Question

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

Answer 1

Answer:

hope that hepls

Step-by-step explanation:

To prove n , n+2 or n+4 is divisible by 3,

Case 1 Let n is divisible by 3

So n = 3k for some positive integer k

Now n+2 = 3k+2 which is not divisible by 3

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

Case 2 n+2 is divisible by 3  so

n+2 =3k

so n =3k-2  not divisible by  3

n+4 = 3k+2 = not divisible by  3

Case 3: Let n+4 is divisible by 3 so n+4 = 3k

n =3k-4=3(k-1)-1=3m-1 not divisible by 3

n+2 = 3k-2 not 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.