Question

Show that a number and its cube leaves the same remainder when divided by 6

Answer 1

Using Euclid's Division Lemma,

a = bq + r

Let b = 6.

Thus,

a = 6q + r

$As\ \ 0 \leq r < b\ \ \ \ \ \Longrightarrow\ \ \ 0 \leq r < 6, \\ \\ r \in \{0, 1, 2, 3, 4, 5\}$

1. Let r = 0.

=> a = 6q + 0

=> a = 6q

Here the number leaves remainder 0 on division by 6.

=> a³ = (6q)³

=> a³ = 216q³

=> a³ = 6 × 36q³

=> a³ = 6m [m = 36q³]

Here its cube also leaves remainder 0.

2. Let r = 1.

=> a = 6q + 1

Here the number leaves remainder 1.

=> a³ = (6q + 1)³

=> a³ = 216q³ + 108q² + 18q + 1

=> a³ = 6(36q³ + 18q² + 3q) + 1

=> a³ = 6m + 1 [m = 36q³ + 18q² + 3q]

Here its cube also leaves remainder 1.

3. Let r = 2.

=> a = 6q + 2

Here the number leaves remainder 2.

=> a³ = (6q + 2)³

=> a³ = 216q³ + 216q² + 72q + 8

=> a³ = 216q³ + 216q² + 72q + 6 + 2

=> a³ = 6(36q³ + 36q² + 12q + 1) + 2

=> a³ = 6m + 2 [m = 36q³ + 36q² + 12q + 1]

Here its cube also leaves remainder 2.

4. Let r = 3.

=> a = 6q + 3

Here the number leaves remainder 3.

=> a³ = (6q + 3)³

=> a³ = 216q³ + 324q² + 162q + 27

=> a³ = 216q³ + 324q² + 162q + 24 + 3

=> a³ = 6(36q³ + 54q² + 27q + 8) + 3

=> a³ = 6m + 3 [m = 36q³ + 54q² + 27q + 8]

Here its cube also leaves remainder 3.

5. Let r = 4.

=> a = 6q + 4

Here the number leaves remainder 4.

=> a³ = (6q + 4)³

=> a³ = 216q³ + 432q² + 288q + 64

=> a³ = 216q³ + 432q² +288q + 60 + 4

=> a³ = 6(36q³ + 72q² + 48q + 10) + 4

=> a³ = 6m + 4 [m = 36q³ + 72q² + 48q + 10]

Here its cube also leaves remainder 4.

6. Let r = 5.

=> a = 6q + 5

Here the number leaves remainder 5.

=> a³ = (6q + 5)³

=> a³ = 216q³ + 540q² + 450q + 125

=> a³ = 216q³ + 540q² + 450q + 120 + 5

=> a³ = 6(36q³ + 90q² + 75q + 20) + 5

=> a³ = 6m + 5 [m = 36q³ + 90q² + 75q + 20]

Here its cube also leaves remainder 5.

Hence Proved!!!