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LCM and GCD can som1 suppose any safe and easy method?
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07 Jun 2011, 00:22
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If the LCM of A and 12 is 36, what are the possible values of A?




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Re: LCM and GCD can som1 suppose any safe and easy method?
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07 Jun 2011, 10:44
dimri10 wrote: If the LCM of A and 12 is 36, what are the possible values of A? Think of what LCM means before going ahead. If I say LCM of two numbers is \(36 (= 4*9 = 2^2 * 3^2)\), it means that at least one of them must have a \(2^2\) and at least one of them must have a \(3^2\) (It is not possible that both numbers have just 3 because then, the LCM would have just 3, not 9) If one number is \(12 (= 2^2 * 3)\), the other number A must have \(3^2\) since 12 doesn't have it. So minimum value of A will be 9. A can also have a 2 or a \(2^2\) so other possible values are \(18 (=9*2)\) and \(36 (= 9*2^2)\) Also, A cannot have any other factors since if it did, then the LCM would have to have that factor too.
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Re: LCM and GCD can som1 suppose any safe and easy method?
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07 Jun 2011, 11:36
Here is what I do in my head (mental tricks): LCM ( 12,36) > (12/ 12,36/ 12) > (1, 3) > 3* 12 or 1*36 > 36 LCM ( 12, 30) > (12/ 6,30/ 6) > (2, 5) > 5* 12 or 2*30 > 60 LCM ( 10, 28) > (10/ 2,28/ 2) > (5, 14) > 14* 10 or 5*28 > 140 GCD (12,36) > (12/ 12,36/ 12) > (1,3) > 12GCD (12, 30) > (12/ 6,30/ 6) > (2,5) > 6GCD (10, 28) > (10/ 2,28/ 2) > (5,14) > 2LCM and GCD have a nice property: LCM(x,y)*GCD(x,y) = xy For example, LCM(12,30)*GCD(12,30) = 60*6 = 12 * 30
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Re: LCM and GCD can som1 suppose any safe and easy method?
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08 Jun 2011, 02:24
i thank both of you, but i did not get walker's approach.can you please sharpen your explenation?
let's say that there are 3 numbers: 440,120 and 80.how can you use your shortcut



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Re: LCM and GCD can som1 suppose any safe and easy method?
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08 Jun 2011, 03:03
LCM ( 80, 120, 440) > (80/ 40, 120/ 40, 440/ 40) > (2, 3, 11) > 3* 11* 80 > 2640 GCD (80, 120, 440) > (80/ 40, 120/ 40, 440/ 40) > (2, 3, 11) > 4040 here is the greatest common divisor.
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If the LCM of a amd 12 is 36 what could be the possible values of a?
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25 Jun 2011, 13:45
If the LCM of a & 12 is 36 what could be the possible values of a? The LCM of 12 and a contains two 2's. Since the LCM contains each prime factor to the power it appears the MOST, we know that a cannot contain more than two 2's. LCM of 12 and a contain two 3's. But 12 only contains one 3. The 3^2 factor in the LCM must have come from prime factorization of a. Thus we know that a contains exactly two 3's. Since a must contain exactly two 3's nd can contain no 2's, one 2 or two 2's a could be 3*3=9 3*3*2=18 3*3*2**2=36. Thus 9,18 and 36 are three values. I am struggling to understand the concept.
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Re: If the LCM of a amd 12 is 36 what could be the possible values of a?
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25 Jun 2011, 14:40
12 = (2^2)*3 a = (2^2 or 2^1 or 2^0)*(3^2)
L.C.M of a & 12 = 36 = (2^2 )*(3^2) = product of Maximum powers of each prime factor
This can be easily understood by comparing 12 with 36 .
12 has one 3 , but L.C.M which is the maximum powers of each prime factor has two 3's
=> a must have two 3's in it = 3^2 1
12 has two 2's. L.c.m which is the maximum powers of each prime factor has two 2's as well.
=> a could have zero 2's or one 2 or two 2's2
clubbing above two statements marked 1 *2 , we have a = (2^2 or 2^1 or 2^0)*(3^2) =(4 or 2 or 1)*9 => a could be 36 or 18 or 9.
Hope its clear now.



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Re: LCM and GCD can som1 suppose any safe and easy method?
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15 Jan 2016, 01:24
VeritasPrepKarishma wrote: dimri10 wrote: If the LCM of A and 12 is 36, what are the possible values of A? Think of what LCM means before going ahead. If I say LCM of two numbers is \(36 (= 4*9 = 2^2 * 3^2)\), it means that at least one of them must have a \(2^2\) and at least one of them must have a \(3^2\) (It is not possible that both numbers have just 3 because then, the LCM would have just 3, not 9) If one number is \(12 (= 2^2 * 3)\), the other number A must have \(3^2\) since 12 doesn't have it. So minimum value of A will be 9. A can also have a 2 or a \(2^2\) so other possible values are \(18 (=9*2)\) and \(36 (= 9*2^2)\) Also, A cannot have any other factors since if it did, then the LCM would have to have that factor too. Thanks Karishma. Your solution is very clear and in simple way. I didn't understand the complicated solution prepared by Manhattan. Their prime columns technique is good but solution explanation was confusing.



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Re: If the LCM of a amd 12 is 36 what could be the possible values of a?
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15 Mar 2018, 14:40
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