|
|
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Updated on: 01 Mar 2018, 01:10
Originally posted by sekharm2389 on 01 Mar 2018, 00:53.
Last edited by Bunuel on 01 Mar 2018, 01:10, edited 1 time in total.
Last edited by Bunuel on 01 Mar 2018, 01:10, edited 1 time in total.
Renamed the topic and edited the question.
Kudos
Bookmarks
A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
47% (02:24) correct
53%
(02:28)
wrong
based on 234
sessions
History
Date
Time
Result
Not Attempted Yet
If P is a prime number and Q is a positive integer, how many factors does P^2*Q have?
(1) The lowest number that has both P^2 and Q^3 as its factors is 5400
(2) P and Q have only one common factor
(1) The lowest number that has both P^2 and Q^3 as its factors is 5400
(2) P and Q have only one common factor
01 Mar 2018, 01:25
Kudos
Bookmarks
If P is a prime number and Q is a positive integer, how many factors does P^2*Q have?
(1) The lowest number that has both P^2 and Q^3 as its factors is 5400.
This means that 5,400 is the least common multiple of P^2 (square of a prime) and Q^3 (perfect cube), so both P^2 and Q^3 are factors of 5,400. Factorize: \(5400 = 2^3*3^3*5^2\). P^2, which is a square of a prime, must be 5^2 (so P must be 5) because in any other case (say if P^2 is 2^2 or 3^2), the remaining multiple will not be a perfect square (Q^3). Therefore, \(P^2 = 5^2\) (P = 5) and \(Q^3 = 2^3*3^3\) (Q = 6).
\(P^2*Q = 5^2*6 = 5^2*2*3\). The number of factors \(= (2 + 1)(1 + 1)(1 + 1) = 12\).
Sufficient.
(2) P and Q have only one common factor. This means that P and Q are co-prime, their only common factor is 1. This is clearly insufficient: for example, P could be 2 and Q could be 3, 3^2, 3^3, ...
Answer: A.
Hope it's clear.
(1) The lowest number that has both P^2 and Q^3 as its factors is 5400.
This means that 5,400 is the least common multiple of P^2 (square of a prime) and Q^3 (perfect cube), so both P^2 and Q^3 are factors of 5,400. Factorize: \(5400 = 2^3*3^3*5^2\). P^2, which is a square of a prime, must be 5^2 (so P must be 5) because in any other case (say if P^2 is 2^2 or 3^2), the remaining multiple will not be a perfect square (Q^3). Therefore, \(P^2 = 5^2\) (P = 5) and \(Q^3 = 2^3*3^3\) (Q = 6).
\(P^2*Q = 5^2*6 = 5^2*2*3\). The number of factors \(= (2 + 1)(1 + 1)(1 + 1) = 12\).
Sufficient.
(2) P and Q have only one common factor. This means that P and Q are co-prime, their only common factor is 1. This is clearly insufficient: for example, P could be 2 and Q could be 3, 3^2, 3^3, ...
Answer: A.
Hope it's clear.
srinjoy1990
Joined: 04 Feb 2016
Last visit: 20 Nov 2018
Posts: 47
Own Kudos:
Given Kudos: 4
Location: India
Concentration: Technology, Marketing
Schools: S.P.Jain Institute (A)
GMAT 1: 650 Q48 V32
GPA: 4
WE:Sales (Computer Software)
09 Mar 2018, 09:15
Kudos
Bookmarks
Hi Bunuel,
Why are we assuming P^2 and Q^3 are the only factors apart from 1 for the number 5400?
I considered multiple possibilities for Q [2,3,6,1] so went with E. I think I am missing something in the reasoning. Can you help?
Tx.
Why are we assuming P^2 and Q^3 are the only factors apart from 1 for the number 5400?
I considered multiple possibilities for Q [2,3,6,1] so went with E. I think I am missing something in the reasoning. Can you help?
Tx.
