GMAT Prep : Prime number / factor question

This topic has expert replies
Senior | Next Rank: 100 Posts
Posts: 57
Joined: Sun Oct 05, 2008 8:27 pm
GMAT Score:710

GMAT Prep : Prime number / factor question

by Skywalker » Sat Aug 28, 2010 5:41 am
If K is a +ve interger, is K the square of an interger ?

1) K is divisible by 4.
2) K is divisible by exactly 4 different prime numbers.

OA E

Note : How do I approach such problems ? What prime numbers should I pick ? Any tricks that I can use ?
StAy HuNgry sTay fOOlisH
Source: — Data Sufficiency |

User avatar
Legendary Member
Posts: 659
Joined: Mon Dec 14, 2009 8:12 am
Thanked: 32 times
Followed by:3 members

by Gurpinder » Sat Aug 28, 2010 6:32 am
Hey,

A square has pairs in its prime tree. Ex. 4= 2,2 So 4 is a perfect square.

Now (A)
K/4=int. K=4-->2,2 So K is a square. But K=8=4,2-->2,2,2 < only one pair. So its not

Therefore, this one is insufficient.

(B)

Clearly insufficient. We don't know anything about the prime numbers.

Together:

Again insufficient.

Multiple of 4 that is divisible by 4 different primes.

Again, square has pair primes. 4 DIFFERENT primes = not pairs.


So (E).

I hope this makes sense.
"Do not confuse motion and progress. A rocking horse keeps moving but does not make any progress."
- Alfred A. Montapert, Philosopher.

Legendary Member
Posts: 2326
Joined: Mon Jul 28, 2008 3:54 am
Thanked: 173 times
Followed by:2 members
GMAT Score:710

by gmatmachoman » Sat Aug 28, 2010 8:46 am
st1 :K is divisible by 4.

case 1 : K =16

K is divisible by 4. Yes. is K the square of an interger : YES

case 2 : K =8

K is divisible by 4. Yes. is K the square of an interger :NO

Inconsistent; insufficient

st2 :K is divisible by exactly 4 different prime numbers.

case 1: K = 2* 3*5*7

yes K is divisible by 4 different prime numbers

Is K the square of an interger ?? NO

case 2: K = 2^2 * 3^2 * 5^2 * 7^2

yes K is divisible by 4 different prime numbers

Is K the square of an interger ?? YES
Inconsistent..Insufficient

combining , again we can have 2 cases ( K = 2^2* 3*5*7 and K= 2^2 * 3^2 * 5^2 * 7^2). Here they are divisible by 4 but one is square of a integer and other in not.

so Pick E