Asked by Drake
                Find the largest integer N for which N−6 evenly divides N^3−6.
            
            
        Answers
                    Answered by
            Count Iblis
            
    N^3−6 is zero Modulo (N-6).  If we compute Modulo (N-6) then obviously:
N-6 = 0 ----->
N = 6
Here and in the following the equals sign means equality modulo N - 6.
We then have:
N^3 -6 = 6^3 - 6 = 210
Therefore:
210 = 0
Reverting back to the ordinary definition of the equals sign, this means that:
210 = k (N-6)
So, N-6 must be the largest possible factor of 210, which is 210 therefore
N = 216.
    
N-6 = 0 ----->
N = 6
Here and in the following the equals sign means equality modulo N - 6.
We then have:
N^3 -6 = 6^3 - 6 = 210
Therefore:
210 = 0
Reverting back to the ordinary definition of the equals sign, this means that:
210 = k (N-6)
So, N-6 must be the largest possible factor of 210, which is 210 therefore
N = 216.
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