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

I'm showing it as something like mathematical induction.

Let n = 3k.

So n = 3k is divisible by 3.

When n + 2 = 3k + 2 is divided by 3, we get remainder 2.

When n + 4 = 3k + 4 is divided by 3, we get remainder 1.

(∵ 3k + 4 = 3k + 3 + 1 = 3(k + 1) + 1)

∴ Only n among n, n + 2 and n + 4 is divisible by 3 here.

Let n = 3k + 1.

So when n = 3k + 1 is divided by 3, we get remainder 1.

n + 2 = 3k + 1 + 2 = 3k + 3 is divisible by 3.

(∵ 3k + 3 = 3(k + 1))

When n + 4 = 3k + 1 + 4 = 3k + 5 is divided by 3, we get remainder 2.

(∵ 3k + 5 = 3k + 3 + 2 = 3(k + 1) + 2)

∴ Only n + 2 among  n, n + 2 and n + 4 is divisible by 3.

Let n = 3k + 2.

n = 3k + 2 leaves remainder 2 on division by 3.

n + 2 = 3k + 2 + 2 = 3k + 4 leaves remainder 1 on division by 3. (Saw earlier)

But  n + 4 = 3k + 2 + 4 = 3k + 6 = 3(k + 2)  is divisible by 3.

∴ Only n + 4 among  n, n + 2 and n + 4 is divisible by 3.

If we take n = 3k + 3, it repeats the procedure of n = 3k, as 3k + 3 = 3(k + 1).

∴ If n is any positive integer, and when we consider n, n + 2 and n + 4, only one among them is divisible by 3.

Hence it's shown!!!

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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.