HMMT 二月 2006 · 冲刺赛 · 第 21 题

HMMT February 2006 — Guts Round — Problem 21

[8] Find the smallest positive integer k such that z + z + z + z + z + z + 1 divides z − 1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . th IX HARVARD-MIT MATHEMATICS TOURNAMENT, 25 FEBRUARY 2006 — GUTS ROUND

解析

Find the smallest positive integer k such that z + z + z + z + z + z + 1 divides k z − 1. Answer: 84 Solution: Let Q ( z ) denote the polynomial divisor. We need that the roots of Q are 7 k -th roots of unity. With this in mind, we might observe that solutions to z = 1 and z 6 = 1 are roots of Q , which leads to its factorization. Alternatively, we note that 11 9 7 4 2 4 2 7 ( z − 1) Q ( z ) = z − z + z − z + z − 1 = ( z − z + 1)( z − 1) √ 1+ i 3 2 Solving for the roots of the first factor, z = = ± cis π/ 3 (we use the notation 2 cis( x ) = cos( x ) + i sin( x )) so that z = ± cis( ± π/ 6). These are primitive 12-th roots of unity. The other roots of Q ( z ) are the primitive 7-th roots of unity (we introduced z = 1 by multiplication.) It follows that the answer is lcm[12 , 7] = 84. 6