Question

show that only one form of the number n, n + 2 and n + 4 is divisible by 3.​

Answer 1

Answer:

-2

Step-by-step explanation:

3n+6=0

3n=-6

3n/3=6/3

n=-2

Answer 2

:

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

So, we have following cases:

Case-I: When n=3q

In this case, we have

n=3q, which is divisible by 3

Now, n=3q

n+2=3q+2

n+2 leaves remainder 2 when divided by 3

Again, n=3q

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

n+4 leaves 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 remainder 1 when divided by 3.

n is divisible by 3

Now, n=3q+1

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

n+2 is divisible by 3.

Again, n=3q+1

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

n+4 leaves 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.

Now, 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 divisible by 3.

Again, n=3q+2

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

n+4 is divisible by 3.

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