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Math Help - Word problem

  1. #1
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    Word problem

    If 6 time j is 1 more than the square of k, where k is an integer, what is the smallest possible value of j?

    How can I show that it is indeed 1/6 ?

    (Converting into symbols: 6j=k^2 + 1 but I'm stuck..)
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  2. #2
    Senior Member MaxJasper's Avatar
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    Re: Word problem

    You should prove that there exist no integer k such that (1+k^2) be divisible by 2 & 3 at the same time. 1+k^2 either is odd or even hence in each case is not divisible either by 2 or 3
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  3. #3
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    Re: Word problem

    Quote Originally Posted by donnagirl View Post
    If 6 time j is 1 more than the square of k, where k is an integer, what is the smallest possible value of j?
    How can I show that it is indeed 1/6 ?
    (Converting into symbols: 6j=k^2 + 1 but I'm stuck..)
    If you use 0 as smallest integer , then k=0, so 6j = 1 : j = 1/6
    If you use 1 as smallest integer , then k=1, so 6j = 2 : j = 1/3
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