Question

show that one and only one out of n,n+2,n+4 is divisible by 3,where n is any positive integer

Answer 1

We applied Euclid Division algorithm on n and 3.
a = bq +r  on putting a = n and b = 3
n = 3q +r  , 0<r<3
i.e n = 3q   -------- (1),n = 3q +1 --------- (2), n = 3q +2  -----------(3)
n = 3q is divisible by 3
(2) n +2  = 3q +1+2 = 3q +3 also divisible by 3
(3)n +4 = 3q + 2 +4 = 3q + 6 is also divisible by 3
Hence n, n+2 , n+4 are divisible by 3.

Answer 2

On dividing n by 3,let q be the quotient and r be the remainder .
Then,n=3q+r,where 0 greater than equal to r and less than 3.
#n=3q+r,where r=0,1and 2
#n=3q or n=3q+1 or n=3q+2
Case1:if n=3q, then n is divisible by 3 but n+2 and n+4 are not divisible by 3.
Case2:if n=3q+1,
then n+2 =3q+1+2 =3q+3 =3(q+1),
which is divisible by 3.
But, here n and n+4 are not divisible by 3.
Case3:if n= 3q+2,
then n+4=3q+2+4=3q+6=3(q+2),which is divisible by 3
But, here also, n and n+2 are not divisible by 3.
Hence,one and only one out of n,n+2,n+4 is divisible by 3.