Question

show that one and only one out of n, n+2,n+4 is Diviseble by 3, where n is here positive integer ​

Answer 1

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

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

→ 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.

$\boxed{\boxed{\bold{Thus \ one\ and \ only \ one \ out \of\ n ,\ n+2,\ n+4\ is\ divisible\ by\ 3.}}}$

                                                     

Answer 2

Proof:

Let 'n' is any positive integer .

.°. n = 3q+r

Where, q is the quotient and r is the remainder

.°. r will vary from 0≤ r <3

.°. The remainders can be 0,1 and 2

.°. n may be in the form of (3q), (3q+1) and (3q+2)

Now, we have 3 cases.

CASE-I

When n = 3q

=> n+2= 3q+2

=> n+4= 3q+4

Here, only n is divisible by 3

CASE II

When n = 3q+1

=> n+2 = 3q+3

=> n+4 = 3q + 5

Here, only n+2 is divisible by 3

CASE III

When n = 3q+2

=> n+2 = 3q+4

=> n+4 = 3q+6

Here, only n+4 is divisible by 3

Thus, in each case one and only one out of n, (n+2) and (n+4) is divisible by 3.

Hence, Proved.