by tommywallach Tue Mar 05, 2013 8:41 pm
Hey Sowjan,
This is a rule you must learn:
For X to be a factor of Y, Y must contain all of X's factors.
We need a number that is not a factor of 11!. For that to be the case, we need a number that has factors which 11! does not have.
11! = 11 * 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2
But we can break this down further, because:
10 = 2 * 5
9 = 3 * 3
8 = 2 * 2 * 2
6 = 2 * 3
4 = 2 * 2
That gives us an extra 3^3, 2^7, and 5.
So 11! = 11 * 7 * 5^2 * 3^4 * 2^8
So we need a number that has something not included in that list.
Obviously, everything from 1-11 is a factor of 11!
12 = 2 * 2 * 3 (11! has all of those)
13 = prime
14 = 2 * 7 (11! has both)
15 = 3 * 5 (11! has both)
16 = 2^4 (11! has that)
17 = prime
18 = 2 * 3 * 3 (11! has those)
19 = prime
20 = 2 * 2 * 5 (11! has that)
21 = 7 * 3 (11! has that)
22 = 11 * 2 (11! has that)
23 = prime
24 = 2 * 2 * 2 * 3 (11! has those)
25 = 5 * 5 (11! has those)
26 = 13 * 2 ---> Uh-oh, 11! doesn't have a 13 in it, and this isn't prime, so it fits!
Now, you might be thinking, "Isn't that crazy hard?" It's pretty hard. I'm not sure you'd get this question on the real GRE. If you did, it would be among your harder questions. But notice that it isn't really difficult, just time-consuming.
Hope that helps!
-t