[Math Revolution GMAT math practice question]
When A and B are positive integers, is AB a multiple of 4?
1) The greatest common divisor of A and B is 6
2) The least common multiple of A and B is 30
When A and B are positive integers, is AB a multiple of 4?
This topic has expert replies
- Max@Math Revolution
- Elite Legendary Member
- Posts: 3991
- Joined: Fri Jul 24, 2015 2:28 am
- Location: Las Vegas, USA
- Thanked: 19 times
- Followed by:37 members
Timer
00:00
Your Answer
A
B
C
D
E
Global Stats
Math Revolution
The World's Most "Complete" GMAT Math Course!
Score an excellent Q49-51 just like 70% of our students.
[Free] Full on-demand course (7 days) - 100 hours of video lessons, 490 lesson topics, and 2,000 questions.
[Course] Starting $79 for on-demand and $60 for tutoring per hour and $390 only for Live Online.
Email to : [email protected]
- fskilnik@GMATH
- GMAT Instructor
- Posts: 1449
- Joined: Sat Oct 09, 2010 2:16 pm
- Thanked: 59 times
- Followed by:33 members
Timer
00:00
Your Answer
A
B
C
D
E
Global Stats
Another beautiful problem, Max. Congrats!Max@Math Revolution wrote:[Math Revolution GMAT math practice question]
When A and B are positive integers, is AB a multiple of 4?
1) The greatest common divisor of A and B is 6
2) The least common multiple of A and B is 30
$$A,B\,\,\, \geqslant 1\,\,\,{\text{ints}}$$
$$\frac{{A \cdot B}}{4}\,\,\,\mathop = \limits^? \,\,\,\operatorname{int} $$
$$\left( 1 \right)\,\,\,GCD\left( {A,B} \right) = 6\,\,\,\, \Rightarrow \,\,\,\left\{ \matrix{
\,A = 6M\,,\,\,\,M \ge 1\,\,{\mathop{\rm int}} \hfill \cr
\,B = 6N\,,\,\,\,N \ge 1\,\,{\mathop{\rm int}} \hfill \cr} \right.\,\,\,\,\,\,\,\,\,{\rm{with}}\,\,\,\,M,N\,\,\,{\rm{relatively}}\,\,\,{\rm{prime}}$$
$$?\,\,\,\,:\,\,\,\,\frac{{A \cdot B}}{4}\,\,\, = \,\,\,\frac{{6M \cdot 6N}}{4}\,\,\, = \,\,\,9MN\,\,\, = \,\,\,\operatorname{int} \,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left\langle {{\text{YES}}} \right\rangle $$
$$\left( 2 \right)\,\,\,LCM\left( {A,B} \right) = 30\,\,\,\,\,\,\left\{ \matrix{
\,{\rm{Take}}\,\,\left( {A,B} \right) = \left( {1,30} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr
\,{\rm{Take}}\,\,\left( {A,B} \right) = \left( {2,30} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \hfill \cr} \right.$$
The correct answer is therefore (A).
This solution follows the notations and rationale taught in the GMATH method.
Regards,
Fabio.
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br
GMAT/MBA Expert
- Brent@GMATPrepNow
- GMAT Instructor
- Posts: 16207
- Joined: Mon Dec 08, 2008 6:26 pm
- Location: Vancouver, BC
- Thanked: 5254 times
- Followed by:1268 members
- GMAT Score:770
Timer
00:00
Your Answer
A
B
C
D
E
Global Stats
Target question: Is AB a multiple of 4?Max@Math Revolution wrote: When A and B are positive integers, is AB a multiple of 4?
1) The greatest common divisor of A and B is 6
2) The least common multiple of A and B is 30
Useful property: If N is divisible by d, we can say that N = dk for some integer k
For example, if N is divisible by 5, we can say that N = 5k for some integer k
Statement 1: The greatest common divisor of A and B is 6
This means that A is divisible by 6, and B is divisible by 6.
Applying the above property, we can write A = 6k for some integer k, and B = 6j for some integer j
So, AB = (6k)(6j) = 36kj = (4)(9)(kj)
Aha! We can see that AB is a multiple of 4
So, the answer to the target question is YES, AB IS a multiple of 4
Since we can answer the target question with certainty, statement 1 is SUFFICIENT
Statement 2: The least common multiple of A and B is 30
There are several values of A and y that satisfy statement B. Here are two:
Case a: A = 10 AND B = 30 (the LCM of 10 and 30 is 30). In this case, AB = (10)(30) = 300. So, the answer to the target question is YES, AB IS a multiple of 4
Case b: A = 1 AND B = 30 (the LCM of 1 and 30 is 30). In this case, AB = (1)(30) = 30. So, the answer to the target question is NO, AB is NOT a multiple of 4
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT
Answer: A
Cheers,
Brent
- Max@Math Revolution
- Elite Legendary Member
- Posts: 3991
- Joined: Fri Jul 24, 2015 2:28 am
- Location: Las Vegas, USA
- Thanked: 19 times
- Followed by:37 members
Timer
00:00
Your Answer
A
B
C
D
E
Global Stats
=>
Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.
The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question.
Asking if AB is a multiple of 4 is equivalent to asking if AB = 4k for some integer k.
Condition 1)
Since A = 6a = 2*3*a and B = 6b = 2*3*b for some integers a and b, AB = 2^2*3^2*ab = 4*3^2ab.
Thus, AB is a multiple of 4 and condition 1) is sufficient.
Condition 2)
If A = 6 and B = 5, then lcm(A,B) = 30, but AB = 30 is not a multiple of 4, and the answer is 'no'.
If A = 6 and B = 10, then lcm(A,B) = 30, and AB = 60 is a multiple of 4. The answer is 'yes'.
Since we don't have a unique solution, condition 2) is not sufficient.
Therefore, A is the answer.
Answer: A
Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.
The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question.
Asking if AB is a multiple of 4 is equivalent to asking if AB = 4k for some integer k.
Condition 1)
Since A = 6a = 2*3*a and B = 6b = 2*3*b for some integers a and b, AB = 2^2*3^2*ab = 4*3^2ab.
Thus, AB is a multiple of 4 and condition 1) is sufficient.
Condition 2)
If A = 6 and B = 5, then lcm(A,B) = 30, but AB = 30 is not a multiple of 4, and the answer is 'no'.
If A = 6 and B = 10, then lcm(A,B) = 30, and AB = 60 is a multiple of 4. The answer is 'yes'.
Since we don't have a unique solution, condition 2) is not sufficient.
Therefore, A is the answer.
Answer: A
Math Revolution
The World's Most "Complete" GMAT Math Course!
Score an excellent Q49-51 just like 70% of our students.
[Free] Full on-demand course (7 days) - 100 hours of video lessons, 490 lesson topics, and 2,000 questions.
[Course] Starting $79 for on-demand and $60 for tutoring per hour and $390 only for Live Online.
Email to : [email protected]