Cube of a positive integer is in the form of 9m or 9m+1 or 9m+8 -10th class-Mathematics- Real Numbers--Exercise 1.1- 4th Problem-Euclid Division Algorithm

Math for class 10-Real Numbers-Exercise 1.1- 4th Problem

Before going to learn the Proof, We know the following

☆Euclid Division Lemma

For any pair of positive integers a and b (a＞b), there exist a unique pair q and r such that a=bq+r,  0≤ r＜b

☆ (a + b)³ = a³ + 3a²b + 3ab² + b³

☆ (a - b)³ = a³ - 3a²b + 3ab² - b³

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Prove that the Cube of a positive integer is in the form of 9m or 9m+1 or 9m+8(m is an integer)

Proof:

By Euclid Division Algorithm, we have

Pair of positive integers a and b (a＞b), there exist a unique pair q and r such that a=bq+r,  0≤ r ＜b

Let "a" be a positive integer and 

Take b=3 (Because 9 is multiple of 3)  

∴ a=3q+r,     0≤ r ＜3

a=3q+r,  value of r may be 0 or 1 or  2

Do Cube on both sides

(a)³=(3q+r)³      [apply (a + b)³ = a³ + 3a²b + 3ab² + b³ formula in RHS]

a³ = (3q)³ + 3(3q)²(r) + 3(3q)(r)² + (r)³    [Here a = 3q and b = r ]

[ Substitute different "r" values]

Case-i: Take r = 0

a³ = (3q)³ + 3(3q)²(0) + 3(3q)(0)² + (0)³

  

a³ = (3q)³ + ( 0 ) + ( 0 ) + ( 0 )

a³ = 3³q³

a³ = 27q³

a³ = 9 x 3 x q³

a³ = 9(3q³) 

a³ = 9m          (Here  m= 3q³  )

Case-ii: Take r = 1

a³ = (3q)³ + 3(3q)²(1) + 3(3q)(1)² + (1)³

a³ = 3³q³ + 3(3²q²)(1) + 3(3q)(1) + (1)

a³ = 27q³ + 3(9q²) + 9q + (1)

a³ = 27q³ + 27q² + 9q + 1

Factorize first three terms of RHS

a³ = 3x3x3xq³ + 3x3x3xq² + 3x3xq + 1

Take (3 x 3) as Common from first three terms

 

a³ = 3 x 3 x (3q³ + 3q² + q) + 1

a³ = 9(3q³ + 3q² + q) + 1

a³ = 9m + 1    (Here  P = 3q³ + 3q² + q )

Case-iii: Take r=2

a³ = (3q)³ + 3(3q)²(2) + 3(3q)(2)² + (2)³

a³ = 3³q³ + 3(3²q²)(2) + 3(3q)(4) + (2)³

a³ = 27q³ + 3(9q²)(2) + 9q(4) + (8)

a³ = 27q³ + (27q²)(2) + 36q + 8

a³ = 27q³ + 54q² + 36q + 8

Factorize the terms of RHS

a³ = 3x3x3xq³ + 3x3x3x2xq² + 3x3x2x2q + 2x2x2

Take (3 x 3) as Common from first three terms

 

a³ = 3 x 3 x (3q³ + 3x2xq² + 2x2xq) + 2x2x2

a³ = 9(3q³ + 6q² + 4q) + 8

a³ = 9m + 8    (Here  P = 3q³ + 6q² + 4q )

From above three cases, we to conclude that

"Cube of every positive integer is in the form of  9m or 9m+1 or 9m+8"

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^{Click the link to see the Proof of " Square of a positive integer is in the form of 5P or 5P+1 or 5P+4"}

Model Problem for Practice:

1. Use Euclid Division lemma to Show that the cube of any positive integer is in the form of  4m or 4m+1 or 4m+3.( for m is an integer)