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1988 AIME Problems/Problem 4

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Problem

Suppose that for . 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 ?

Solution

Since then

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

So . We now just need to find an example where : suppose and ; 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 .

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
NEW! Hard Problems DVD
A documentary about the 2006 US IMO team. Features many current and past AoPS members!
Click here for more details and to order
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