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∑n​xk​∣∣∣∣∣∣​≤k=1∑n​∣xk​∣<n(1)
since ∣∑k=1n​xk​∣≥0 and ∣xk​∣<1 for 1≤k≤n. Since it is given that
k=1∑n​∣xk​∣−∣∣∣∣∣∣​k=1∑n​xk​∣∣∣∣∣∣​=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