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

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Math Expert
Joined: 02 Sep 2009
Posts: 58443
If x is a positive integer such that (x-1)(x-3)(x-5)....(x-93) < 0, ho  [#permalink]

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29 Jan 2016, 03:14
1
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Difficulty:

95% (hard)

Question Stats:

40% (02:25) correct 60% (01:44) wrong based on 172 sessions

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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

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Math Expert
Joined: 02 Aug 2009
Posts: 7995
Re: If x is a positive integer such that (x-1)(x-3)(x-5)....(x-93) < 0, ho  [#permalink]

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29 Jan 2016, 22:46
2
1
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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Joined: 12 Mar 2015
Posts: 82
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WE: Information Technology (Computer Software)
Re: If x is a positive integer such that (x-1)(x-3)(x-5)....(x-93) < 0, ho  [#permalink]

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30 Jan 2016, 07:11
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

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
Intern
Joined: 23 Jun 2017
Posts: 10
Re: If x is a positive integer such that (x-1)(x-3)(x-5)....(x-93) < 0, ho  [#permalink]

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25 Jul 2017, 05:09
2
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
Intern
Joined: 23 Aug 2018
Posts: 2
If x is a positive integer such that (x-1)(x-3)(x-5)....(x-93) < 0, ho  [#permalink]

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26 Aug 2018, 04:06
1
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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If x is a positive integer such that (x-1)(x-3)(x-5)....(x-93) < 0, ho   [#permalink] 26 Aug 2018, 04:06
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