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sowjanyaangara
Prospective Students
 
Posts: 10
Joined: Sat Mar 02, 2013 3:38 am
 

number properties, 3 edition ,chapter 10 ,#6, page 179

by sowjanyaangara Tue Mar 05, 2013 4:40 pm

6) what is the smallest positive integer that is not prime and is not factor of 11!

when come to numbers 25 and 26. Prime factorization of 25(5*5), 26(13*2). For number 25 (5*5), 5 appeared only once in 11! .what about other 5? will 25 consider as smallest possible value?. For number 26 (13*2), 2 appeared in 11!. so how could i differentiate between number 25 and 26. please explain.

Thanks
tommywallach
Manhattan Prep Staff
 
Posts: 1917
Joined: Thu Mar 31, 2011 11:18 am
 

Re: number properties, 3 edition ,chapter 10 ,#6, page 179

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