Events & Promotions
|
|
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.
Apr 2610:00 AM EDT
-11:00 AM EDT
The Target Test Prep course represents a quantum leap forward in GMAT preparation, a radical reinterpretation of the way that students should study. Try before you buy with a 5-day, full-access trial of the course for FREE!
Apr 2711:00 AM EDT
-12:00 PM EDT
Prefer video-based learning? The Target Test Prep OnDemand course is a one-of-a-kind video masterclass featuring 400 hours of lecture-style teaching by Scott Woodbury-Stewart, founder of Target Test Prep and one of the most accomplished GMAT instructors
Apr 2808:00 AM PDT
-11:00 AM PDT
Whether you are just beginning your MBA journey or fine-tuning your path to a 700+ score, GMAT Day is designed to give you a competitive edge.
Updated on: 06 Mar 2026, 19:38
Originally posted by MartyMurray on 21 Sep 2025, 09:12.
Last edited by MartyMurray on 06 Mar 2026, 19:38, edited 3 times in total.
Last edited by MartyMurray on 06 Mar 2026, 19:38, edited 3 times in total.
Kudos
Bookmarks
C
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:
36% (01:54) correct
64%
(01:45)
wrong
based on 89
sessions
History
Date
Time
Result
Not Attempted Yet
If a is a positive integer and 18a has 12 different factors, what is the maximum number of distinct prime factors that a could have?
(A) 0
(B) 1
(C) 2
(D) 3
(E) 4
Source: Marty Murray Coaching
(A) 0
(B) 1
(C) 2
(D) 3
(E) 4
Source: Marty Murray Coaching
21 Sep 2025, 10:14
Kudos
Bookmarks
18 has 6 factors--> \(18 = 2*3^2\) --> number of factors \(= (1+1)(2+1) = 6\)
18a has 12 factors which means that the product of (exponents+1) of each distinct prime factor must be equal to 12.
\(a =\) (any prime other than 2 and 3)^1; \(2^2\); \((2*3)\)
Example: (1) a = 5
\(18a = 18*5\) --> number of factors \(= 6 * (1+1) = 12\) (remember that 18 has 6 factors)
(2) \(a = 2^2\)
\(18a = (2^3*3^2)\) --> number of factors \(= (3+1) * (2+1) = 12\)
(3) \(a = (2*3)\)
\(18a = (2^2*3^3)\) --> number of factors \(= (2+1) * (3+1) = 12\)
So the maximum possible number of distinct prime factors that a could have is 2.
Answer is C.
18a has 12 factors which means that the product of (exponents+1) of each distinct prime factor must be equal to 12.
\(a =\) (any prime other than 2 and 3)^1; \(2^2\); \((2*3)\)
Example: (1) a = 5
\(18a = 18*5\) --> number of factors \(= 6 * (1+1) = 12\) (remember that 18 has 6 factors)
(2) \(a = 2^2\)
\(18a = (2^3*3^2)\) --> number of factors \(= (3+1) * (2+1) = 12\)
(3) \(a = (2*3)\)
\(18a = (2^2*3^3)\) --> number of factors \(= (2+1) * (3+1) = 12\)
So the maximum possible number of distinct prime factors that a could have is 2.
Answer is C.
MartyMurray
22 Sep 2025, 05:40
Kudos
Bookmarks
If a is a positive integer and 18a has 12 different factors, what is the maximum number of distinct prime factors that a could have?
We can answer this question using the following strategy for finding the number of different factors of a number:
- Prime factorize the number.
- Find the powers of the prime factors.
- Add \(1\) to each power.
- Multiply the resulting values to arrive at the number of factors of the number.
Using this approach for \(18\), we get the following:
\(18 = 2^1 × 3^2\)
\(1 + 1 = 2\)
\(2 + 1 = 3\)
\(2 × 3 = 6\)
So, \(18\) has \(6\) factors.
Now, let's consider ways in which \(18a\) could have \(12\) factors.
If \(a\) is a prime number that is not \(2\) or \(3\), meaning that \(a\) has \(1\) distinct prime factor, we get the following number of factors for \(18a\):
\(18a = 2^1 × 3^2 × a^1\)
\(1 + 1 = 2\)
\(2 + 1 = 3\)
\(1 + 1 = 2\)
\(2 × 3 × 2 = 12\)
So, \(a\) could have \(1\) distinct prime factor.
Let's now see whether there is a way in which \(a\) could have \(2\) distinct prime factors:
Try \(a = 6\).
If \(a = 6\), then \(18a = 2^1 × 3^2 × 2 × 3 = 2^2 × 3^3\).
\(2 + 1 = 3\)
\(3 + 1 = 4\)
\(3 × 4 = 12\)
So, if \(a = 6\), then \(18a\) has \(12\) factors, and \(a\) has \(2\) distinct prime factors.
We can also see that, if we add another distinct prime factor \(p\) to \(a\) so that \(a = 2 × 3 × p\), we'll get the following for \(18a\).
\(18a = 2^2 × 3^3 × p^1\)
\(2 + 1 = 3\)
\(3 + 1 = 4\)
\(1 + 1 = 1\)
\(3 × 4 × 2 = 24\)
So, if \(a\) has \(3\) distinct prime factors, \(18a\) has \(24\), rather than \(12\), different factors.
So, \(3\) doesn't work, and we can tell that \(4\) will not work either.
So, the maximum number of distinct prime factors that \(a\) could have is \(2\).
(A) 0
(B) 1
(C) 2
(D) 3
(E) 4
Answer: C
We can answer this question using the following strategy for finding the number of different factors of a number:
- Prime factorize the number.
- Find the powers of the prime factors.
- Add \(1\) to each power.
- Multiply the resulting values to arrive at the number of factors of the number.
Using this approach for \(18\), we get the following:
\(18 = 2^1 × 3^2\)
\(1 + 1 = 2\)
\(2 + 1 = 3\)
\(2 × 3 = 6\)
So, \(18\) has \(6\) factors.
Now, let's consider ways in which \(18a\) could have \(12\) factors.
If \(a\) is a prime number that is not \(2\) or \(3\), meaning that \(a\) has \(1\) distinct prime factor, we get the following number of factors for \(18a\):
\(18a = 2^1 × 3^2 × a^1\)
\(1 + 1 = 2\)
\(2 + 1 = 3\)
\(1 + 1 = 2\)
\(2 × 3 × 2 = 12\)
So, \(a\) could have \(1\) distinct prime factor.
Let's now see whether there is a way in which \(a\) could have \(2\) distinct prime factors:
Try \(a = 6\).
If \(a = 6\), then \(18a = 2^1 × 3^2 × 2 × 3 = 2^2 × 3^3\).
\(2 + 1 = 3\)
\(3 + 1 = 4\)
\(3 × 4 = 12\)
So, if \(a = 6\), then \(18a\) has \(12\) factors, and \(a\) has \(2\) distinct prime factors.
We can also see that, if we add another distinct prime factor \(p\) to \(a\) so that \(a = 2 × 3 × p\), we'll get the following for \(18a\).
\(18a = 2^2 × 3^3 × p^1\)
\(2 + 1 = 3\)
\(3 + 1 = 4\)
\(1 + 1 = 1\)
\(3 × 4 × 2 = 24\)
So, if \(a\) has \(3\) distinct prime factors, \(18a\) has \(24\), rather than \(12\), different factors.
So, \(3\) doesn't work, and we can tell that \(4\) will not work either.
So, the maximum number of distinct prime factors that \(a\) could have is \(2\).
(A) 0
(B) 1
(C) 2
(D) 3
(E) 4
Answer: C
