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

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.

According to Euclid’s division Lemma,

Let the positive integer = n

And b=3

n =3q+r, where q is the quotient and r is the remainder

0<r<3 implies remainders may be 0, 1 and 2

Therefore, n may be in the form of 3q, 3q+1, 3q+2

When n=3q

n+2=3q+2

n+4=3q+4

Here n is only divisible by 3

When n = 3q+1

n+2=3q=3

n+4=3q+5

Here only n+2 is divisible by 3

When n=3q+2

n+2=3q+4

n+4=3q+2+4=3q+6

Here only n+4 is divisible by 3

So, we can conclude that one and only one out of n, n + 2 and n + 4 is divisible by 3.

Hence Proved