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Hint: First all, we know that any positive integer can be written in the form of 3q, 3q + 1 and 3q + 2. So take n = 3q, 3q + 1, 3q + 2. From this, find the values of (n + 2) and (n + 4). Then check which numbers are divisible by 3.

Complete step-by-step answer:

Here, we have to show that one and only one out of (n) (n + 2) or (n + 4) is divisible by 3, where n is any positive integer. We know that according to Euclid’s division lemma, any natural number ‘b’ can be written as b = aq + r where r = 0, 1, 2, 3, …..(a – 1).

Here ‘a’ is the divisor, ‘q’ is the quotient and ‘r’ is the remainder. By substituting a = 3, we get,

b = 3q + r

where r = 0, 1, 2….

So, we get b = (3q + 0) or (3q + 1) and (3q + 2)

Thus, we get that any natural number or positive integer can be written in the form of 3q, 3q + 1 and 3q + 2.

First of all, let us take, n = 3q……(i)

This means that n is divisible by 3.

By adding 2 on both sides of the equation (i), we get,

n + 2 = 3q + 2 which is not divisible by 3.

By adding 4 on both the sides of equation (i), we get,

n + 4 = 3q + 4 = 3(q + 1) + 1 which is not divisible by 3.

Hence, in this case, we get that only n is divisible by 3 among (n), (n + 2) and (n + 4).

Now, let us take n = 3q + 1…..(ii)

This means that n is not divisible by 3.

By adding 2 on both the sides of equation (ii), we get,

n + 2 = 3q + 1 + 2 = 3q + 3

Or, n + 2 = 3 (q + 1) which is divisible by 3.

By adding 4 on both the sides of the equation (ii), we get,

n + 4 = 3q + 1 + 4 = 3q + 5

Or, n + 4 = 3 (q + 1) + 2 which is not divisible by 3.

Hence, in this case, we get that only (n + 2) is divisible by 3 among (n), (n + 2) and (n + 4).

Now, let us take, n = 3q + 2….(iii)

This means that n is not divisible by 3.

By adding 2 on both the sides of equation (iii), we get,

n + 2 = 3q + 2 + 2 = 3q + 4

Or, n + 2 = 3 (q + 1) + 1 which is not divisible by 3.

By adding 4 on both the sides of equation (iii), ,we get,

n + 4 = 3q + 2 + 4 = 3q + 6

Or, n + 4 = 3 (q + 2) which is divisible by 3.

Hence, in this case, we get that only (n + 2) is divisible by 3 among n, (n + 2) and (n + 4).

Therefore, we get that one and only one out of (n), (n + 2) and (n + 4) is divisible by 3 where n is any positive integer.

Note:

Students should remember Euclid’s division lemma which is the base of these types of questions. Also, students can assume the same value of n and verify if their answer is correct or not. Suppose we take n = 1, so we get n, (n + 2) and (n + 4) as 1, 3 and 5 respectively. So, we get that (n + 2) = 3 is the only number out of (n), (n + 2) and (n + 4) which is divisible by 3. Similarly, we can take other values of n and check.

Complete step-by-step answer:

Here, we have to show that one and only one out of (n) (n + 2) or (n + 4) is divisible by 3, where n is any positive integer. We know that according to Euclid’s division lemma, any natural number ‘b’ can be written as b = aq + r where r = 0, 1, 2, 3, …..(a – 1).

Here ‘a’ is the divisor, ‘q’ is the quotient and ‘r’ is the remainder. By substituting a = 3, we get,

b = 3q + r

where r = 0, 1, 2….

So, we get b = (3q + 0) or (3q + 1) and (3q + 2)

Thus, we get that any natural number or positive integer can be written in the form of 3q, 3q + 1 and 3q + 2.

First of all, let us take, n = 3q……(i)

This means that n is divisible by 3.

By adding 2 on both sides of the equation (i), we get,

n + 2 = 3q + 2 which is not divisible by 3.

By adding 4 on both the sides of equation (i), we get,

n + 4 = 3q + 4 = 3(q + 1) + 1 which is not divisible by 3.

Hence, in this case, we get that only n is divisible by 3 among (n), (n + 2) and (n + 4).

Now, let us take n = 3q + 1…..(ii)

This means that n is not divisible by 3.

By adding 2 on both the sides of equation (ii), we get,

n + 2 = 3q + 1 + 2 = 3q + 3

Or, n + 2 = 3 (q + 1) which is divisible by 3.

By adding 4 on both the sides of the equation (ii), we get,

n + 4 = 3q + 1 + 4 = 3q + 5

Or, n + 4 = 3 (q + 1) + 2 which is not divisible by 3.

Hence, in this case, we get that only (n + 2) is divisible by 3 among (n), (n + 2) and (n + 4).

Now, let us take, n = 3q + 2….(iii)

This means that n is not divisible by 3.

By adding 2 on both the sides of equation (iii), we get,

n + 2 = 3q + 2 + 2 = 3q + 4

Or, n + 2 = 3 (q + 1) + 1 which is not divisible by 3.

By adding 4 on both the sides of equation (iii), ,we get,

n + 4 = 3q + 2 + 4 = 3q + 6

Or, n + 4 = 3 (q + 2) which is divisible by 3.

Hence, in this case, we get that only (n + 2) is divisible by 3 among n, (n + 2) and (n + 4).

Therefore, we get that one and only one out of (n), (n + 2) and (n + 4) is divisible by 3 where n is any positive integer.

Note:

Students should remember Euclid’s division lemma which is the base of these types of questions. Also, students can assume the same value of n and verify if their answer is correct or not. Suppose we take n = 1, so we get n, (n + 2) and (n + 4) as 1, 3 and 5 respectively. So, we get that (n + 2) = 3 is the only number out of (n), (n + 2) and (n + 4) which is divisible by 3. Similarly, we can take other values of n and check.