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