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The greatest common factor of positive integers m and n is 12. What is

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The greatest common factor of positive integers m and n is 12. What is [#permalink]

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The greatest common factor of positive integers m and n is 12. What is the greatest common factor of (2m^2, 2n^2)?

A. 2
B. 12
C. 24
D. 144
E. 288

Kudos for a correct solution.
[Reveal] Spoiler: OA

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Re: The greatest common factor of positive integers m and n is 12. What is [#permalink]

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Bunuel wrote:
The greatest common factor of positive integers m and n is 12. What is the greatest common factor of (2m^2, 2n^2)?

A. 2
B. 12
C. 24
D. 144
E. 288

Kudos for a correct solution.


METHOD-1

Given : GCD of (m and n) = 12 = 2^2*3

i.e. m and n are both multiples of 2^2*3

i.e. m^2 and n^2 will both be multiples of (2^2*3)^2 = 2^4*3^2

i.e. 2m^2 and 2n^2 will both be multiples of 2(2^2*3)^2 = 2^5*3^2 = 288

Answer: Option E

METHOD-2

Let, m = 12 and n = 24
i.e. GCD of m and n = 12

2m^2 = 288
2n^2 = 1152
i.e. GCD of 2m^2 and 2n^2 = 288

Answer: Option E
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The greatest common factor of positive integers m and n is 12. What is [#permalink]

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New post 08 Jul 2015, 04:31
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Bunuel wrote:
The greatest common factor of positive integers m and n is 12. What is the greatest common factor of (2m^2, 2n^2)?

A. 2
B. 12
C. 24
D. 144
E. 288

Kudos for a correct solution.


GCD (m,n)=12

Thus \(m = 2^2*3*p\)
and \(n = 2^2*3*q\), with p,q co-primes

Now \(2m^2 = 2^5*3^2*p^2\)
and \(2n^2 = 2^5*3^2*q^2\)

Thus GCD (\(2m^2, 2n^2\)) = \(2^5*3^2\) = 288. Thus E is the correct answer.

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Re: The greatest common factor of positive integers m and n is 12. What is [#permalink]

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New post 13 Jul 2015, 02:24
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Bunuel wrote:
The greatest common factor of positive integers m and n is 12. What is the greatest common factor of (2m^2, 2n^2)?

A. 2
B. 12
C. 24
D. 144
E. 288

Kudos for a correct solution.


800score Official Solution:

Suppose we factorize m and n into prime factors. The greatest common factor of positive integers m and n is 12 so the only prime factors m and n have in common are 2, 2 and 3. (12 = 2 × 2 × 3).

If we factorize m² into prime factors we will get each of prime factors of m twice. The same happens to n². So the only prime factors m² and n² would have in common are 2, 2, 2, 2 and 3, 3.

If we factorize 2m² into prime factors we will get the same prime factors as for m² and one more prime factor “2”. The same happens to n². So the only prime factors 2m² and 2n² would have in common are 2, 2, 2, 2, 2 and 3, 3. By factoring those we get the greatest common factor of 2m² and 2n².

2 × 2 × 2 × 2 × 2 × 3 × 3 = 288.

The answer is (E).
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: The greatest common factor of positive integers m and n is 12. What is [#permalink]

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New post 22 Jul 2015, 09:30
Hello,

I have a question. Can we also use the relationship between LCM and GCF to find the solution?

So, we would say:
m*n =12*x, where x is the LCM. Then,
m^2*n^2 = (12x)^2
m^2*n^2 = (12x)^2 = 144x^2. Finally,

2(m)^2 * 2 (n)^2 = 2 (144x^2)
2(m)^2 * 2 (n)^2 = 228*2x^2.

So, we end up with 228, which is E.

Is this correct?

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Re: The greatest common factor of positive integers m and n is 12. What is [#permalink]

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Re: The greatest common factor of positive integers m and n is 12. What is [#permalink]

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New post 16 Aug 2017, 09:34
Hello from the GMAT Club BumpBot!

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The greatest common factor of positive integers m and n is 12. What is [#permalink]

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New post 16 Aug 2017, 14:03
Bunuel wrote:
The greatest common factor of positive integers m and n is 12. What is the greatest common factor of (2m^2, 2n^2)?

A. 2
B. 12
C. 24
D. 144
E. 288

Kudos for a correct solution.

Kinda dorky, but I'm all for simple if it works. I just wrote out, in stages, what factors m and n had to have.

LCM of m and n is 12
12 = 2 * 2 * 3

m: 2, 2, 3
n: 2, 2, 3

Variables squared? Each number has exactly one copy, so

m\(^2\): 2, 2, 3, 2, 2, 3
n\(^2\): 2, 2, 3, 2, 2, 3

Then each term * 2?

2m\(^2\): 2, 2, 3, 2, 2, 3, 2
2n\(^2\): 2, 2, 3, 2, 2, 3, 2

Both have 2\(^5\), 3\(^2\). (32 * 9) = 288

Answer E

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The greatest common factor of positive integers m and n is 12. What is   [#permalink] 16 Aug 2017, 14:03
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