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Math Help - GCDs and LCMs

  1. #1
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    GCDs and LCMs

    Natural numbers a,b,c are such that :

    i) GCD of a,b is 2^33^25^47^1
    ii) GCD of b,c is 2^43^25^5
    iii) GCD of a,c is 2^33^25^4
    iv) Product of abc is 2^1^13^1^05^1^77^2

    Then find the least common multiple (LCM) of a,b,c.

    please help me. thank you.
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  2. #2
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    Quote Originally Posted by amey View Post
    Natural numbers a,b,c are such that :

    i) GCD of a,b is 2^33^25^47^1
    ii) GCD of b,c is 2^43^25^5
    iii) GCD of a,c is 2^33^25^4
    iv) Product of abc is 2^1^13^1^05^1^77^2

    Then find the least common multiple (LCM) of a,b,c.

    please help me. thank you.
    It's obvious (I hope) that each of the numbers a, b, c must be of the form 2^?3^?5^?7^?, and that the lcm will be the product of those four primes each raised to the highest of the powers that occur in a, b or c.

    You can deal with each of those four primes separately. Take 7 first, because that's the easiest. Each of a, b contains a 7 (to some nonzero power), but the product abc only contains 7 to the power 2. Therefore a, b each contain 7 to the power 1, and c does not contain 7 at all. So the highest power of 7 in the three numbers is 7^1, and hence the lcm will have a factor of 7.

    Now think about the powers of 3. Each pair of numbers has 3^2 in its gcd. So each of the three numbers contains 3 to the power at least 2. But if any pair of numbers contained 3 to a power greater than 2 then so would the gcd of those two numbers. Hence two of the numbers a, b, c must contain 3 to the power exactly 2. But the product abc contains 3^{10}. So if two of the numbers have 3^2 then the third one must have 3^6 (because 2+2+6=10). Thus the lcm must have 3 to the power 6.

    Now do a similar analysis of the powers of 2 and 5, and you will find that the lcm is 2^?3^65^?7^1 (where I have left the two ?s for you to find).
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  3. #3
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    Thanks. I got the remaining ones as 2^4 and 5^8.
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