Bunuel wrote:
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. 46
C. 23
D. 22
E. 21
HI,
We should always try to gather info from the Q stem within initial 15-20 seconds and then work towards a solution..
More than often the info will guide us to the solution
Lets see what all can we make of the inequality..
(x-1)(x-3)(x-5)....(x-93) < 0...
following info can be deduced..
1) there are (93-1)/2 + 1= 47 terms multiplied with each other..
2) we require odd numbers of -ive terms in these 47 terms to get our answer.
next we divide the x values in three parts..
1) x<1..
there is no value of a positive integer <1, so 0 values of x possible
2) x between 1 and 93 inclusive..
--- if x is an odd number, say 1,3,5,etc, one of the terms will become 0 and the Left Hand Side will become 0..---- if x is an even number, two scenarios.. when we take multiples of 4 as the value of x, example 4, then x-1 and x-3 become positive and rest 47-2=45 terms are -ive, so the product will be <0... But if we tke x as even numbers, not multiples of 4, say 6, x-1 ,x-3, and x-5 will become positive and rest 47-3=44 terms are negative, so the product in these cases will be >0..
3) x>93,
all 47 terms are positive, so product >0..
so we see only values of x, which make product<0, are multiples of 4 between 1 and 93..
lowest value=4... highest= 92
number of terms=92-4/4 +1 =22 +1=23
C
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