Question

Use Euclid�s division lemma to show that the cube of any positive integer is of the form 9m,9m+1or9m+8.

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Answer 1

SOLUTION :-

Let us consider two positive numbers a and b where b= 3

We know that According to Euclid’s Division Lemma

a = bq + r { condition for r is (0 ≤ r < b)}

a = 3q + r —————(i) {b=3}

so r is an integer which lies in between o and 3

Hence r can be either 0, 1 or 2.

Case 1: When r = 0, the equation (i) becomes

a = 3q

On cubing both the sides, we get

a³= (3q)³= 27 q³ = 9 (3q³) = 9m

a³= 9m {where m = 3q³}

Case 2: When r = 1, the equation (i) becomes

a = 3q + 1

On cubing both the sides, we get

a³ = (3q + 1)³ {using (a+b)³= a³+ b³+ 3a²b + 3ab²}

= (3q)³+ 13 + 3 × 3q × 1(3q + 1)

= 27q³ + 27q² + 9q + 1

= 9 ( 3q³+ 3q²+ q) + 1

a³ = 9m + 1

Where m = ( 3q³ + 3q² + q)

Case 3: When r = 2, the equation becomes

a = 3q + 2

On cubing both the sides, we get

a³ = (3q + 2)³

= 27q³+ 54q²+ 36q + 8

= 9 (3q³ + 6q² + 4q) + 8

a³= 9m + 8

Where m = (3q³ + 6q²+ 4q)

So a can be any of the form 9m or 9m + 1 or, 9m + 8.

Hence proved!!

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Answer 2

$\huge\mathfrak\green{Answer:}$

Given:

• We have been given three numbers 9m, 9m + 1 and 9m + 8.

To Show:

• We need to show that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8 using Euclid's Division lemma.

Solution:

Let us assume a positive integer x such that it is in the form of 3q, (3q + 1) or

(3q + 2).

Now, For x = 3q

x³ = (3q)³

= 27q³

= 9(3q³)

= 9m ________(1)

Putting 3q³ = m, where m is any integer.

For x = 3q + 1, we have

x³ = (3q + 1)³

= 27q³ + 27q² + 9q + 1

= 9(3q³ + 3q² + q) + 1

= 9m + 1 ______(2)

Putting (3q³ + 3q² + q) = m, where m is any integer.

Now, for x = 3q + 2,

x³ = (3q + 2)³

= 27q³ + 54q² + 36q + 8

= 9(3q³ + 6q² + 4q) + 8

= 9m + 8 ______(3)

Putting (3q³ + 6q² + 4q) = m, where m is any integer.

Hence, the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8 where m is any integer.