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

(Solution 1)
(Solution 2)
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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 01:23, 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 (???dubious)

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 $n > 19$.

See also

1988 AIME (ProblemsAnswer KeyResources)
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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