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Math Help - Simple Number Theory

  1. #1
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    Simple Number Theory

    Positive integers 30, 72, and N have the property that the product of any two of them is divisible by the third. What is the smallest possible value of N?

    Note I have not yet taken a Number Theory course.

    I think I have found the solution using a bit of reasoning and some luck. N=60? I figured N could not be smaller than the gcd of 30 and 72, and could not be greater than their product. I also found a pattern for (30N)/72. Inputting 10, 15, 20, 30 for N gave a result of 25/6, 25/4, 25/3, 25/2, respectively. I figured that this converged to 25/1, which would then be my solution. N=60 indeed yields 25/1.

    However, I feel that this is closer to luck than anything else, and also it is not very elegant. Can someone show me another way of doing this, perhaps something more elegant?


    -F
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  2. #2
    Senior Member roninpro's Avatar
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    Your reasoning seems okay.

    After noting that \gcd(30,72)=6\leq N\leq 30\cdot 72=2160, I would probably list all of my possibilities at this point: 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 27, 30, 36, 40, 45, 48, 54, 60, 72, 80, 90, 108, 120, 135, 144, 180, 216, 240, 270, 360, 432, 540, 720, 1080, 2160.

    Now, 30 divides 72N. Since 30=2\cdot 3\cdot 5 and 72=2^3\cdot 3^2, we must conclude that N is a multiple of 5. This leaves 10, 15, 20, 30, 40, 45, 60, 80, 90, 120, 135, 180, 240, 270, 360, 540, 720, 1080, 2160.

    On the other hand, 72 divides 30N. By similar reasoning as before, N must be a multiple of 2^2\cdot 3=12. Eliminating the bad possibilities leaves 60 as the smallest number satisfying all three conditions.
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  3. #3
    Junior Member NowIsForever's Avatar
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    8|72 so for 8 to divide 30N, 4|N. 5|30, thus 5|N, since 5∤72. 3|N since 9|72 and 9∤30. Thus N must be at minimum 235 = 60.
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  4. #4
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    Consider the three numbers

    ab ~,~ bc ~,~ ca

    the set satisfies the requirement because

     (ab)(bc) = b^2 (ca)

     (bc)(ca) = c^2 (ab)

     (ca)(ab) = a^2 (bc)


    Let  ab = 30 ,  bc = 72 so  N = ca .

    To minimize  N = ca , we have to maximize  b so obviously what we are looking for is the greatest common divisor of  30 and  72 which is  6


    Therefore  N = 5(12) = 60
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  5. #5
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    Hello, FlacidCelery!

    Positive integers 30, 72 and N have the property
    that the product of any two of them is divisible by the third.
    What is the smallest possible value of N ?

    We have: . \begin{array}{ccccc}A &=&30 &=& 2\cdot3\cdot5 \\ B &=& 72 &=& 2^3\cdot3^2 \end{array}


    A times N is divisible by B: . \frac{2\cdot3\cdot5\cdot N}{2^3\cdot3^2} is an integer.
    This reduces to: . \frac{5N}{12} . . . Hence. N is a multiple of 12.


    B times N is divisible by A: . \frac{2^3\cdot3^2\cdot N}{2\cdot3\cdot5} is an integer.
    This reduces to: . \frac{2^2\cdot3\cdot N}{5} . . . Hence, N is a multiple of 5.


    The least number which is a multiple of 12 and a multiple of 5 is: . N \:=\:60

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  6. #6
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    Quote Originally Posted by Soroban View Post
    Hello, FlacidCelery!


    We have: . \begin{array}{ccccc}A &=&30 &=& 2\cdot3\cdot5 \\ B &=& 72 &=& 2^3\cdot3^2 \end{array}


    A times N is divisible by B: . \frac{2\cdot3\cdot5\cdot N}{2^3\cdot3^2} is an integer.
    This reduces to: . \frac{5N}{12} . . . Hence. N is a multiple of 12.


    B times N is divisible by A: . \frac{2^3\cdot3^2\cdot N}{2\cdot3\cdot5} is an integer.
    This reduces to: . \frac{2^2\cdot3\cdot N}{5} . . . Hence, N is a multiple of 5.


    The least number which is a multiple of 12 and a multiple of 5 is: . N \:=\:60




    I really like this solution, it seems the more obvious to me. Thanks a lot everyone.
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