Question

Prove that only one of the numbers n-1, n+1 or n+3 is divisible by 3,where n is any positive integers.

Answer 1

let n be any positive integer and b=3
applying Euclid's division lemma
n=bq+r
n=3q+r. 0>=r<3
as we know,
any positive integer is in the form of 3q,3q+1and 3q+2
case1...
when n=3q
n-1=3q-1(not divisible by 3)
n+1=3q+1(not divisble by 3)
n+3=3q+3=3(q+1) ...(divisible by 3)
here only one and only one out is divisible by 3 that is n+3
case 2...
when n=3q+1
n+1n-1=3q+1-1=3q (divisible by 3)
n+1=3q+1+1=3q+2(not divisible by 3)
n+3=3q+1+3=3q+4(not divisble by 3)
here again only one out is divisible by 3 that is n-1
similarly in third case only one and only one out is divisible by 3 that is n+1