HMMT 十一月 2011 · GEN 赛 · 第 10 题

HMMT November 2011 — GEN Round — Problem 10

[ 8 ] Let r , r , . . . , r be the distinct complex roots of the polynomial P ( x ) = x − 7. Let 1 2 7 ∏ K = ( r + r ), i j 1 ≤ i<j ≤ 7 that is, the product of all numbers of the form r + r , where i and j are integers for which 1 ≤ i < j ≤ 7. i j 2 Determine the value of K .

解析

[ 8 ] Let r , r , . . . , r be the distinct complex roots of the polynomial P ( x ) = x − 7. Let 1 2 7 ∏ K = ( r + r ), i j 1 ≤ i<j ≤ 7 that is, the product of all numbers of the form r + r , where i and j are integers for which 1 ≤ i < j ≤ 7. i j 2 Determine the value of K . 7 Answer: 117649 We first note that x − 7 = ( x − r )( x − r ) · · · ( x − r ), which implies, replacing 1 2 7 7 x by − x and taking the negative of the equation, that ( x + r )( x + r ) · · · ( x + r ) = x + 7. Also note 1 2 7 that the product of the r is just the constant term, so r r · · · r = 7. i 1 2 7 General Test Now, we have that 7 ∏ 7 2 2 2 · 7 · K = ( 2 r ) K i i =1 7 ∏ ∏ 2 = 2 r ( r + r ) i i j i =1 1 ≤ i<j ≤ 7 ∏ ∏ ∏ = ( r + r ) ( r + r ) ( r + r ) i j i j i j 1 ≤ i = j ≤ 7 1 ≤ i<j ≤ 7 1 ≤ j<i ≤ 7 ∏ = ( r + r ) i j 1 ≤ i,j ≤ 7 7 7 ∏ ∏ = ( r + r ) . i j i =1 j =1 7 ∏ However, note that for any fixed i , ( r + r ) is just the result of substuting x = r into ( x + r )( x + i j i 1 j =1 r ) · · · ( x + r ). Hence, 2 7 7 ∏ 7 7 ( r + r ) = r + 7 = ( r − 7) + 14 = 14 . i j i i j =1 7 2 6 Therefore, taking the product over all i gives 14 , which yields K = 7 = 117649. General Test