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​

Answer 1

Step-by-step explanation:

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.

Answer 2

We know that By using Euclid's division lemma

a = bq+r, 0≤r≤b

Now, take b = 3

Let r = 0, 1, 2

for 'n' is any positive integer

Now we can write n = a

a = bq+r

n = 3q+r

n = 3q

➸ Let r = 0

➸ Let r = 1

n = 3q+1

➸Let r = 2

n = 3q+2

Case (i) :- n = 3q

from statement

➳ n

3q

➳ n+2

3q+2

➳ n+4

3q+4

∴ Clearly only one of the integer is n, n+2, n+4 is divisible by 3.

Case (ii) :- n = 3q+1

➳ n

3q+1

➳ n+2

(3q+1)+2

3q+3

➳ n+4

(3q+1)+4

3q+5

∴ Clearly only one of the integer is n, n+2, n+4 is divisible by 3.

Case (iii) :- n = 3q+2

➳ n

3q+2

➳ n+2

(3q+2)+2

3q+4

➳ n+4

(3q+2)+4

3q+6

3(q+2)

∴ Clearly only one of the integer is n, n+2, n+4 is divisible by 3.

Tq