1988 AIME Problems/Problem 4
Problem
Suppose that 
for 
. Suppose further that
What is the smallest possible value of 
?
Solution
Since then
![\[|x_1| + |x_2| + \dots + |x_n| = 19 + |x_1 + x_2 + \dots + x_n| < n.\]](http://latex.artofproblemsolving.com/4/7/c/47c7c53b12e318497f61229adebb8de89eecc87d.png)
So 
. We now just need to find an example where 
: suppose 
and 
; then on the left hand side we have 
. On the right hand side, we have 
, and so the equation can hold for 
.
Solution 2 (Motivating solution)
First off, one can test 
and find that the 
is 
and the RHS is 
Similarly testing 
yields 
on the LHS and 
on the RHS. It seems for every negative we gain twice of that negative on the LHS. However, when we test something like 
we find the LHS to be 
and the RHS to be What happened? There were more negatives than positives. Why does this mean that the LHS doesn't grow? There aren't enough positives to "cancel out!" Therefore if for every negative we need a positive for it to cancel out to grow. We can make the grow by approximately 
every time we choose a negative and a positive, so if we have 
numbers, namely 
positive and negative we can obtain the desired answer.
Solution 3
Straight off, we notice that the RHS must be greater than or equal to 19, because an absolute value only gives nonnegative values. It then becomes clear that 
. If each 
were equal to 1, then 
. However, 
, so there must be at least one extra term to satisfy the inequality. Therefore, 
.
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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