Question

Using edl to prove that one and only one out of n, n+2 and n+4 is divisible by 3

Answer 1

Answer:

Step-by-step explanation:

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

as per euclid division lemma a=bq+r

in the given situation b = 1 q = n

as per the rule 0<=r<b

so r should be less than b

as b=1 r should be 0

so n is the one and only factor of 3

n+2 has remainder 2 , n+4 has remainder 4 so they are not divisible through euclid division lemma