m is non-integer -> m =a*b where a and b are primes in which one of them is not divisible by 11!. We must minimize a,b -> a=2, b=13 -> m=26.

25 also a non prime and neither hidden in n?

25 can be factor from 10 and 5

I didn't get it

let's take it the long way first so you could see it clearly

11! = 11*10*9*8*7*6*5*4*3*2*1

list of all non-prime number

1-11 are very obvious
12 - 3*4
14 - 7*2
15 - 5*3
16 - 8*2
18 - 9*2
20 - 10*2
21 - 7*3
22 - 11*2
24 - 8*3
25 - 5*5(factor out from 10 = 5*2)
26 - 2*13

since we do not have any factor of 13 in 11! 26 is not a factor of 11!

GIVEN: x = (11)(10)(9)(8)(7)(6)(5)(4)(3)(2)(1)

We can clearly see that x is divisible by 11, 10, 9, . . . 3, 2 and 1

So, let's start checking integers that are greater than 11

Is 12 a factor of x?
x = (11)(10)(9)(8)(7)(6)(5)(4)(3)(2)(1)
= (11)(10)(9)(8)(7)(12)(5)(4)(3)(1)
12 is clearly a factor of x

13 is prime, so we can skip that.

Is 14 a factor of x?
x = (11)(10)(9)(8)(7)(6)(5)(4)(3)(2)(1)
= (11)(10)(9)(8)(14)(6)(5)(4)(3)(1)
14 is clearly a factor of x

At this point, we can see the pattern.

15 = (5)(3). Since 5 and 3 are both in the product 11!, we know that 15 is a factor of x
16 = (8)(2). Since 8 and 2 are both in the product 11!, we know that 16 is a factor of x
17 is prime - skip
18 = (6)(3)
19 is prime - skip
20 = (4)(5)
21 = (7)(3)
22 = (11)(2)
23 is prime - skip
24 = (3)(8)
25. (5)(5) [aside: one 5 is hiding in the number 10]

26 = (2)(13) HOLD ON! There is no 13 hiding in the product 11!
So, 26 cannot be a factor of 11!

Answer: 26

Cheers,
Brent

The prime factorization of 11! = 2^7 * 3^4 *5 *7 *11. From this we see that 13 which is the next prime number is not a factor, multiply by 2 to give 26.

is there any other way of solving this problem aside from listing all the numbers from 13-26 and then checking ?

In this case no. Sometimes is not possible if not to inspect the question fully.

However, this is a question for practice at the highest level of the GRE. Difficult to imagine a question like this on the real GRE. Could be