SOLUTION: You play the following game with your worst enemy, who wishes to keep your score down. A referee gets to pick a positive integer n. Then your enemy gets to pick n different real n

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Question 935840: You play the following game with your worst enemy, who wishes to keep your score down. A
referee gets to pick a positive integer n. Then your enemy gets to pick n different real numbers.
You get to multiply different pairs of these numbers together, as many different pairs as you
want. You get one point for each pair whose product is nonnegative. So for instance, if n = 3
and your enemy picks 0, −8 and −π, you would pick all possible pairs, {0, −8}, {0, −π} and
{−8, −π}, because all 3 products (0, 0, 8π) are nonnegative. You would get 3 points.
Determine with proof the highest score you can be sure to get against a smart enemy. Your
answer will depend on n.

Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
The product of a pair of numbers with the same sign (both positive or both negative) is positive.
The product of zero and any number is zero, which is non-negative.
A smart enemy will not pick zero, because that will give you a lot of non-negative products.
He/she will have to pick positive and negative numbers.
If n is an even number, n=2k ,
a smart enemy would pick k negative numbers and k positive numbers would give you k%28k-1%29=highlight%28k%5E2-k%29 positive products,
from k%28k-1%29%2F2 pairs of positive numbers,
and k%28k-1%29%2F2 pairs of positive numbers.
Splitting the negative and positive choices unevenly,
let's say k%2Ba positive numbers and k-a negative numbers, with a%3C=k,
would give you %28k%2Ba%29%28k%2Ba-1%29%2F2 positive products,
and %28k-a%29%28k-a-1%29%2F2 negative products,
for a total of k%5E2-k%2Ba%5E2,
so a smart enemy would not make such a choice.
If one more number could be picked, n=2k%2B1 ,
a smart enemy would choose k positive and k negative numbers,
plus an additional number.
The additional number could be positive or negative,
and paired with the k numbers of the same sign,
would give you k additional positive products,
for a total of k%5E2-k%2Bk=highlight%28k%5E2%29 points.
The formulas work even in the case of
n=1=2%2A0%2B1 , where k=0 and your score would be k%5E2=0%5E2=0 ,
and n=2=2%2A1 , where k=1 and your score would be k%5E2k=1%5E2-1=1-1=0 .