Factors

This topic has expert replies
Junior | Next Rank: 30 Posts
Posts: 21
Joined: Sun Sep 18, 2011 9:25 pm
Thanked: 1 times

Factors

by tdkk123 » Mon Sep 19, 2011 2:25 am
How many prime factors does x have?
(1) X is factor of 7200
(2) 180 is factor of x

OA C

How do you combine both statements?

User avatar
Legendary Member
Posts: 1309
Joined: Mon Apr 04, 2011 5:34 am
Location: India
Thanked: 310 times
Followed by:123 members
GMAT Score:750

by cans » Mon Sep 19, 2011 3:44 am
a) x is factor of 7200.
if x=2, 1 prime factor. if x=6, 2 prime factors
Insufficient
b)180 is factor of x.
x=180*11 or x=180*11*13.. insufficient

a&b) 7200/180 = 40. the factors of 40 (2,5) are already in 180.
thus sufficient
IMO C
If my post helped you- let me know by pushing the thanks button ;)

Contact me about long distance tutoring!
[email protected]

Cans!!

User avatar
Legendary Member
Posts: 504
Joined: Tue Apr 19, 2011 1:40 pm
Thanked: 114 times
Followed by:11 members

by knight247 » Mon Sep 19, 2011 3:58 am
(1) x is a factor of 7200
7200=2^5*3^2*5^2

Basically x contains any of the prime numbers 2,3 or 5 in some combination. Could contain all of them, could contain two of them or all three of them. example x could be 8 i.e. 2^3 or x could be 15 i.e. 3*5 etc. No fixed value of x can be deduced. INSUFFICIENT

(2)180 is a factor of x meaning
x=2^2*3^2*5*unknown
if unknown=1 then x has 3 prime factors
if unknown=7 then x has 4 prime factors etc. We get different answers so INSUFFICIENT

Combining both,
From(1) we have that, x contains only the prime factors 2,5,3 in some combination and NO OTHER PRIME example if x contained a 7 then 7200 wouldn't be divisible by x. Also, from statement (2) x=2^2*3^2*5*unknown and unknown can't contain any other prime as this would make 7200 indivisible by x. So unknown has to contain some other power of 2,3,5 which makes 7200 perfectly divisible by x. So this narrows down the prime factorisation of x to 2 AND 3 AND 5 with different/same powers for each. So x has three prime factors. Hence C