Cubes And Multiples Of 7
Prove that for any number that is not a multiple of seven, then its cube will be one more or one less than a multiple of 7.
Given n is any number that is not evenly divisible by 7, let n = 7a + b, where b = 1,2,3,4,5,6.
n3 = (7a + b)(7a + b)(7a + b)
Clearly the only term under expansion that does not have at least one multiple of 7 will be b3 and as b = 1,2,3,4,5,6 we get b3 = 1,8,27,64,125,216.
b3 1,1,-1,1,-1,-1 mod 7, respectively.
Hence n3 1 mod 7, where n is not a multiple of 7.
n3 -1,0,1 mod 7, where nN.