Question

☺✌☺heya_______________________________________________




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

Answer 1

heya frnd here is ur answer..........
on dividing n by 3 , let q be the quotient and r be the remainder.
then, n = 3q + r, where 0 is less than or equal to r < 3
=> n = 3q + r where r = 0,1,2
=> n = 3q or n = 3q + 1 or n = 3q + 2.
case 1. if n = 3q, then n is divisible by 3.
case 2. if n =3q + 1 , then ( n + 2 ) = 3q + 3 = 3 ( q + 1 ) , which is divisible by 3.
case 3. when n = 3q + 2 , then ( n + 4 ) = 3q + 6 = 3 ( q + 2 ), which is divisible by 3.
so, in this case, ( n + 4 ) is divisible by 3.
hence, one and only one out of n, n+2, n+4 is divisible by 3.
hope it will help you.... ☺☺

Answer 2

let q be the quotient and r be the remainder when n is divisible by 3.
therefore a=bq+r
n=3q+r where r=0,1,2
n=3q or n= 3q+1 or n=3q+2

in first case =if n=3q then n is divisible by 3 n+2 and n +4 are not divisible by3.

in second case = if n=3q +1 then n+2 =3q+3 =3( q+1) which is divisible by 3 and n+4=3q+5 which is not divisible by 3

in third case = if n = 3q+2 then n +2=3q+4 which is divisible by 3. and (n+4)=3q +6=3(q+2)which is divisible by,3 .

so only (n +4) is divisible by 3 .
hence one and only one out of n, (n+2),(n +4)is divisible by 3.