Question

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

Answer 1

Let n be any positive integer.
By E.D.L
n=3q+r. (where q is quotient and r is remainder)
= n=3q+r where r=0,1,2
n=3q, 3q+1, 3q+2

Case 1 = when n=3q
n is divisible by 3
n+2= 3q+2
it is not divisible by 3
n+4= 3q+4
it is not divisible by 3
Case 2 = when n=3q+1
n=3q+1 it is not divisible by 3
n+1= (3q+1)+2.
3q+3 it is divisible by 3
n+4=3q+1+3+1. it is not divisible by 3
Case 3 = when n = 3q+2
n= 3q+2 it is not divisible by 3
n+2=(3q+2)+2 it is not divisible by 3
n+4=(3q+2)+4
3q+6 it is divisible by 3

Answer 2

▶ Question :-

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

▶ 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