Question

Q.12. Show that only one of the numbers n, nd2 and 4 is divisible by 3.

Answer 1

Answer:

We know that any positive integer is of the form 3q or 3q + 1 or 3q + 2 for some integer q & one and only one of these possibilities can occur

Case I : When n = 3q

In this case, we have,

n=3q, which is divisible by 3

n=3q

= adding 2 on both sides

n + 2 = 3q + 2

n + 2 leaves a remainder 2 when divided by 3

Therefore, n + 2 is not divisible by 3

n = 3q

n + 4 = 3q + 4 = 3(q + 1) + 1

n + 4 leaves a remainder 1 when divided by 3

n + 4 is not divisible by 3

Thus, n is divisible by 3 but n + 2 and n + 4 are not divisible by 3

Case II : When n = 3q + 1

In this case, we have

n = 3q +1

n leaves a reaminder 1 when divided by 3

n is not divisible by 3

n = 3q + 1

n + 2 = (3q + 1) + 2 = 3(q + 1)

n + 2 is divisible by 3

n = 3q + 1

n + 4 = 3q + 1 + 4 = 3q + 5 = 3(q + 1) + 2

n + 4 leaves a remainder 2 when divided by 3

n + 4 is not divisible by 3

Thus, n + 2 is divisible by 3 but n and n + 4 are not divisible by 3

Case III : When n = 3q + 2

In this case, we have

n = 3q + 2

n leaves remainder 2 when divided by 3

n is not divisible by 3

n = 3q + 2

n + 2 = 3q + 2 + 2 = 3(q + 1) + 1

n + 2 leaves remainder 1 when divided by 3

n + 2 is not divsible by 3

n = 3q + 2

n + 4 = 3q + 2 + 4 = 3(q + 2)

n + 4 is divisible by 3

Thus, n + 4 is divisble by 3 but n and n + 2 are not divisible by 3