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

   n,  n + 2,  n + 4   

All numbers can be classified in to three groups:
   n =  3 k  multiples of 3
      = 3 k + 1    ie., when divided by 3,  they give a remainder 1
      = 3 k + 2  ie., when divided by 3, give a remainder 2.

If  n = 3 k ,  then  it is divisible by 3.
   then  n + 2 = 3 k +2  or n+4 = 3k + 4 is not divisible by 3.

If n = 3 k + 1,  then  n+2= 3k +3  divisible by 3.
                 then n+4 = 3 k + 5  is not divisible by 5.

If n = 3 k + 2 ,  then  n + 4 = 3 k + 6  is divisible by 3.
                 then n + 2 = 3 k  + 4 is not divisible by 3.


so exactly one of them only is divisible by 3 for any n.

Answer 2

Step-by-step explanation:

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.