Question

show that the numbers n, n+2, n+4 only one of them are divisible by 3

Answer 1

Hey

Here is your answer,

We applied Euclid Division algorithm on n and 3.

a = bq +r on putting a = n and b = 3

n = 3q +r , 0i.e n = 3q -------- (1),n = 3q +1 --------- (2), n = 3q +2 -----------(3)

n = 3q is divisible by 3
or n +2 = 3q +1+2 = 3q +3 also divisible by 3
or n +4 = 3q + 2 +4 = 3q + 6 is also divisible by 3

Hence n, n+2 , n+4 are divisible by 3.

Hope it helps you!

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.