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1988 AIME Problems/Problem 4

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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

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).... 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.

See also

1988 AIME (ProblemsResources)
Preceded by
Problem 3
Followed by
Problem 5
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
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