Question

show that 1 and only 1 out of n,(n+2),(n+4) is divisible by 3. while n is natural number.​

Answer 1

Solution:

let n be any positive integer and b=3

n =3q+r

where q is the quotient and r is the remainder

0_ <r<3

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

let n be any positive integer

where b=3

using euclids division lemma

n=bq+r , where r=0,1,2

when r=0 , n=3q

when r=1 , n=3q+1

when r=2 , n=3q+2

case 1- when n =3q then n is divisible by 3

case 2- when n =3q+1 , then n+2=3q+3 which is divisble by 3

case 3- when n=3q+2,then n+4=3q+6 which is also divisible by 3

hence prooved