PUMaC 2019 · 团队赛 · 第 7 题

PUMaC 2019 — Team Round — Problem 7

For all sets A of complex numbers, let P ( A ) be the product of the elements of A . Let S = z { } 1 1 1 , 2 , 9 , 99 , 999 , , , let T be the set of nonempty subsets of S (including S ), and let 2 z z z z z ∑ f ( z ) = 1 + P ( s ). Suppose f ( z ) = 6125000 for some complex number z . Compute the s ∈ T z product of all possible values of z . 2

解析

For all sets A of complex numbers, let P ( A ) be the product of the elements of A . Let S = z { } 1 1 1 , 2 , 9 , 99 , 999 , , , let T be the set of nonempty subsets of S (including S ), and let 2 z z z z z ∑ f ( z ) = 1 + P ( s ). Suppose f ( z ) = 6125000 for some complex number z . Compute the s ∈ T z product of all possible values of z . Proposed by: Kapil Chandran Answer: 48 n ∏ Fact: if A = { a , . . . , a } , then f ( A ) = ( a + 1). This is easily checked by expanding and 1 n i i =1 ∏ observing that the monomial a corresponds precisely and uniquely to the subset { a | i ∈ i i i ∈ I I } ⊂ A for any I ⊂ { 1 , . . . , n } . 1 2 2 Write y = . We wish to solve 6125000 = 2 · 3 · 10 · 100 · 1000( y +1)( y +1) = 6000000( y +1)( y + z 6125000 − 6000000 1 1). By Vieta’s, the product of all possible values of y is = , so reciprocating 6000000 48 gives 48 as the final answer. 2