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07 Dec 2022, 09:30
Kudos
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B
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:
49% (02:39) correct
51%
(02:51)
wrong
based on 85
sessions
History
Date
Time
Result
Not Attempted Yet
If positive integer \(A = m^3n^2\), where m and n are distinct prime numbers, is 54 a factor of A?
(1) 25mn is the least common multiple of 15 and 50
(2) The greatest common factor of \(15m^3\) and \(14n^2\) is 6
(1) 25mn is the least common multiple of 15 and 50
(2) The greatest common factor of \(15m^3\) and \(14n^2\) is 6
07 Dec 2022, 10:14
Kudos
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1. LCM of 15 and 50 is 150=25mn
mn= 150/25=6
m and n are distinct prime numbers.
So, CASE 1. If m= 3, n=2
A= 3^3*2^2= 108= 54*2. 54 is a factor of A
CASE 2. If m= 2, n=3
A= 2^3*3^2= 72 . 54 is not a factor of A .
Statement I alone is not sufficient.
Reject A and D.
2. GCF of 15m^3 and 14n^2 is 6.
15m^3= 5*3*m^3
14n^2= 2*7*n^2
GCF can be 6 only if m=2 and n =3
15m^3= 5*3*2*2*2
14n^2= 2*7*3*3
So, A= 2*2*2*3*3= 72 and 54 is not a factor of 72.
Statement II is clearly sufficient.
Answer should be B IMO
Posted from my mobile device
mn= 150/25=6
m and n are distinct prime numbers.
So, CASE 1. If m= 3, n=2
A= 3^3*2^2= 108= 54*2. 54 is a factor of A
CASE 2. If m= 2, n=3
A= 2^3*3^2= 72 . 54 is not a factor of A .
Statement I alone is not sufficient.
Reject A and D.
2. GCF of 15m^3 and 14n^2 is 6.
15m^3= 5*3*m^3
14n^2= 2*7*n^2
GCF can be 6 only if m=2 and n =3
15m^3= 5*3*2*2*2
14n^2= 2*7*3*3
So, A= 2*2*2*3*3= 72 and 54 is not a factor of 72.
Statement II is clearly sufficient.
Answer should be B IMO
Posted from my mobile device
07 Dec 2022, 14:05
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Bunuel
\(54 = 3^3 * 2^1\)
Inference : We need to find out if A consists of at least three 3's and one 2's
Statement 1
25mn is the least common multiple of 15 and 50
From statement 1, we can conclude the product of m and n is 6. However, we don't have any additional information to find out whether m = 2 and n = 3 or m = 3 and n = 2.
Given this ambiguity, statement 1 is not sufficient and we can rule out A and D.
Statement 2
The greatest common factor of \(15m^3\) and \(14n^2\) is 6
As m and n are distinct prime, they do not share any common factor other than 1. As the GCF of \(15m^3\) and \(14n^2\) is 6, m must be 2 and n must be 3.
As we have a single value of m and n, the statement is sufficient.
Option B
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