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

Revision as of 21:20, 15 February 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 $n = 020$ .

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 18: / Line 18: @@
 ===Solution 2===
-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>.
 == 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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