Question

show that exactly one of the numbers n, n + 2 or n + 4 is divisible by 3

Answer 1

Answer:

Step-by-step explanation:

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

Answer 2

Let n = 3k, 3k + 1 or 3k + 2.

(i) When n = 3k :  n is divisible by 3.

n + 2 = 3k + 2

⇒ n + 2 is not divisible by 3.

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

⇒ n + 4 is not divisible by 3.

When n = 3k + 1  n is not divisible by 3.

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

= 3k + 3 = 3 (k + 1)

⇒ n + 2 is divisible by 3.

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

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

⇒ n + 4 is not divisible by 3.

When n = 3k + 2  n is not divisible by 3.

n + 2 = (3k + 2) + 2

= 3k + 4

= 3 (k + 1) + 1

⇒ n + 2 is not divisible by 3.

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

= 3k + 6

= 3(k + 2)

⇒ n + 4 is divisible by 3. 1

Therefore, exactly one of the numbers n, n + 2 or  n + 4 is divisible by 3.