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# If m is a positive integer and m^2 is divisible by 48, then

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If m is a positive integer and m^2 is divisible by 48, then [#permalink]  10 Oct 2011, 11:42
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If m is a positive integer and m^2 is divisible by 48, then the largest positive integer that must divide m is?

(A) 3
(B) 6
(C) 8
(D) 12
(E) 16

Can someone please explain this one? Knewton's explanation was not useful to me. Thx
OA
[Reveal] Spoiler:
D
[Reveal] Spoiler: OA
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Re: Division question [#permalink]  10 Oct 2011, 11:56
Well m^2 is divisible by 48

if we brake 48 we will get 4*4*3

now we know that m is an integer therefore, m^2 is completely divisible by 48 then to get the square number that will be divided by 48 we need to multiply 48 by 3
so that we will get 4*4*3*3 ....
square root of the above number will be 4*3 = 12
thus m = 12 and the greatest number that will divide 12 is 12 itself ...thus the answer.
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Re: Division question [#permalink]  10 Oct 2011, 12:05
arcanis2000 wrote:
Q: If m is a positive integer and m^2 is divisible by 48, then the largest positive integer that must divide m is?

(A) 3
(B) 6
(C) 8
(D) 12
(E) 16

Can someone please explain this one? Knewton's explanation was not useful to me. Thx
OA
[Reveal] Spoiler:
D

M^2 is divisible by 48 so M^2 must be multiple of 48.
If the value of M is Multiples of 12 then it will satisfy the condition. If we If M is 12 or 24 or 36 then it ans is D but if M = 48 then answer should be 16.

Is the question right? Or am i missing some thing?
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Re: Division question [#permalink]  10 Oct 2011, 12:07
vyassaptarashi wrote:
Well m^2 is divisible by 48

if we brake 48 we will get 4*4*3

now we know that m is an integer therefore, m^2 is completely divisible by 48 then to get the square number that will be divided by 48 we need to multiply 48 by 3
so that we will get 4*4*3*3 ....
square root of the above number will be 4*3 = 12
thus m = 12 and the greatest number that will divide 12 is 12 itself ...thus the answer.

This explanation looks fine but why M shouldn't be 48. There is no mention in the question about M's maximum limit.
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Re: Division question [#permalink]  10 Oct 2011, 12:22
1
KUDOS
Well, the largest integer, which will divide m, has asked in the question. This means there will be no other integer that will divide m is larger than what the answer is.
Also, It is not asked us that which of the following will be the largest integer that will divide m....

Thus m can not be 48 because the largest integer that divides 48 will be 48 ...which is not in the options, hence can not be correct. 16 is also not correct because it is the largest number in the options but larger that 16 divisors are also present.
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Re: If m is a positive integer and m^2 is divisible by 48, then [#permalink]  10 Oct 2011, 14:45
ok, I will try to explain this as best as possible.
the following is a tree diagram, or columns. hope it is easy to see it

m^2____m__ 2 2
____m__ 2 2 3

so, 2 * 2* 3= 12

In another words,
find out the prime factors of 48, ( 2,2,2,2,3)

, you know that M^2 is m repeated twice, so you

should distribute the the prime factors evenly

among the 2 m's, the 3 will be left over so you

should just put it out there under one of the m's.

Now, for every columns take one common number

which is 2 for the first columns, 2 for the second

column and 3 in the last column. multiplying the

prime factors ( 2*2*3) = 12...

best of luck
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Re: If m is a positive integer and m^2 is divisible by 48, then [#permalink]  10 Oct 2011, 16:37
m^2 is divisible by 48

=> 48 is a factor of m^2

=> 2^4*3 is a factor of m^2

to make the above number a perfect square m^2 must have another 3 as a factor

=> (2^4)*(3^2) must be a factor of m^2

=> (2^2)*(3^1) = 12 must be a factor of m.

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Re: If m is a positive integer and m^2 is divisible by 48, then [#permalink]  10 Oct 2011, 18:31
its mentioned indirectly by saying m has to be an integer.

if m^2 is not a perfect square then m will not be an integer.

for m to be in an integer, m^2 has to be a perfect square.

lets say m^2=48*3 = 144 (perfect square ) => m =12 (integer)

lets say m^2 = 48(not a perfect square) = > m= sqrt(48) = 2*sqrt(12) (not an integer)

arcanis2000 wrote:

@Spidy001: The perfect square is what Knewton gave as part of the answer. In what part of this question does it suggest that M^2 has to be a perfect square? What am I missing?

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Re: If m is a positive integer and m^2 is divisible by 48, then [#permalink]  09 Feb 2012, 00:25
I dont Get it. Factors of 48 are= 2,2,2,2 and 3 and we know that 48 is a factor of m. Therefor 48 is also a factor of m^2.
So (2^4)*3 is a factor of m and m^2. But how do you come to the solution of 12? I don't see the link between (2^4)*3 and (2^2)*(3^2). Can someone explain please?

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Re: If m is a positive integer and m^2 is divisible by 48, then [#permalink]  09 Feb 2012, 00:49
m^2 is a divisor of 48. M =/= divisor of 48.

Find the prime factors of 48, [2,2,2,2,3]. When you root square a number you double the amount of it's divisor, so if you root them number it should be half. So divide in half all the repeating multiples. You know there must be a 3, since 3 is a prime and no 2 integer forms a multiple of 3. Also half the 2s are gone so you have a two 2s left. 3*2*2 = 12.
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Re: If m is a positive integer and m^2 is divisible by 48, then [#permalink]  09 Feb 2012, 02:25
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M3tm4n wrote:
I dont Get it. Factors of 48 are= 2,2,2,2 and 3 and we know that 48 is a factor of m. Therefor 48 is also a factor of m^2.
So (2^4)*3 is a factor of m and m^2. But how do you come to the solution of 12? I don't see the link between (2^4)*3 and (2^2)*(3^2). Can someone explain please?

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If m is a positive integer and m^2 is divisible by 48, then the largest positive integer that must divide m is?
(A) 3
(B) 6
(C) 8
(D) 12
(E) 16

m^2 is a positive perfect square divisible by 48, the least such perfect square is 144=48*3=12^2. Thus the least value of m is 12, which means that m in any case must be divisible by 12.

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Hope it helps.
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Re: If m is a positive integer and m^2 is divisible by 48, then [#permalink]  09 Feb 2012, 02:36
Am i right if i say:
M^2 must be a multiple of 48. 48 in prime factors: 2^4 * 3^1. Thus at least the interger must have one 3 and of course a few 2´s

Greated integer which is divisible by m and an factor of m^2 would be 12...
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Re: If m is a positive integer and m^2 is divisible by 48, then [#permalink]  09 Feb 2012, 02:43
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MSoS wrote:
Am i right if i say:
M^2 must be a multiple of 48. 48 in prime factors: 2^4 * 3^1. Thus at least the interger must have one 3 and of course a few 2´s

Greated integer which is divisible by m and an factor of m^2 would be 12...

Algebraic way: $$m^2=48*k=2^4*3*k$$, where $$k$$ is some positive integer. $$m=\sqrt{2^4*3*k}=2^2*\sqrt{3k}$$ --> the least value of $$k$$ for which $$m$$ is an integer (hence the least value of $$m$$) is for $$k=3$$ --> $$m=2^2*\sqrt{3*3}=12$$, hence $$m$$ in any case is divisible by 12..

Hope it's clear.
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Re: If m is a positive integer and m^2 is divisible by 48, then [#permalink]  09 Feb 2012, 02:46
yeah it is clear! Thanks, i wish i could explain it always like you did now.
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Re: If m is a positive integer and m^2 is divisible by 48, then [#permalink]  09 Feb 2012, 06:01
Bunnel thanks for the explanation
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Re: If m is a positive integer and m^2 is divisible by 48, then [#permalink]  24 Aug 2014, 12:43
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Re: If m is a positive integer and m^2 is divisible by 48, then [#permalink]  21 Jul 2015, 12:36
Bunuel wrote:
MSoS wrote:
Am i right if i say:
M^2 must be a multiple of 48. 48 in prime factors: 2^4 * 3^1. Thus at least the interger must have one 3 and of course a few 2´s

Greated integer which is divisible by m and an factor of m^2 would be 12...

Algebraic way: $$m^2=48*k=2^4*3*k$$, where $$k$$ is some positive integer. $$m=\sqrt{2^4*3*k}=2^2*\sqrt{3k}$$ --> the least value of $$k$$ for which $$m$$ is an integer (hence the least value of $$m$$) is for $$k=3$$ --> $$m=2^2*\sqrt{3*3}=12$$, hence $$m$$ in any case is divisible by 12..

Hope it's clear.

Thanks alot for the help here. I have a quick doubt, the question is asking for the largest value of the largest value of m. Why are we finding the least value of m. What am I missing here.

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Re: If m is a positive integer and m^2 is divisible by 48, then [#permalink]  21 Jul 2015, 13:08
neeraj609 wrote:
Bunuel wrote:
MSoS wrote:
Am i right if i say:
M^2 must be a multiple of 48. 48 in prime factors: 2^4 * 3^1. Thus at least the interger must have one 3 and of course a few 2´s

Greated integer which is divisible by m and an factor of m^2 would be 12...

Algebraic way: $$m^2=48*k=2^4*3*k$$, where $$k$$ is some positive integer. $$m=\sqrt{2^4*3*k}=2^2*\sqrt{3k}$$ --> the least value of $$k$$ for which $$m$$ is an integer (hence the least value of $$m$$) is for $$k=3$$ --> $$m=2^2*\sqrt{3*3}=12$$, hence $$m$$ in any case is divisible by 12..

Hope it's clear.

Thanks alot for the help here. I have a quick doubt, the question is asking for the largest value of the largest value of m. Why are we finding the least value of m. What am I missing here.

The question is asking for the largest values from the given options. You are supposed to find which one of the 5 options gives you the correct answer. You are confusing between finding the "maximum possible value" when no options are given and "maximum possible value" out of the given options. For our use, the least value worked . If lets say the least did not work, we could have looked for a multiple of 12 (say 24 or 26 or 48 etc.).
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Re: If m is a positive integer and m^2 is divisible by 48, then [#permalink]  21 Jul 2015, 16:18
Thanks alot for the response. I was also able find the explanation from on of the another post from Bunnel.
Re: If m is a positive integer and m^2 is divisible by 48, then   [#permalink] 21 Jul 2015, 16:18
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