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

Given, n is positive integer and

If ,n =3q+r

where q is the quotient and r is the remainder

0<r<3

so the remainders can be 0,1 and 2

so n can be written in the form of 3q, 3q=1,3q+2

Situation-1

IF n=3q

n+4=3q+4

n+2=3q+2

In this case n is only divisible by 3

CASE 2

if n = 3q+1

n+4=3q+5

n+2=3q+3=3*(q+1)

Only n+2 is divisible by 3

Situation 3

If n=3q+2

n+2=3q+4

n+4=3q+2+4

=3q+6

Only n+4 is divisible by 3

So,in all three cases only one is divisible.

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

THANKS