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  1. #1
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    operations

    Gary performed the following three operations: 15 x, 21 x and 6 x. If he never got a remainder and x is an integer greater than 1, what is the value of x?
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  2. #2
    Senior Member Dinkydoe's Avatar
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    From these assumptions follows:

    ax = 15
    bx = 21
    cx = 6

    Where x> 1, a,b,c are natural numbers.

    Now we can write out the possible values of x
    x|6 gives possible values x= 2,3,6
    x|21 gives possible values x = 3,7,21
    x|15 gives possible values x = 3,5,15

    What is the only possible number x in all 3 lists?
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  3. #3
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    Quote Originally Posted by sri340 View Post
    Gary performed the following three operations: 15 x, 21 x and 6 x. If he never got a remainder and x is an integer greater than 1, what is the value of x?
    Or \frac{e}{f}=g\ \Rightarrow\ \frac{e}{g}=f

    Hence x=\frac{15}{a}=\frac{21}{b}=\frac{6}{c}

    The prime factors of 6 are 1, 2 and 3.
    15 and 21 are divisible by 3 but not by 2.
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