By using Euclid Division Lemma, Show that one and only one out of n, n+ 1,n+ 2 and n+ 3 is divisible by 4. About the author Emma

Step-by-step explanation: Solution: let n be any positive integer and b=3 where is the quotient and r is the n =3q+r remainder 0_ <r<3 so the remainders may be 0,1 and 2 so n may be in the form of =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 l 56 HENCE IT IS JUSTIFIED THAT ONE AND ONLY ONE AMONG n,n+2,n+4 IS DIVISIBLE BY 3 IN EACH CASE Reply

Step-by-step explanation:Solution:

let n be any positive integer and b=3 where is the quotient and r is the

n =3q+r

remainder

0_ <r<3

so the remainders may be 0,1 and 2 so n may be in the form of =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

l 56

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