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Does the integer g have a factor f such that 1 < f < g ?

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Does the integer g have a factor f such that 1 < f < g ?  [#permalink]

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New post 12 Feb 2017, 06:43
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Does the integer g have a factor f such that 1 < f < g ?
1. g>3!
2. 11!+11>=g>=11!+2

ans is B but why?
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Posts: 50544
Re: Does the integer g have a factor f such that 1 < f < g ?  [#permalink]

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New post 12 Feb 2017, 06:50
1
gupta87 wrote:
Does the integer g have a factor f such that 1 < f < g ?
1. g>3!
2. 11!+11>=g>=11!+2

ans is B but why?


This is a copy of bunch of questions already present on the forum:
http://gmatclub.com/forum/if-x-is-an-in ... 00670.html
http://gmatclub.com/forum/does-the-inte ... 26735.html
http://gmatclub.com/forum/for-any-integ ... 68575.html
http://gmatclub.com/forum/does-integer- ... 65983.html
http://gmatclub.com/forum/if-z-is-an-in ... 28732.html
http://gmatclub.com/forum/does-p-have-a ... 88773.html
https://gmatclub.com/forum/dose-positiv ... 90858.html

Hope this helps.
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Re: Does the integer g have a factor f such that 1 < f < g ?  [#permalink]

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New post 13 May 2018, 04:45
gupta87 wrote:
Does the integer g have a factor f such that 1 < f < g ?
1. g>3!
2. 11!+11>=g>=11!+2

ans is B but why?



This problem is essentially saying 11>=g>=2.i.e g is between 11 and 2 (inclusive).
g can take 2, 3, 4, 5, 6,7,8,9,10,11
There are integer for ex: 6 for which the above condition holds. Hence the answer is B.
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Re: Does the integer g have a factor f such that 1 < f < g ?  [#permalink]

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New post 13 May 2018, 23:46
2
gupta87 wrote:
Does the integer g have a factor f such that 1 < f < g ?
1. g>3!
2. 11!+11>=g>=11!+2

ans is B but why?


It is evident that statement 1 ALONE is not sufficient.

So, let us focus on statement 2 ALONE.
11!+11>=g>=11!+2

Essentially, what the statement says is that 'g' is an integer that lies between (11! + 2) and (11! + 11). 'g' takes 10 different values.
What we have to determine is whether all of the values of 'g' have at least 1 factor other than 1 and g.

11! is divisible by all positive integers from 2 to 11.

So, if g = (11! + 2), because 11! is divisible by 2 and 2 is divisible by 2, (11! + 2) will be divisible by 2. So, we can infer that there exists a factor f for g such that 1 < f < g. In this case f = 2.
We can extend the same argument for all numbers from (11! + 3) to (11! + 11).
So, we can conclude that there will exist at least one factor f, such that 1 < f < g if g lies between (11! + 2) and (11! + 11)
Statement 2 ALONE is sufficient.
Choice B
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Re: Does the integer g have a factor f such that 1 < f < g ? &nbs [#permalink] 13 May 2018, 23:46
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