Problem Solving Question: Polynomials / Inequalities

Q-51 Practice / Problem Solving / Algebra / Inequalities / Q1
This polynomials question combines concepts from sequences and series, and number properties. The focus is on positive and negative numbers and odd and even numbers. A formula based approach may not work for hard GMAT questions - as is evident from this question.

Question

  • PS
  • If x is a positive integer such that (x-1)(x-3)(x-5)....(x-93) < 0, how many values can x take? A. 47
    B. 23
    C. 46
    D. 21
    E. 22

 

  • Correct Answer
    Choice B. x can take 23 values

Explanatory Answer

Algebra polynomials GMAT Problem Solving question
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Detailed Solution

If x takes any value greater than 93, the expression will definitely be positive.

Therefore, the set of values that x takes should be from the set of positive integers upto 93

The expression(x-1)(x-3)(x-5)....(x-93) has a total of 47 terms.

When x = 1, 3, 5,7 ....93 the value of the express will be zero. i.e., for odd values of x, the expression will be zero.

We need to evaluate whether the expression will be negative for all even numbers upto 93

Let x = 2; first term (x-1) > 0; remaining 46 terms < 0.
The product of one positive number and 46 negative numbers will be positive (product of even number of negative terms will be positive).
So, x = 2 does not satisfy the condition.

Let x = 4; (x-1) and (x-3) are positive; the remaining 45 terms < 0.
The product of two positive numbers and 45 negative numbers will be negative. So, x = 4 satisfies the condition.

Let x = 6; first 3 terms (x-1), (x-3), and (x-5) are positive and the remaining 44 terms are negative.
Their product > 0. So, x = 6 does not satisfy the condition.

Let x = 8; first 4 terms positive; remaining 43 terms < 0. Their product < 0. x = 8 satisfies the condition.

Extrapolating what we have oberved with these 4 terms, we could see a pattern.
The expression takes negative values when x = 4, 8, 12, 16 ..... 92 (multiples of 4)

92, the last value that x can take is the 23rd multiple of 4. Hence, number of such values of x = 23