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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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12 Feb 2017, 07:43
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45% (medium)

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54% (01:12) correct 46% (01:53) wrong based on 101 sessions

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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
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Joined: 02 Sep 2009
Posts: 58427
Re: Does the integer g have a factor f such that 1 < f < g ?  [#permalink]

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12 Feb 2017, 07:50
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Manager
Joined: 08 Nov 2015
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Re: Does the integer g have a factor f such that 1 < f < g ?  [#permalink]

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13 May 2018, 05:45
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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14 May 2018, 00:46
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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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Does the integer g have a factor f such that 1 < f < g ?  [#permalink]

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18 Aug 2019, 17:50
1
The question is really asking whether g is prime.

If g is prime then it will only have factors g and 1. If g is not prime then it will have factors besides g and 1.

Statement 1 is insufficient as there are multiple answers that could be prime/ non-prime.

Statement 2 is sufficient because
11!+11 and all the integers between up to 11!+2 all contain 11! which has all the factors of 11! i.e. 11x10x9x8...x3x2x1 such that adding any factor in range of 11! will only make the final number contain more factors of any given added number.

This plays out as follows:
If g=11!+11 then we can factor out 11, so g would be multiple of 11, thus not a prime;
If g=11!+10 then we can factor out 10, so g would be multiple of 10, thus not a prime;
If g=11!+9 then we can factor out 10, so g would be multiple of 9, thus not a prime;
If g=11!+8 then we can factor out 10, so g would be multiple of 8, thus not a prime;
If g=11!+7 then we can factor out 10, so g would be multiple of 7, thus not a prime;
If g=11!+6 then we can factor out 10, so g would be multiple of 6, thus not a prime;
If g=11!+5 then we can factor out 10, so g would be multiple of 5, thus not a prime;
If g=11!+4 then we can factor out 10, so g would be multiple of 4, thus not a prime;
If g=11!+3 then we can factor out 10, so g would be multiple of 3, thus not a prime;
If g=11!+2 then we can factor out 10, so g would be multiple of 2, thus not a prime;

Thus all numbers we can factor out contain a factor other than g and 1; thus g is not prime.

Excellent learnings from Brian, here: https://gmatclub.com/forum/if-x-is-an-i ... ml#p777801

Thus, statement 2 tells us g has more than g and 1 as its factors. Thus statement 2 is sufficient.
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Does the integer g have a factor f such that 1 < f < g ?  [#permalink]

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18 Aug 2019, 19:07
gupta87 wrote:
Does the integer g have a factor f such that 1 < f < g ?

(1) g > 3!
(2) 11! + 11 >= g >= 11! + 2

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

(1) g > 3!
g>6
g= 7 NO
g=8 YES
NOT SUFFICIENT

(2) 11! + 11 >= g >= 11! + 2
If g= 11! + k where $$2 \leq k \leq 11$$
Then g will be divisible by k
SUFFICIENT

IMO B

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Does the integer g have a factor f such that 1 < f < g ?   [#permalink] 18 Aug 2019, 19:07
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