Question

Show that exactly one of the numbers n, n+2, or n+4 is divisible by 3.

Answer 1

let n be any positive integer and b=3
n =3q+r
where q is the quotient and r is the remainder
0_ so the remainders may be 0,1 and 2
so n may be in the form of 3q, 3q=1,3q+2


CASE-1

IF N=3q
n+4=3q+4
n+2=3q+2
here n is only divisible by 3

CASE 2
if n = 3q+1
n+4=3q+5
n+2=3q=3
here only n+2 is divisible by 3

CASE 3
IF n=3q+2
n+2=3q+4
n+4=3q+2+4
=3q+6
here only n+4 is divisible by 3



HENCE IT IS JUSTIFIED THAT ONE AND ONLY ONE AMONG n,n+2,n+4 IS DIVISIBLE BY 3 IN EACH CASE .

Answer 2

assume n = 3q + r where r = 0,1,2

when r =0   n = 3q divisible by 3

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

n+4 = 3q + 4 which is not divisible

when r = 1  n = 3q + 1 which is not divisible

n+2 = 3q + 3 = 3[q+1] which is divisible

n+4 = 3q + 5 which is not divisible

when r = 2  n = 3q + 2 which is not divisible

n+2 = 3q + 4 which is divisible

n+4 = 3q +6 = 3[q + 2] which is divisible