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M14-04

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New post 15 Sep 2014, 23:50
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  45% (medium)

Question Stats:

59% (00:54) correct 41% (01:02) wrong based on 101 sessions

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New post 15 Sep 2014, 23:50
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New post 05 Sep 2016, 19:13
Hello, what is the shortest working out steps to find the common divisor?
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New post 07 May 2017, 00:39
unleesooj wrote:
Hello, what is the shortest working out steps to find the common divisor?


I did prime factorization in order to calculate the number of factors:

\(48=2^4*3^1\)

\(factors=5*2=10\)

and then divided 48 by ascending numbers:
\(48/1\)
\(48/2\)
\(48/3\)
\(48/4\)
\(...\)

This took me 2:15 for this question. Nevertheless, the very first step is not necessary, it was helping me not to forget any factor.
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New post 07 May 2017, 02:17
36=1,2,3,4,6,9,12,18,36
48=1,2,3,4,6,8,12,16,24,48
common sum=1+2+3+4+6+12=28
ans: E
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New post 16 Jun 2017, 10:59
1. Find the GCD of the numbers
2. Factors of GCD are also factors of the two numbers -> List out all the factors of the GCD.
3. Sum them up.
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New post 23 Aug 2018, 05:36
Quote:
1. Find the GCD of the numbers
2. Factors of GCD are also factors of the two numbers -> List out all the factors of the GCD.
3. Sum them up.


Bunuel
Could you please confirm whether the above method should hold true for general cases? IMO the GCD should have all the common factors of a list of numbers. And there is that general formula to calculate the sum of all factors of a number, using which we can calculate the sum of factors of GCD. This should ideally provide a generic method to arrive at the sum of common factors for all numbers?
Please correct me if I am mistaken.

Thanks
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New post 23 Aug 2018, 21:52
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shubhajit wrote:
Quote:
1. Find the GCD of the numbers
2. Factors of GCD are also factors of the two numbers -> List out all the factors of the GCD.
3. Sum them up.


Bunuel
Could you please confirm whether the above method should hold true for general cases? IMO the GCD should have all the common factors of a list of numbers. And there is that general formula to calculate the sum of all factors of a number, using which we can calculate the sum of factors of GCD. This should ideally provide a generic method to arrive at the sum of common factors for all numbers?
Please correct me if I am mistaken.

Thanks

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Yes, that approach is generally true.
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New post 17 Nov 2018, 01:09
Hi guys,
Any shorter way to do this. This took me 3.5+ minutes. listed down all factors by brute force :(

Please guide
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New post 18 Nov 2018, 03:24
ShankSouljaBoi wrote:
Hi guys,
Any shorter way to do this. This took me 3.5+ minutes. listed down all factors by brute force :(

Please guide


Hi,

1.Write down prime factorization of both numbers.

\(48 = 3 * 2^4\)
\(36 = 3^2 * 2^2\)

2. Next find GCD of the numbers.
GCD \(= 3 * 2^2\) (To find GCD just write down the common prime factors. Then choose their lowest powers.)

3.Write down all the numbers u can make from this factorization.
\(1, 2, 3, 3*2 , 3*2*2 , 2*2\)
Sum them up! And you have the answer. This can be done within 2 minutes easily if u just be careful and practice step 3!

Hope it helps!
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M14-04 &nbs [#permalink] 18 Nov 2018, 03:24
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