Question

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

Answer 1

n is an integer.

So,it is of the form 3k,3k+1 or 3k+2 where k is an integer.

If n=3k,then  

n+1=3k+1,not divisible by 3

n+2=3k+2,not divisible by 3

If n=3k+1(not divisible by 3),then  

n+1=3k+2,not divisible by 3

n+2=3k+3=3(k+1), divisible by 3

If n=3k+2(not divisible by 3),then  

n+1=3k+3=3(k+1),divisible by 3

n+2=3k+4=3(k+1)+1,not divisible by 3

Hence by seeing all the above cases, we can safely say that:

out of n,n+1 and n+2,only one is divisible by 3.

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.