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20 Feb 2023, 07:43
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D
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:
48% (02:08) correct
52%
(02:28)
wrong
based on 64
sessions
History
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Time
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If A and B are positive integers and A^3 is divisible by 24, then is A*B^2 a multiple of 216?
(1) B is a multiple of 6
(2) B is divisible by 30
(1) B is a multiple of 6
(2) B is divisible by 30
Pure algebraic questions like this one are no longer part of the GMAT Data Sufficiency syllabus.
For relevant Data Sufficiency questions, please check the Data Sufficiency forum:
Data Sufficiency Forum
Data Sufficiency Forum
20 Feb 2023, 08:20
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Bunuel
Given:
- A and B are positive Integers
- \(A^3\) is a multiple of 24 i.e. \(2^3*3\) i.e. A is divisible by 2 as well as by 3
Question: is \(A*B^2\) a multiple of \(216 = 2^6*3^3\)?
Statement 1: B is a multiple of 6
i.e. A = 6x and B = 6y i.e. \(A*B^2 = 6x*(6y)^2 = 216xy^2\) i.e. divisible by 216 hence
SUFFICIENT
Statement 2: B is divisible by 30
i.e. B is divisible by 30 i.e. B is divisible by 6
i.e. A = 6x and B = 6y i.e. \(A*B^2 = 6x*(6y)^2 = 216xy^2\) i.e. divisible by 216 hence
SUFFICIENT
Answer: Option D
13 Mar 2023, 05:02
Kudos
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Bunuel
Solution:
Pre Analysis:
- We are given that A^3 is divisible by 24
- This means we can write \(A^3=24P\) where P is a positive integer
\(⇒A^3=2^3\times 3\times P\) - We can assume \(P=3^2\times Q^3\) where Q is another positive integer
\(⇒A^3=2^3\times 3^3\times Q^3\)
\(⇒A=6Q\) - We are asked if A*B^2 a multiple of 216 or not
- Or if \(\frac{A*B^2}{216}=\frac{6\times Q\times B^2}{2^3\times 3^3}=\frac{Q\times B^2}{2^2\times 3^2}\) is an integer or not
Statement 1: B is a multiple of 6
- According to this statement, \(B=6R\) where R is a positive integer
- So, \(\frac{Q\times B^2}{2^2\times 3^2}=\frac{Q\times 2^2\times 3^2\times R^2}{2^2\times 3^2}=QR^2=integer\)
- Thus, statement 1 alone is sufficient and we can eliminate options B, C and E
Statement 2: B is divisible by 30
- According to this statement, \(B=30R\) where R is a positive integer
- So, \(\frac{Q\times B^2}{2^2\times 3^2}=\frac{Q\times 2^2\times 3^2\times 5^2\times R^2}{2^2\times 3^2}=25QR^2=integer\)
- Thus, statement 2 alone is also sufficient
Hence the right answer is Option D
