N is a natural number such that when N3 divided by 9 ,it leaves remainder a.It can be concluded that
a)a is a perfect square b) a is a perfect cube c) both of these d ) none of these
a is a perfect cube.
by the Euclid's Lemma any number can be represented as either of 3k , 3k+1 or 3k+2 form
case 1: let N = 3k
which is divisible by 9. thus remainder a = 0
case 2: let N = 3k+1
thus when is divided by 9, remainder a = 1, which is a perfect cube.
case 3: let N = 3k+2
thus when is divided by 9, remainder a = 8 which is a perfect cube.
hope this helps you.
by the Euclid's Lemma any number can be represented as either of 3k , 3k+1 or 3k+2 form
case 1: let N = 3k
which is divisible by 9. thus remainder a = 0
case 2: let N = 3k+1
thus when is divided by 9, remainder a = 1, which is a perfect cube.
case 3: let N = 3k+2
thus when is divided by 9, remainder a = 8 which is a perfect cube.
hope this helps you.