If x is a positive integer such that (x-1)(x-3)(x-5)....(x-93)

Question

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

chetan2u · Accepted Answer

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

hwgmat2015 · Answer

total number of terms = 47
So even terms = 93-47=46

We cannot subtract odd terms as answer will be Zero

out of 46 even terms we have to find out terms for which answer is less than Zero

For even term 2 the overall answer will be positive as we will have 46 negative terms . 
But for even term 4 we get 45 negative terms. So we now have a pattern (4,8,12,16..92) i.e. multiples of 4

So 23 terms.

IMO: C

Varun121595 · Answer

Let me explain this in much more easier way
We do this question by plugging method
Here no. of terms are 47
X is a positive integer so,
Let x=1 
so 0 is not less than 0
let x=2 
1* rest negative terms are even so result will be >0
Let x=3
2*0*....=0
Let x=4 
3*1* rest 45 terms are negative so answer is <0
therefore this is true for x=4 and its multiples 
until 92

GMATTaker99887 · Answer

If x is a positive integer such that (x-1)(x-3)(x-5)....(x-93) < 0, how many values can x take

If x >= 94, then the overall product is positive. If x is an odd number less than 94, then the overall product is 0. Therefore, given that x must be a non-odd positive integer less than 94, x must be an even number.

If x = 2, then (x-1) is positive and the other factors are negative, so what matters is how many remaining factors e.g. (x - 3), (x - 5) are negative.

2n - 1 = 93
2n = 94
n = 47

There are 47 factors. If (x - 1) is positive, then 47 - 1 = 46 factors are negative, and when you multiply an even number of negative numbers with a positive number, you get a positive number as a result. Therefore, 2 can't be a solution.

For the next even number, 4, (x-1) and (x-3) are positive, leaving 47 - 2 = 45 negative factors. Since an odd number of negative numbers multiplied with a positive number results in a negative number, 4 is a solution.

It's clear that for each even x, the left hand remains positive while the product of the remaining factors alternates between positive and negative. Therefore, half of the even numbers below 47 are solutions. (47-1)/2 = 23.

There are 23 solutions.

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SaquibHGMATWhiz · Answer

Solution: METHOD 1 (logic) Firstly, value of x has to be less than 93 . Because if lets say x=94 , then all the terms x-1, x-3, x-5, ...., x-93 will be positive and their product will also be positive Secondly, values of x cannot be odd . Because if lets say x=5 , then x-5=0 and the whole product will become 0 So, x is even less than 93 but not all even numbers x cannot be even numbers like 2, 6, 10 etc because these would make odd numbers of terms negative and thus the whole product positive Since there are a total 47 (odd) number of terms, therefore for the product to be negative, we will need even number of terms to be negative The possible values of x are 4, 8, 12....92 which is an AP We can write 92=4+(n-1)4 or n=23 METHOD 2 (methodical and number line) Upon plugging the zero points and marking positive and negative ranges alternately, we get negative.png We can clearly see that the possible values of x are 4, 8, 12....92 which is an AP We can write 92=4+(n-1)4 or n=23 Hence the right answer is Option C

Kinshook · Answer

If x is a positive integer such that (x-1)(x-3)(x-5)....(x-93) < 0, how many values can x take?

(x-1)(x-3)(x-5)....(x-93)
Total terms = (93-1)2 + 1 = 47

x = 2; x-1 > 0 ; other 46 terms < 0; (x-1)...(x-93) > 0
x = 4; (x-1)(x-3)> 0; other 45 terms <0; (x-1)...(x-93) < 0
x = 6; (x-1)(x-3)(x-5) > 0; other 44 terms <0 ; (x-1)...(x-93) > 0

For x = 4, 8, 12.... 92, (x-1)...(x-93) < 0
Number of values x can take = (92-4)/4 + 1 = 23

IMO C

If x is a positive integer such that (x-1)(x-3)(x-5)....(x-93) < 0, ho

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