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

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
or n +2  = 3q +1+2 = 3q +3 also divisible by 3
or n +4 = 3q + 2 +4 = 3q + 6 is also divisible by 3
Hence n, n+2 , n+4 are divisible by 3.

Solution: 
By Euclid’s division lemma, we have 
a = bq + r;  0 ≤ r < b 
For a = n and b = 3, we have 
n = 3q + r, …(i) 
where q is an integer 
and 0 ≤ r < 3, i.e. r = 0, 1, 2. 
Putting r = 0 in (i), we get 
n = 3q 
So, n is divisible by 3. 
n + 2 = 3q + 2 
so, n + 2 is not divisible by 3. 
n + 4 = 3q + 4 
so, n + 4 is not divisible by 3. 
Putting r = 1 in (i), we get 
n = 3q + 1 
so, n is not divisible by 3. 
n + 2 = 3q + 3 = 3(q + 1) 
so, n + 2 is divisible by 3. 
n + 4 = 3q + 5

so, n + 4 is not divisible by 3. 
Putting r = 2 in (i), we get 
n = 3q + 2 
so, n is not divisible by 3. 
n + 2 = 3q + 4 
so, n + 2 is not divisible by 3. 
n + 4 = 3q + 6 = 3(q + 2) 
so, n + 4 is divisible by 3 
Thus for each value of r such that 0 ≤ r < 3  only one out of n, n + 2 and n + 4 is divisible by 3.