Math, asked by Gun05, 1 year ago

For any positive odd integer n, prove that n3-n is divisible by 6

Answers

Answered by sivaprasath
9
Solution:

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Given:

To Prove:

 For any positive integer n, n³ - n is divisible by 6,.

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As we know that,

According to euclid's division algorithm,

Every integer can be expressed in the form,

=>  a = bq + r  (b > r ≥ 0)

=> Possible values of r are {0,1,2,3,4,5} ,. 6q + 6 = 6(q+1) + 0, 6q+7 = 6(q+1)+1...

 If n = 6q,

=> n³ - n

=> (6q)³ - 6q

=> 216q³ - 6q

=> 6q(36q² - 1) .. Multiple of 6 -> Divisible by 6,

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If n = 6q + 1,

=> n³ - n = (6q + 1)³ - (6q + 1)

=> ((6q)³ + 3(6q)²(1) + 3(6q)(1)² + 1³) - 6q - 1

=> (216q³ + 108q² + 18q² + 1) - 6q - 1

=> 216q³ + 108q² + 18q - 6q + 1 - 1 

=> 216q³ + 108q² + 12q

=> 6q(36q² + 18q + 2) ...Multiple of 6,...Divisible by 6

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 If n = 6q + 2

=> n³ - n = (6q + 2)³ - (6q + 2)

=> ((6q)³ + 3(6q)²(2) + 3(6q)(2)² +(2)³ ) - 6q - 2

=> (216q³ + 216q² + 72q + 8) - 6q - 2

=> 216q³ + 216q² - 66q + 6

=> 6(36q³ + 36q² - 11q + 1) ... Multiple of 6, .. Divisible by 6,.

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If n = 6q + 3,..

=> n³ - n = (6q + 3)³ - (6q + 3)

=> ((6q)³ + 3(6q)²(3) + 3(6q)(3)² + (3)³ ) - (6q + 3)

=> (216q³ + 324q² + 162q + 27 ) - 6q - 3

=> 216q³ + 324q² + 156q + 24

=> 6(36q³ + 504q² + 26q + 4)...Multiple of 6,..Divisible by 6..

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If n = 6q + 4,.

=> n³ - n = (6q + 4)³ - (6q + 4)

=> ((6q)³ + 3(6q)²(4) + 3(6q)(4)² + 4³ ) - 6q - 4

=> (216q³ + 432q² + 288q + 64 ) - 6q - 4

=> 216q³ + 432q² + 282q + 60

=> 6(36q³ + 72q² + 46q + 10) ..Multiple of 6,..Divisible by 6,..

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If n = 6q + 5

=> n³ - n = (6q + 5)³ - (6q + 5),..

=> ((6q)³ + 3(6q)²(5) + 3(6q)(5)² + (5)³) - 6q - 5

=> (216q³ + 540q² + 450q + 125) - 6q - 5

=> 216q³ + 540q² + 444q + 120

=> 6 (36q³ + 90q² + 74q + 20)...Multiple of 6,..Divisible by 6,.

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∴ For any value for n, n³ -  n is divisible by 6,..

                                             ∴  Hence , proved,.

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                                                    Hope it Helps !!

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Answered by Anonymous
0

Answer:

n3 – n = n(n2 – 1) = n(n+1)(n – 1) = (n – 1)n(n+1) = product of three consecutive positive integers.

Now, we have to show that the product of three consecutive positive integers is divisible by 6.

We know that any positive integer n is of the form 3q, 3q + 1 or 3q + 2 for some positive integer q.

Now three consecutive positive integers are n, n + 1, n + 2.

Case I. If n = 3q.

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

But we know that the product of two consecutive integers is an even integer.

∴ (3q + 1) (3q + 2) is an even integer, say 2r.

⟹ n(n + 1) (n + 2) = 3q × 2r = 6qr, which is divisible by 6.

Case II. If n = 3n + 1.

∴ n(n + 1) (n + 2) = (3q + 1) (3q + 2) (3q + 3)

= (even number say 2r) (3) (q + 1)

= 6r (q + 1),

which is divisible by 6.

Case III. If n = 3q + 2.

∴ n(n + 1) (n + 2) = (3q + 2) (3q + 3) (3q + 4)

= multiple of 6 for every q

= 6r (say),

which is divisible by 6.

Hence, the product of three consecutive integers is divisible by 6.

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