Question

Show that one and only one out of n, n + 2 or, n+4 is divisible by 3, where n is any positive
integer​

Answer 1

Answer:

Step-by-step explanation:

The given number are n, n+2 and n+4

Here, let a = n, b = n+2 and c = n+4

So, order of triplet is (a,b,c) = (n,n+2,n+4), where n is any positive integer.

Now, if n = 1

(a,b,c) = (1,3,5)

if n = 2

(a,b,c) = (2,4,6)

if n = 3

(a,b,c) = (3,5,7)

if n = 4

(a,b,c) = (4,6,8)

if n = 5

(a,b,c) = (5,7,9)

Here, in the above triplets, it is observed that only one number in each triplet is divisible by 3. Hence, one and only one out of n,n+2 and n+4 is divisible by 3. [Proved]

Answer 2

Step-by-step explanation:

Euclid's division Lemma any natural number can be written as: .

where r = 0, 1, 2,. and q is the quotient.

thus any number is in the form of 3q , 3q+1 or 3q+2.

case I: if n =3q

n = 3q = 3(q) is divisible by 3,

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

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

case II: if n =3q + 1

n = 3q + 1 is not divisible by 3.

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

n + 4 = 3q + 1 + 4 = 3q + 5 = 3(q + 1) + 2 is not divisible by 3.

case III: if n = 3q + 2

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

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

n + 4 = 3q + 2 + 4 = 3q + 6 = 3(q + 2) is divisible by 3.

thus one and only one out of n , n+2, n+4 is divisible by 3.

Hence, it is solved

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