Question

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

Answer 1

Answer:

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

Answer:

Let n be any positive integer

n=bq+r

n=3q+r

where,r=0,1,2,3

if r=0

n=3q

n+2=3q+2

n+4=3q+4

n=3q is divisible by 3.

r=1

n=3q+1

n+2=3q+1+2=3q+3

n+4=3q+1+4=3q+5

n=3q+3 is divisible by 3.

r=2

n=3q+2

n=3q+2+2=3q+4

n=3q+2+4=3q+6

n=3q+6 is divisible by 3