots + x_n| < n$

So $n \ge 20$ . We now just need to find an example where $n = 20$ : suppose $x_{2k-1} = \frac{19}{20}$ and $x_{2k} = -\frac{19}{20}$ ; then on the left hand side we have $\left|\frac{19}{20}\right| + \left|-\frac{19}{20}\right| + \dots + \left|-\frac{19}{20}\right| = 20\left(\frac{19}{20}\right) = 19$ . On the right hand side, we have $19 + \left|\frac{19}{20} - \frac{19}{20} + \dots - \frac{19}{20}\right| = 19 + 0 = 19$ , and so the equation can hold for $n = \boxed{020}$ .

Solution 2

Let $|x_1 + x_2 + \dots + x_n| = 0$ and $|x_1| + |x_2| + \dots + |x_n| = 19$ . Then the smallest value of $n = \boxed {20}$ because $|x_i| < 1$ , and therefore $n > 19$ .

@@ Line 12: / Line 12: @@
 <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 = 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===

1988 AIME (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
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