Question

Use Division Algorithm to show that the square of any postive integer is of the form 9m, 9m+1 or 9m+8.

Chapter: Real Numbers
Class 10: Secondary School​

Answer 1

Step-by-step explanation:

⇒ EUCLID'S ALGORITHM: a = bq + r (0 ≤ r < b)

                                        a = 9q + r (0 ≤ r < 9) --- (1)

Given: To prove that any cube positive integer is of the form 9m, 9m+1 or 9m+8.

b = 9, r = 0,1,2,3,4,5,6,7

If r = 0, put in Eq (1),

a = bq + r

a = 9q + r

If r = 1, put in Eq (1),

 a = 9q + 1

If r = 2, put in Eq (1)

 a = 9q + 2

∴ We have proved Cube of an Integer is in the form of 9m (or) 9m + 1 (or) 9m + 8 .

We get, a = 9q, a = 9q + 1, a = 9q + 8

= a = 9q³

= a³= (9q³)

= a³ = 9³q³ -----> a (9²q²) when 9²q³ ∈  $I$

= a³ = 9m

= a³ = 9m + 1

= a³ = (9m + 1)³ (a + 3)³ - a² - b² + 3a²b + 3ab

= a³ = (9m)³ + 1³ + 3 (9m)² + 39 m²

= a³ = 9³m³ + 1 + 39m² + 3.9

= a³ = 9³m³ + 6.9m² + 3.9m + 1

= a³ = 9 (9²m³ 1.3m²) + 1

= a²m³ + 3m² + 3m ∈  $I$

= a³ = 9m + 1

= a = 9m + 2 |  a³ = (9m + 2)³

We know,

a = 9m + 2

a³ = (9m + 2)³ (a+b)³ = a³ + b³ + 3a²b + 3ab²

a³ = 9³m³ + 8 + 3 (9m) 2 + 3 (9m) 2²

a³ = 9³m³ + 8 + 3 (9m) + 2 + 3 (9m) 4

a³ = 9³m³ + 6 x 9²m² + 12 (9m) + 8

a³ = 9m (9m² + 6.9m + 12) + 8

∴ 9m² + 54. + 12 ∈ i

a³ = 9m + 8

∴ The Cube of any positive integer are in the form of 9m (or) 9m + 1 (or) 9m + 8.

This Question is done using b = 9, if you want to do your Sum using b = 3 then, Check this Link below:

brainly.in/question/10344818

Answer 2

Answer:

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