Answer to Question #142693 in Combinatorics | Number Theory for mkami

Question

120 numbers are written on the board. It is known that among all their pairwise
products there are exactly 2000 negative numbers. What is the largest number of
zeroes that could be written on the board?

Answer 1

The number pairwise numbers from the 120 numbers is $\tbinom{120}{2}= 7140$ ways

Let $A := \text{\textbraceleft}x_i \ge 0 : 1\le i \le 100 \text{\textbraceright}$ and $B := \text{\textbraceleft}y_j < 0 : 1\le j \le 20 \text{\textbraceright}$

It is obvious that |A| + |B| = 120 and the number pairwise numbers from each of the set is $\tbinom{100}{2}+\tbinom{20}{2} = 4950 +190=5140$. When each $x_i \in A$ is used to multiply each $y_j \in B$ we are going to have $2000$ negative numbers.

Then $\tbinom{100}{2}+\tbinom{20}{2} +2000 = 7140 =\tbinom{120}{2}$. This will only hold if $0 \notin A$ .

Therefore there are no zeros to be written on the board.