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Math Help - LCM involving algebra

  1. #1
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    LCM involving algebra

    Three lighthouses flash at intervals of 12,45 and x minutes respectively. If all of them flash at the same time at 12 hours interval, find the smallest / largest possible of x.

    I have prime factorized the numbers - please advise how to proceed??
    12 = 2^2 x 3
    45 = 3^2 x 5
    12 hours = 720 mins = 2^4 x 3^2 x 5
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  2. #2
    Member kalyanram's Avatar
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    Re: LCM involving algebra

    Hi Drdj,
    Firstly lets observe the numbers
    12 = 2^2.3
    45 = 3^2.5
    720 = 2^4.3^2.5
    so we observe that the LCM lacks a term that supplies 2^4 i.e if had taken the LCM of just 12,45 we would have ended up in 180 = 2^2.3^2.5
    so the least value of x = 16 in that case we have all the factors we would need.

    Now other feasible values of x are those that have 2^4 along with other factors like 48 = 2^4.3, 90 = 2^4.5,....,and last(and definitely not the least ) 720 itself so the maximum values of x = 720.

    Kalyan.
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  3. #3
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    Re: LCM involving algebra

    Quote Originally Posted by kalyanram View Post
    Hi Drdj,
    Firstly lets observe the numbers
    12 = 2^2.3
    45 = 3^2.5
    720 = 2^4.3^2.5
    so we observe that the LCM lacks a term that supplies 2^4 i.e if had taken the LCM of just 12,45 we would have ended up in 180 = 2^2.3^2.5
    so the least value of x = 16 in that case we have all the factors we would need.

    Now other feasible values of x are those that have 2^4 along with other factors like 48 = 2^4.3, 90 = 2^4.5,....,and last(and definitely not the least ) 720 itself so the maximum values of x = 720.

    Kalyan.
    Just a question.. why the minimum value of x can't be be 1?
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  4. #4
    Member kalyanram's Avatar
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    Re: LCM involving algebra

    Well take x = 1 and compute the LCM. You will see lcm(12,45,1) = 180 and not 720 which is actually required.

    Kalyan.
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