Problem:
Suppose that ∣xi∣<1 for i=1,2,…,n. Suppose further that
∣x1∣+∣x2∣+…+∣xn∣=19+∣x1+x2+…+xn∣.
What is the smallest possible value of n?
Solution:
If n is a positive integer, then
k=1∑n∣xk∣−∣∣∣∣∣∣k=1∑nxk∣∣∣∣∣∣≤k=1∑n∣xk∣<n(1)
since ∣∑k=1nxk∣≥0 and ∣xk∣<1 for 1≤k≤n. Since it is given that
k=1∑n∣xk∣−∣∣∣∣∣∣k=1∑nxk∣∣∣∣∣∣=19(2)
it follows that 19<n. Thus the answer is 20 if there is a solution to (2) with n=20. One such solution is
xk={.95−.95 if k is odd if k is even.
The problems on this page are the property of the MAA's American Mathematics Competitions