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The greatest common factor of positive integers m and n is 12. What is
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08 Jul 2015, 02:43
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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.
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Re: The greatest common factor of positive integers m and n is 12. What is
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08 Jul 2015, 02:50
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. METHOD1Given : GCD of (m and n) = 12 = 2^2*3i.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 METHOD2Let, 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
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08 Jul 2015, 03:31
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 coprimes 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
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13 Jul 2015, 01:24
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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Re: The greatest common factor of positive integers m and n is 12. What is
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22 Jul 2015, 08:30
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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The greatest common factor of positive integers m and n is 12. What is
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16 Aug 2017, 13:03
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? 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, 3n\(^2\): 2, 2, 3, 2, 2, 3Then each term * 2? (The last step. We must find 2m\(^2\) and 2n\(^2\)) 2m\(^2\): 2, 2, 3, 2, 2, 3, 22n\(^2\): 2, 2, 3, 2, 2, 3, 2Both 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
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18 Sep 2018, 10:20
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. 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



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Re: The greatest common factor of positive integers m and n is 12. What is
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23 Sep 2018, 15:32
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 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
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Re: The greatest common factor of positive integers m and n is 12. What is
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12 Oct 2019, 16:01
GMATinsight wrote: 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. METHOD1Given : GCD of (m and n) = 12 = 2^2*3i.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 METHOD2Let, 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 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?



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Re: The greatest common factor of positive integers m and n is 12. What is
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13 Oct 2019, 00:40
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. 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
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Re: The greatest common factor of positive integers m and n is 12. What is
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29 Mar 2020, 06:23
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. 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
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Re: The greatest common factor of positive integers m and n is 12. What is
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30 Mar 2020, 01:50
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.




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