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If positive integer A = m^3*n^2, where m and n are distinct prime numb

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If positive integer A = m^3*n^2, where m and n are distinct prime numb  [#permalink]

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If positive integer A = m^3*n^2, where m and n are distinct prime numbers, is A divisible by 72?


(1) 25mn is a multiple of 15

(2) 6m^2 is divisible by 12

Originally posted by baru on 23 Mar 2019, 11:56.
Last edited by Bunuel on 24 Mar 2019, 01:55, edited 2 times in total.
Renamed the topic and edited the question.
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Re: If positive integer A = m^3*n^2, where m and n are distinct prime numb  [#permalink]

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New post Updated on: 23 Mar 2019, 12:43
baru wrote:
If positive integer A =\(m3\)\(n2\), where m and n are distinct prime numbers, is A divisible by 72?
(1) 25mn is a multiple of 15
(2) 6\(m2\) is divisible by 12


baru could you please edit the question so the powers are written properly?

As this is a question dealing with integers and divisibility, it is almost guaranteed to have to do with prime factorization.
Then we'll go the Precise route of prime factorizing 72 to see what we need to do.

72 = 36*2 = 6^2 *2 = (2^3)(3^2)
So the answer to the question is yes only if m= 2 and n = 3.
(1) if 25mn is a multiple of 15 then 5mn is a multiple of 3 so (at least) one of m,n must be 3. This is not enough.
(2) if 6m^2 is divisible by 12 then m^2 is divisible by 2 and therefore m = 2. without knowledge on n, this is not enough.

Combined, we know that m = 2 and n = 3, exactly what we need.
Sufficient.

(C) is our answer.
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Originally posted by DavidTutorexamPAL on 23 Mar 2019, 12:30.
Last edited by DavidTutorexamPAL on 23 Mar 2019, 12:43, edited 1 time in total.
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Re: If positive integer A = m^3*n^2, where m and n are distinct prime numb  [#permalink]

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New post 23 Mar 2019, 12:40
baru wrote:
If positive integer A =\(m3\)\(n2\), where m and n are distinct prime numbers, is A divisible by 72?
(1) 25mn is a multiple of 15
(2) 6\(m2\) is divisible by 12


From
1) 25mn is a multiple of 15 so either m=3 or n=3 since m and n are prime numbers. If m=3 A = \(m^3n^2\) = \(27n^2\) is not divisible by 72=\(3^2*2^3\)
if n=3 then A= \(9m^3\) can be divisible by 72 if m=2 otherwise not divisible by 72. Hence not sufficient.

2) 6\(m^2\) is divisible by 12 so m=2. A=\(8n^2\) . But we don't know the value of n. Hence not sufficient.

1)+2)
Since from (2) we get m=2
Hence from (1) n = 3
Sufficience.

Answer (C)
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Re: If positive integer A = m^3*n^2, where m and n are distinct prime numb  [#permalink]

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New post 23 Mar 2019, 13:06
DavidTutorexamPAL wrote:
baru wrote:
If positive integer A =\(m3\)\(n2\), where m and n are distinct prime numbers, is A divisible by 72?
(1) 25mn is a multiple of 15
(2) 6\(m2\) is divisible by 12


baru could you please edit the question so the powers are written properly?

As this is a question dealing with integers and divisibility, it is almost guaranteed to have to do with prime factorization.
Then we'll go the Precise route of prime factorizing 72 to see what we need to do.

72 = 36*2 = 6^2 *2 = (2^3)(3^2)
So the answer to the question is yes only if m= 2 and n = 3.
(1) if 25mn is a multiple of 15 then 5mn is a multiple of 3 so (at least) one of m,n must be 3. This is not enough.
(2) if 6m^2 is divisible by 12 then m^2 is divisible by 2 and therefore m = 2. without knowledge on n, this is not enough.

Combined, we know that m = 2 and n = 3, exactly what we need.
Sufficient.

(C) is our answer.


Done ..Sorry for the trouble
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Re: If positive integer A = m^3*n^2, where m and n are distinct prime numb   [#permalink] 23 Mar 2019, 13:06
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If positive integer A = m^3*n^2, where m and n are distinct prime numb

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