Question

the sum of all possible remainders which can be obtained when the cube of a natural number is divided by 9 is?​

Answer 1

Answer

9

Explanation

An integer 'a' can be written as the form of a = bq + r where b, q and r are integers and 0 ≤ r < b.

When we keep b as 3, we get that r can be 0, 1 or 2

Thus a natural number 'n' can be written as

n = 3q or n = (3q + 1) or n = (3q + 2)

Case 1, n = 3q

n = 3q

Cubing both sides, we get

n³ = (3q)³

⇒ n³ = 27q³

⇒ n³ = 9(3q³)

⇒ n³ = 9m (for some integer m = 3q³)

Since n³ = 9m, we can conclude that when cube of a natural number of the form 3q is divided by 9, the remainder comes as 0.

Case 2, n = 3q + 1

n = 3q + 1

⇒ n³ = (3q + 1)³

⇒ n³ = 27q³ + 27q² + 9q + 1  (using (a + b)³ = a³ + 3a²b + 3ab² + b³)

⇒ n³ = 9(3q³ + 3q² + q) + 1

⇒ n³ = 9m + 1 (for some integer m = 3q³ + 3q² + q)

Here, we can see that the remainder is 1.

Case 3, n = 3q + 2

n = 3q + 2

⇒ n³ = (3q + 2)³

⇒ n³ = 27q³ + 27q² + 18q + 8

⇒ n³ = 9(3q³ + 3q² + 2q) + 8

⇒ n³ = 9m + 8 (for some integer m = 3q³ + 3q² + 2q)

Here, the remainder is 8

As we have seen that in all cases, the possible remainders are 0, 1 and 8, so the sum of all possible remainders = 0 + 8 + 1 = 9