Maths Olympiad Prep

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Combinatorics Difficulty 5.3 AIME, harder Prove it Soviet Union

Problem:
What is the largest possible value of a1a2a3a1990|\ldots |a_1 - a_2| - a_3| - \ldots - a_{1990}|, where a1,a2,,a1990a_1, a_2, \ldots, a_{1990} is a permutation of 1,2,3,,19901, 2, 3, \ldots, 1990?

Solution

Solution:
Answer 19891989

Since abmax(a,b)|a - b| \leq \max(a, b), a trivial induction shows that the expression does not exceed max(a1,a2,,a1990)=1990\max(a_1, a_2, \ldots, a_{1990}) = 1990. But for integers, ab|a - b| has the same parity as a+ba + b, so a trivial induction shows that the expression has the same parity as a1+a2++a1990=19901991/2a_1 + a_2 + \ldots + a_{1990} = 1990 \cdot 1991 / 2, which is odd. So it cannot exceed 19891989. That can be attained by the permutation 2,4,5,3,6,8,9,7,,4k+2,4k+4,4k+5,4k+3,,1984+2,1984+4,1984+5,1984+3,1990,12, 4, 5, 3, 6, 8, 9, 7, \ldots, 4k + 2, 4k + 4, 4k + 5, 4k + 3, \ldots, 1984 + 2, 1984 + 4, 1984 + 5, 1984 + 3, 1990, 1. Because we get successively 2,3,0;6,2,7,0;10,2,11,0;;4k+2,2,4k+3,0;;1986,2,1987,0;1990,19892, 3, 0; 6, 2, 7, 0; 10, 2, 11, 0; \ldots; 4k + 2, 2, 4k + 3, 0; \ldots; 1986, 2, 1987, 0; 1990, 1989.

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