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.

therefore 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.

therefore b3 congruent 1,1,-1,1,-1,-1 mod 7, respectively.

Hence n3 congruent plus or minus1 mod 7, where n is not a multiple of 7.


n3 congruent -1,0,1 mod 7, where nis in the setN.

Problem ID: 24 (Nov 2000)     Difficulty: 3 Star

Only Show Problem