Question

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

Answer 1

Case 1:

We assume that n is divisible by 3

=> n = 3p where p is an integer
=> n+2 = 3p+2, which cannot be expressed in the form 3*(a number)
ie n+2 is not divisible by 3

n=3p

=> n+4=3p + 4
=> = 3p + 3 +1
= 3(p+1) +1
= 3q + 1, where q is an integer

Clearly 3q+1 also cannot be expressed in the form 3*(a number)
ie n+4 is not divisible by 3

Case 2:

We assume that n+2 is divisible by 3

=> n+2= 3x where x is an integer
=>n = 3x-2, which cannot be expressed in the form 3*(a number)
ie n is not divisible by 3

n+2=3x

=> n+4=3x + 2

Clearly 3x+2 also cannot be expressed in the form 3*(a number)
ie n+4 is not divisible by 3

Case 3:

We assume that n+4 is divisible by 3

n+4 = 3y where p is an integer
=>n+2 = 3y-2, which cannot be expressed in the form 3*(a number)
ie n+2 is not divisible by 3

n+4=3y

n=3y- 4
= 3y - 3 -1
= 3(y-1) -1
= 3z - 1, where z is an integer

Clearly 3z+1 also cannot be expressed in the form 3*(a number)
ie n+4 is not divisible by 3

Hence exactly one out of n, n+2 and n+4 is divisible by 3.