Question

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

Answer 1

Let n = 3k, 3k + 1 or 3k + 2.

(i) When n = 3k: 
n is divisible by 3. 
n + 2 = 3k + 2 ⇒ n + 2 is not divisible by 3. 
n + 4 = 3k + 4 = 3(k + 1) + 1 ⇒ n + 4 is not divisible by 3.

(ii) When n = 3k + 1: 
n is not divisible by 3. 
n + 2 = (3k + 1) + 2 = 3k + 3 = 3(k + 1) ⇒  n + 2 is divisible by 3.
n + 4 = (3k + 1) + 4 = 3k + 5 = 3(k + 1) + 2 ⇒ n + 4 is not divisible by 3.

(iii) When n = 3k + 2: 
n is not divisible by 3. 
n + 2 = (3k + 2) + 2 = 3k + 4 = 3(k + 1) + 1 ⇒ n + 2 is not divisible by 3. 
n + 4 = (3k + 2) + 4 = 3k + 6 = 3(k + 2) ⇒ n + 4 is divisible by 3. 

Hence exactly one of the numbers n, n + 2 or n + 4 is divisible by 3.

Answer 2

Step-by-step explanation:

Euclid's division Lemma any natural number can be written as: .

where r = 0, 1, 2,. and q is the quotient.

∵ Thus any number is in the form of 3q , 3q+1 or 3q+2.

→ Case I: if n =3q

⇒n = 3q = 3(q) is divisible by 3,

⇒ n + 2 = 3q + 2 is not divisible by 3.

⇒ n + 4 = 3q + 4 = 3(q + 1) + 1 is not divisible by 3.

→ Case II: if n =3q + 1

⇒ n = 3q + 1 is not divisible by 3.

⇒ n + 2 = 3q + 1 + 2 = 3q + 3 = 3(q + 1) is divisible by 3.

⇒ n + 4 = 3q + 1 + 4 = 3q + 5 = 3(q + 1) + 2 is not divisible by 3.

→ Case III: if n = 3q + 2

⇒ n =3q + 2 is not divisible by 3.

⇒ n + 2 = 3q + 2 + 2 = 3q + 4 = 3(q + 1) + 1 is not divisible by 3.

⇒ n + 4 = 3q + 2 + 4 = 3q + 6 = 3(q + 2) is divisible by 3.

Thus one and only one out of n , n+2, n+4 is divisible by 3.

Hence, it is solved.