Question

4) Show that cube of any positive integer as of the
form 3m or 3m+1 or 3mt8 for some integer m​

Answer 1

cube of

any positive integer is either of form 9q ,9q+1 or 9q+8 for some integer Q

Answer 2

Answer:

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Let us consider a positive integer a

Divide the positive integer a by 3, and let r be the reminder and b be the quotient such that

a = 3b + r……………………………(1)

where r = 0,1,2,3…..

Case 1: Consider r = 0

Equation (1) becomes

a = 3b

On squaring both the side

a² = (3b)²

a² = 9b²

a²= 3 × 3b²

a²= 3m

Where m = 3b²

Case 2: Let r = 1

Equation (1) becomes

a = 3b + 1

Squaring on both the side we get

a² = (3b + 1)²

a² = (3b)² + 1 + 2 × (3b) × 1

a²= 9b² + 6b + 1

a² = 3(3b² + 2b) + 1

a² = 3m + 1

Where m = 3b²+ 2b

Case 3: Let r = 2

Equation (1) becomes

a = 3b + 2

Squaring on both the sides we get

a²= (3b + 2)2

a² = 9b² + 4 + (2 × 3b × 2)

a² = 9b²+ 12b + 3 + 1

a² = 3(3b²+ 4b + 1) + 1

a²= 3m + 1

where m = 3b²+ 4b + 1

∴ square of any positive integer is of the form 3m or 3m+1.

Hence proved.

error its square not cube

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