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Math Help - lowest common multiple

  1. #1
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    lowest common multiple

    The lowest common multiple of 2, x and 21 is 126. Find the smallest possible value of x
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  2. #2
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    126 = 2\cdot3\cdot21.
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  3. #3
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    that was what i did!! hi-5

    but wait the answer is not 3 but 9.18 or 63
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  4. #4
    Newbie Exotique's Avatar
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    Quote Originally Posted by maddas View Post
    126 = 2\cdot3\cdot21.
    The lowest common multiple of 2, 3 and 21 is 42.

    You just need to find the smallest number that is a factor of 126, but which 2 OR 21 are not factors, and which that number is not a factor of 2 OR 21 . The smallest number that fits those criteria is 63.
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  5. #5
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    I didn't quite understand that... perhaps a working would be nice

    and with reference to "and which that number is not a factor of 2 OR 21" , isnt 21 a factor of 63?
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  6. #6
    Newbie Exotique's Avatar
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    Quote Originally Posted by Punch View Post
    and with reference to "and which that number is not a factor of 2 OR 21" , isnt 21 a factor of 63?
    Yes, but 2 isn't. Hence the OR.

    I don't know if I could show any working, I really just worked it out by common sense and logic. Put it this way; it can't be 3 because the lowest common multiple of 2, 3 and 21 is 42, not 126. Maybe the easiest way to try and explain it is to divide the LCM by 2, which gives you 63. Since 2 doesn't go into 63, the next multiple of 63 which is divisible by both 2 and 21 is 126, so x = 63.

    You could divide 126 by 3, which gives 42, but the LCM of 2, 21 and 42 is 42. Similarly you could keep dividing 126 by integers until you reach another number which isn't divisible by both 2 and 21, and that would be your x, but the only number which satisfies the criteria I gave in the previous post is 63 (well, so does 126 itself, but you are asked to find the smallest possible number for x).

    I hope this makes it a little clearer. I'll hunt around for a formula or I might just have to make one up.


    EDIT:

    Here's a way to check that the number you've found is correct.

    lcm(a,b)=\frac{|a\cdot b|}{gcd(a,b)} where gcd is the greatest common divisor of a and b. Here, choose one of your given factors (2 or 21), and the factor you'd like to check (in this case, 63). It's best to test with both given factors, as we'll see.
    So we get:
    lcm(2,63)=\frac{2\cdot63}{1}=2\cdot63=126

    However if you used 21 instead of 2, you would get:
    lcm(21,63)=\frac{21\cdot63}{21}=63

    So you see why it's important to test both factors.

    This formula is originally designed to find the LCM, not to work the other way around, but it does provide a good way to test your answer.

    Hope this helps.
    Last edited by Exotique; April 25th 2010 at 07:29 AM.
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  7. #7
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    well , nevertheless the smallest value of x is 9. LCM of 2,9,21 is 126
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  8. #8
    MHF Contributor undefined's Avatar
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    Quote Originally Posted by luhter View Post
    well , nevertheless the smallest value of x is 9. LCM of 2,9,21 is 126
    I agree, the correct answer is 9.

    126 = 2^1\cdot3^2\cdot7^1

    The relevant primes are 2, 3, and 7.

    2=2^1\cdot3^0\cdot7^0

    21=2^0\cdot3^1\cdot7^1

    So we can write

    x=2^a\cdot3^b\cdot7^c

    and

    \text{lcm}(2,x,21)=2^{\max(1,0,a)}\cdot3^{\max(0,1  ,b)}\cdot7^{\max(0,1,c)}

    We need x = 9 in order to get max(0,1,b) = 2.
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