Show that one and only out of n , n+2 , n+4 Is divisible by 3 where n is any positive integer.
Show that one and only out of n , n+2 , n+4 Is divisible by 3 where n is any positive integer.
By Euclid's division Lemma any natural number can be written as: b = aq + r
where r = 0, 1, 2,.........(a-1). and q is the quotient.
put a = 3: b = 3q+r and r = 0,1,2.
thus any number is in the form of 3q , 3q+1 or 3q+2.
case I: if n =3q
n is divisible by 3,
n+2 = 3q+2 is not divisible by 3, as it leaves remainder 2.
n+4 = 3q+4 = 3(q+1)+1 is not divisible by 3, as it leaves remainder 1.
case II: if n =3q+1
n = 3q+1 is not divisible by 3 (leaves remainder 1)
n+2 = 3q+1+2=3q+3 = 3(q+1) is divisible by 3.
n+4 = 3q+1+4 = 3q+5 = 3(q+1)+2 is not divisible by 3 (remainder 2)
case III: if n = 3q+2
n =3q+2 is not divisible by 3. (leaves remainder 2)
n+2 = 3q+2+2 =3q+4 = 3(q+1)+1 is not divisible by 3.( leaves remainder 1)
n+4 = 3q+2+4 = 3q+6 = 3(q+2) is divisible by 3.
Thus one and only one out of n , n+2, n+4 is divisible by 3.
By Euclid's division Lemma any natural number can be written as: b = aq + r
where r = 0, 1, 2,.........(a-1). and q is the quotient.
put a = 3: b = 3q+r and r = 0,1,2.
thus any number is in the form of 3q , 3q+1 or 3q+2.
case I: if n =3q
n is divisible by 3,
n+2 = 3q+2 is not divisible by 3, as it leaves remainder 2.
n+4 = 3q+4 = 3(q+1)+1 is not divisible by 3, as it leaves remainder 1.
case II: if n =3q+1
n = 3q+1 is not divisible by 3 (leaves remainder 1)
n+2 = 3q+1+2=3q+3 = 3(q+1) is divisible by 3.
n+4 = 3q+1+4 = 3q+5 = 3(q+1)+2 is not divisible by 3 (remainder 2)
case III: if n = 3q+2
n =3q+2 is not divisible by 3. (leaves remainder 2)
n+2 = 3q+2+2 =3q+4 = 3(q+1)+1 is not divisible by 3.( leaves remainder 1)
n+4 = 3q+2+4 = 3q+6 = 3(q+2) is divisible by 3.
Thus one and only one out of n , n+2, n+4 is divisible by 3.
