PK 2015 Q24

It is certainly possible to underline two numbers. In the list below both 1 and −1 are equal to the product of the other nine numbers \(1, -1, -2, \tfrac{1}{2}, 3, \tfrac{1}{3}, 4, \tfrac{1}{4}, 5, \tfrac{1}{5}\).
However, it is not possible to underline three numbers. For supposing it was possible to underline the numbers \(a\), \(b\), \(c\). Let \(N\) be the product of the other seven numbers. Then we have
\[a = b \times c \times N \ldots (1) \qquad b = a \times c \times N \ldots (2) \qquad c = a \times b \times N \ldots (3)\]
Substituting (1) into (2) gives \(b = (b \times c \times N) \times c \times N = b \times c^2 \times N^2\) so \(c^2 \times N^2 = 1\). Thus \(c \times N = 1\) or \(c \times N = -1\). But if \(c \times N = 1\), then (1) becomes \(a = b\), contradicting that \(a\), \(b\) are distinct. Hence \(c \times N = -1\), and (1) becomes \(a = -b\). Substituting this into (3) gives \(c = (-b) \times b \times N\). Multiplying both sides by \(c\) gives \(c^2 = -b^2 \times c \times N = -b^2 \times (-1) = b^2\). Hence either \(c = b\) (contradicting that they are distinct) or \(c = -b = a\) (contradicting that \(a\), \(c\) are distinct).
Hence there is no way that \(a\), \(b\), \(c\) can be distinct numbers and also underlined. (If the numbers are allowed to be the same, it is possible to underline them all by choosing the numbers to be all equal to one).