If x is a positive integer such that (x-1)(x-3)(x-5)....(x-93) < 0, how many values can x take?
Pick one:
A) 47	B) 23	C) 46	D) 21	E) 22
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) &gt; 0; remaining 46 terms &lt; 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 &lt; 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 &gt; 0. So, x = 6 does not satisfy the condition.   Let x = 8; first 4 terms positive; remaining 43 terms &lt; 0. Their product &lt; 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   Answer 23
<result>B</result>
If x is a positive integer such that (x-1)(x-3)(x-5)....(x-93) < 0, how many values can x take?