Question

Show that one and only one out of n, (n + 2) and (n + 4) is divisible by 3, where n is any positive integer. (CBSE 2008)
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Answer 1

When dividing any integer by 3, the remainder can be 0,1 or 2. In other words it can be of the form of $3k,3k+1$ or $3k+2$ where $k\in\mathbb{Z}$

If $n=3k$ then $n+2=3k+2$ and $n+4=3k+4$. Only $n$ is divisible by 3.

If $n=3k+1$ then $n+2=3k+3=3(k+1)$ and $n+4=3k+5$. Only $n+2$ is divisible by 3.

If $n=3k+2$ then $n+2=3k+4$ and $n+4=3k+6=3(k+2)$. Only $n+4$ is divisible by 3.

Answer 2

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......

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