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For positive integers A and x, A^3 = 48x Find the least value of x
[#permalink]
29 May 2020, 08:43
29 May 2020, 08:43
1
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Expert Reply
Show timer
00:00
A
B
C
D
E
Difficulty:
25%
(medium)
Question Stats:
74% (01:17) correct
26%
(01:56)
wrong
based on 76
sessions
For positive integers A and x, \(A^3 = 48x\)
Find the least value of x
A) 4
B) 9
C) 36
D) 48
E) \(48^2\)
Find the least value of x
A) 4
B) 9
C) 36
D) 48
E) \(48^2\)
Source: https://www.GMATinsight.com
For positive integers A and x, A^3 = 48x Find the least value of x
[#permalink]
Updated on: 29 May 2020, 09:21
Updated on: 29 May 2020, 09:21
1
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Expert Reply
Solution
Given
In this question, we are given that
- • For positive integers A and x, \(A^3 = 48x\)
To find
We need to determine
- • The least value of x
Approach and Working out
\(48 = 2^4 * 3^1\)
Hence, if 48x is a perfect cube, x must have at least \(2^2\) and \(3^2\) (so that the powers of 2 and 3 becomes a multiple of 3)
- • Therefore, least value of x = \(2^2 * 3^2 = 36\)
Thus, option C is the correct answer.
Correct Answer: Option C
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Originally posted by EgmatQuantExpert on 29 May 2020, 09:09.
Last edited by EgmatQuantExpert on 29 May 2020, 09:21, edited 1 time in total.
Last edited by EgmatQuantExpert on 29 May 2020, 09:21, edited 1 time in total.
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Re: For positive integers A and x, A^3 = 48x Find the least value of x
[#permalink]
29 May 2020, 09:17
29 May 2020, 09:17
GMATinsight wrote:
For positive integers A and x, \(A^3 = 48x\)
Find the least value of x
A) 4
B) 9
C) 36
D) 48
E) \(48^2\)
Find the least value of x
A) 4
B) 9
C) 36
D) 48
E) \(48^2\)
Source: https://www.GMATinsight.com
48 ; 2^4*3
for A^3 to be a +ve integer ; value of x has to be 2^2*3^2 ; 36
OPTION C
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Joined: 08 Jul 2010
Status:GMAT/GRE Tutor l Admission Consultant l On-Demand Course creator
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Re: For positive integers A and x, A^3 = 48x Find the least value of x
[#permalink]
31 May 2020, 22:28
31 May 2020, 22:28
2
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Expert Reply
GMATinsight wrote:
For positive integers A and x, \(A^3 = 48x\)
Find the least value of x
A) 4
B) 9
C) 36
D) 48
E) \(48^2\)
Find the least value of x
A) 4
B) 9
C) 36
D) 48
E) \(48^2\)
Source: https://www.GMATinsight.com
OFFICIAL EXPLANATION:
CONCEPT: A perfect cube \(A^3\) must have all exponents of its unique prime factors a multiple of 3
\(A^3 = 48x = 2^4*3*x\)
the exponent of 2 is 4 which needs to be increased by 2 to become 6 i.e. multiple of 3 i.e. x must have \(2^2\)
the exponent of 3 is 1 which needs to be increased by 2 to become 3 i.e. multiple of 3 i.e. x must have \(3^2\)
i.e. \(x_{min} = 2^2*3^2\) to make the \(48 x\) a perfect cube
i.e. \(x_{min} = 36\)
Answer: Option C
gmatclubot
Re: For positive integers A and x, A^3 = 48x Find the least value of x [#permalink]
31 May 2020, 22:28
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