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

Question

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

ENGRTOMBA2018 · Accepted Answer

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.

GMATinsight · Answer

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

Bunuel · Answer

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).

pacifist85 · Answer

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?

generis · Answer

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? Just take the prime factors and list them again:

m = (2 * 2 * 3)
m ^2  = (m * m) = 
(2 * 2 * 3) * (2 * 2 * 3)

After squaring both m and n we have:

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

Then each term * 2? (The last step. We must find  2 m ^2  and  2 n ^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

dave13 · Answer

if GCF of m and n is 12 i.e. \frac{m+n}{12} ----> \frac{2m^2}{GCF} + \frac{2n^2}{GCF} \frac{(2m^2 + 2n^2)}{GCF} factorize \frac{2 (m^2 + n^2)}{GCF} hence GCF = 2(m^2)= 2(12^2) = 2*144 = 288 is my reasoning correct ?

have a fantastic day everyone

ScottTargetTestPrep · Answer

Notice that 2m^2 = 2 x m x m and 2n^2 = 2 x n x n. We see that the greatest factor (gcf) of (2, 2) = 2, the gcf of first pair of (m, n) = 12 and the gcf of the second pair of (m, n) = 12. Therefore, the gcf of (2m^2, 2n^2) = 2 x 12 x 12 = 288.

Answer: E

jamalabdullah100 · Answer

In this question, you know the GCD of m and n is 12, so to find GCD for 2m^2 and 2n^2, why not just square 12 and double it? So 12*12 = 144, 144*2= 288? Isn't this an easier way to obtain the answer?

Pritishd · Answer

One of the points that may concern many would be the presence of other prime factors that  m  and  n  would contain that we have no clue about. Let's simulate that scenario by adding a couple of unknown prime factors and understand whether those unknown prime factors actually matter at all.

We are given that the HCF/GCD of  m  and  n  is 1 2  which means that the only common prime factors between  m  and  n  are two 2's and one 3 ( 12 = 2 * 2* 3 )

Say,
 m = 2 * 2 * 3 * 5 * 7 * 17 
 n = 2 * 2 * 3 * 13 * 11

The GCD of  m  and  n  from the above examples is 12. Now lets square both  m  and  n  as our ultimate goal is to find the GCD of  2m^2  and  2n^2 .
 m^2 = (2 * 2 * 3 * 5 * 7)^2 = 2^2 * 2^2 * 3^2 * 5^2 * 7^2 * 17^2 
 n^2 = (2 * 2 * 3 * 13 * 11)^2 = 2^2 * 2^2 * 3^2 * 13^2 * 11^2

The GCD of  m^2  and  n^2  from the above examples is  2^2 * 2^2 * 3^2  which is  12^2 . This shows that in GCD (where we take the lowest power of the common primes) the prime factors that are NOT common do not matter.

When we multiply  2  to  m^2  and  n^2  we get another prime factor that is common to both  m^2  and  n^2

Hence GCD of  2m^2  and  2n^2  is  2 * 2^2 * 2^2 * 3^2 = 2^5 * 3^2 = 32 * 9 =288

Ans  E

Kinshook · Answer

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

Asked: What is the greatest common factor of (2m^2, 2n^2)?

gcd (m,n) = 12
gcd (m^2,n^2) = 144
gcd (2m^2,2n^2) = 288

IMO E

Ravixxx · Answer

12 is the greatest common factor of m and n. 
It means that m = (12*x) and n= (12*x).

2m^2 = 2(12*x)^2
 = 2(144x^2)
 = 288x^2

2n^2 = 2(12*x)^2
 = 2(144x^2)
 = 288x^2

For m and n = 288 * x^2
It means that the greatest common factor is 288.

bumpbot · Answer

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