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

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.

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.

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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how can we sure that f is an integer?

In the question its not mentioned that whether f is an integer. If f is 2.987 then it will never be divisible by any given number.

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