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 the largest positive integer that must divide m is?

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

Is there a strategic approach to this question? Can any experts help?

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by EconomistGMATTutor » Thu Nov 02, 2017 4:05 pm
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

Is there a strategic approach to this question? Can any experts help?
Hi ardz24,
Let's take a look at your question.

The question states:
"If m is a positive integer and m^2 is divisible by 48, then the largest positive integer that must divide m is? "

The largest positive integer that must divide m will be m itself.
One way to solve this question is to check each of the option if it is true or not.
For that purpose, square each of the given options and find out if it is divisible by 48 or not.

Let's start from option A.
Let m = 3 then m^2 = 9
9 is not divisible by 48 because it is less than 48.

Option B
Let m = 6 then m^2 = 36
36 is not divisible by 48 because it is less than 48.

Option C
Let m = 8 then m^2 = 64
64 is not divisible by 48 because it gives a decimal quotient when we try to divide 64 by 48.

Option D
Let m = 12 then m^2 = 144
144 is divisible by 48 because 144/48 = 3

Hence, the largest positive integer that must divide m is 12.
Therefore, Option D is correct.

Hope it helps.
I am available if you'd like any follow up.
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by Scott@TargetTestPrep » Fri Nov 01, 2019 6:50 pm
BTGmoderatorAT wrote: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

Is there a strategic approach to this question? Can any experts help?
We are given that m^2/48 = integer or (m^2)/(2^4)(3^1) = integer.

However, since m^2 is a perfect square, we need to make 48, or (2^4)(3^1), a perfect square. Since all perfect squares consist of unique prime, each raised to an even exponent, the smallest perfect square that divides into m^2 is (2^4)(3^2) = 144.

Thus, m^2/144 = integer

Since m^2 is divisible by 144, we see that the largest value that must divide m is 12.

Answer: D

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by [email protected] » Tue Nov 26, 2019 12:50 pm
Hi All,

This question has a number of logic 'shortcuts' built into it, so if you were low on time you could logically guess the correct answer. Here's how you could do it... To start, you have to note that the answer choices are numbers and one of them IS the LARGEST positive integer that will divide into M.

Since M is an integer and M^2 is divisible by 48, you should be able to logically deduce that M will be divisible by both 2 (since 48 is even) and 3 (you can use the 'rule of 3' to quickly determine that fact). So we need a number that is divisible by BOTH 2 and 3. With that information alone, we can eliminate Answer choices A, C and E. The prompt asks for the LARGEST positive integer, so between the two remaining answers, and since 48 has so many 2s in it - and that the remaining answers are 6 and 12 - it's highly likely that the correct answer is 12 (which you can prove using the other approaches offered here).

Final Answer: D

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by Scott@TargetTestPrep » Thu Nov 28, 2019 4:21 pm
BTGmoderatorAT wrote: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

Is there a strategic approach to this question? Can any experts help?
We are given that m^2/48 = integer, or (m^2)/(2^4)(3^1) = integer.

However, since m^2 is a perfect square, we need to make 48, or (2^4)(3^1), a perfect square. Since all perfect squares consist of unique primes, each raised to an even exponent, the smallest perfect square that divides into m^2 is (2^4)(3^2) = 144.

Thus, m^2/144 = integer

Since m^2 is divisible by 144, we see that the largest value that must divide m is 12.

Answer: D

Scott Woodbury-Stewart
Founder and CEO
[email protected]

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