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Algebra Difficulty 3.3 AMC 10/12 Find the answer

For how many positive integers xx is (x2)(x4)(x6)(x2016)(x2018)0(x-2)(x-4)(x-6) \cdots(x-2016)(x-2018) \leq 0?

A number or a short expression. Spacing and $ signs are ignored.

Solution

We count the positive integers xx for which the product (x2)(x4)(x6)(x2016)(x2018)(x-2)(x-4)(x-6) \cdots(x-2016)(x-2018) equals 0 and is less than 0 separately. The product equals 0 exactly when one of the factors equals 0. This occurs exactly when xx equals one of 2,4,6,,2016,20182,4,6, \ldots, 2016,2018. These are the even integers from 2 to 2018, inclusive, and there are 20182=1009\frac{2018}{2}=1009 such integers. The product is less than 0 exactly when none of its factors is 0 and an odd number of its factors are negative. We note further that for every integer xx we have x2>x4>x6>>x2016>x2018x-2>x-4>x-6>\cdots>x-2016>x-2018. When x=1x=1, we have x2=1x-2=-1 and so all 1009 factors are negative, making the product negative. When x=3x=3, we have x2=1,x4=1x-2=1, x-4=-1 and all of the other factors are negative, giving 1008 negative factors and so a positive product. When x=5x=5, we have x2=3,x4=1x-2=3, x-4=1 and x6=1x-6=-1 and all of the other factors are negative, giving 1007 negative factors and so a negative product. This pattern continues giving a negative value for the product for x=1,5,9,13,,2013,2017x=1,5,9,13, \ldots, 2013,2017. There are 1+201714=5051+\frac{2017-1}{4}=505 such values (starting at 1, these occur every 4 integers). When x2019x \geq 2019, each factor is positive and so the product is positive. Therefore, there are 1009+505=15141009+505=1514 positive integers xx for which the product is less than or equal to 0.

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Source: Omni-MATH, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.