"to show that for all sets of 6 consecutive naturals {k,k+1,k+2,k+3,k+4,k+5}, where k is a natural that is NOT divisible by 5, then 5 will divide exactly one of the elements of the set (and will do the job).Therefore, all that remains is to prove the result for all {5m,5m+1,5m+2,5m+3,5m+4,5m+5}, where m is a natural.It is also trivial to prove that 7 will do the job for such a set if and only if" . . . .