0 0 votes Each of the numbers $x_1, x_2,\dots, x_n, n > 4,$ is equal to $1$ or $–1.$ Suppose, $x_1x_2x_3x_4 + x_2x_3x_4x_5 + x_3x_4x_5x_6 + \dots + x_{n–3}x_{n–2}x_{n–1}x_n + x_{n–2}x_{n–1}x_nx_1+ x_{n–1}x_nx_1x_2 + x_nx_1x_2x_3= 0$, then, $n$ is even. $n$ is odd. $n$ is an odd multiple of $3.$ $n$ is prime Quantitative Aptitude cat2000 quantitative-aptitude algebra + – go_editor 14.2k points 1.2k views answer comment Share Follow Print 0 reply Please log in or register to add a comment.