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.
06 Sep 2024, 01:30
Kudos
Bookmarks
C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
61% (02:38) correct
39%
(02:13)
wrong
based on 198
sessions
History
Date
Time
Result
Not Attempted Yet
149 is a 3-digit positive integer, product of whose digits is 1x4x9 = 36. How many 3-digit positive integers exist whose product of digits is 36?
A. 9
B. 12
C. 21
D. 27
E. 45
A. 9
B. 12
C. 21
D. 27
E. 45
06 Sep 2024, 02:01
Kudos
Bookmarks
The factor pairs of 36:
\(36*1\), \(18*2\), \(12*3\), \(9*4\) and \(6*6\)
Focusing on the first three as they have double digit values:
36*1: The 36*1 can be broken down to a product of three digits while keeping 1, in the following ways: \(9*4*1\) & \(6*6*1\).
\(9*4*1\) can be arranged in \(3! = 6\) different ways.
\(6*6*1\) can be arranged in \(\frac{3!}{2!1!} = 3 \) ways.
18*2: The 18*2 can be broken down to a product of three digits while keeping 2 in the following ways: \(9*2*2\) & \(6*3*2\).
\(9*2*2\) can be arranged in \(\frac{3!}{2!1!} = 3 \) ways.
\(6*3*2\) can be arranged in \(3! = 6\) different ways.
12*3: The 12*3 can be broken down to a product of three digits while keeping 3, in the following ways: \(9*4*1\) & \(4*3*3\).
\(4*3*3\) can be arranged in \(\frac{3!}{2!1!} = 3 \) ways.
Total Ways: \(6+3+3+6+3 = 21\)
ANSWER C
\(36*1\), \(18*2\), \(12*3\), \(9*4\) and \(6*6\)
Focusing on the first three as they have double digit values:
36*1: The 36*1 can be broken down to a product of three digits while keeping 1, in the following ways: \(9*4*1\) & \(6*6*1\).
\(9*4*1\) can be arranged in \(3! = 6\) different ways.
\(6*6*1\) can be arranged in \(\frac{3!}{2!1!} = 3 \) ways.
18*2: The 18*2 can be broken down to a product of three digits while keeping 2 in the following ways: \(9*2*2\) & \(6*3*2\).
\(9*2*2\) can be arranged in \(\frac{3!}{2!1!} = 3 \) ways.
\(6*3*2\) can be arranged in \(3! = 6\) different ways.
12*3: The 12*3 can be broken down to a product of three digits while keeping 3, in the following ways: \(9*4*1\) & \(4*3*3\).
\(4*3*3\) can be arranged in \(\frac{3!}{2!1!} = 3 \) ways.
Total Ways: \(6+3+3+6+3 = 21\)
ANSWER C
General Discussion
Breaking 36 into its prime factor form => 2 * 2 * 3 * 3
We know have to fill 3 blanks _ _ _ with 3 digits whose product is 36. From the prime factor form we know we can form single digits by multiplying at most 2 prime factors.
Let us also consider 1 into the factored form; 36 = 1 * 2 * 2 * 3 * 3
1 _ _ => the other two blanks can have two 6's or one 4 and one 9; 3 ways to arrange 166 and 6 ways to arrange 149
2 _ _ => the other two blanks can have one 2 and one 9 or one 3 and one 6; 3 ways to arrange 229 and 6 ways to arrange 236
3 _ _ => the other two blanks can have one 3 and one 4 (case with one 6 and one 2 is already considered above); 3 ways to arrange 334
3 + 6 + 3 + 6 + 3 = 21 ways or 21 numbers
Answer C.
We know have to fill 3 blanks _ _ _ with 3 digits whose product is 36. From the prime factor form we know we can form single digits by multiplying at most 2 prime factors.
Let us also consider 1 into the factored form; 36 = 1 * 2 * 2 * 3 * 3
1 _ _ => the other two blanks can have two 6's or one 4 and one 9; 3 ways to arrange 166 and 6 ways to arrange 149
2 _ _ => the other two blanks can have one 2 and one 9 or one 3 and one 6; 3 ways to arrange 229 and 6 ways to arrange 236
3 _ _ => the other two blanks can have one 3 and one 4 (case with one 6 and one 2 is already considered above); 3 ways to arrange 334
3 + 6 + 3 + 6 + 3 = 21 ways or 21 numbers
Answer C.
