Find any positive integer n prove that n3-n is divisible by 6
Answers
Here's the proof :-
Let n be any positive integer Which when divided by 6 gives q as quotient and r as remainder.
By Euclid's Division lemma
a = bq + r
where, 0 ≤ r < b
So,
n = 6q + r
where , r = 0 , 1 , 2 , 3 , 4 , 5
____________________
n = 6q
n = 6q + 1
n = 6q + 2
n = 6q + 3
n = 6q + 4
n = 6q + 5
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• CASE - 1
n = 6q
n³ - n = ( 6q )³ - 6q
= 216q³ - 6q
= 6 ( 36q³ - q )
= 6m [ Where m = 36q³ - q )
Therefore , it is divisible by 6
________________________
• CASE - 2
n = 6q + 1
n³ - n = ( 6q + 1 )³ - ( 6q + 1 )
{ Using identity :- ( a + b )³ = a³ + b³ + 3a²b + 3ab² }
= 216q³ + 1 + 108q² + 18q - 6q - 1
= 6 ( 36q³ + 18q² + 2q )
= 6m [ Where m = 36q³ + 18q² + q ]
Therefore , it is divisible by 6 .
_____________________
• CASE - 3
n = 6q + 2
n³ - n = ( 6q + 2 )³ - ( 6q + 2 )
= 216q³ + 8 + 216q² + 72q - 6q - 2
= 216q³ + 216q² + 66q + 6
= 6 ( 36q³ + 36q² + 11q + 1 )
= 6m [ Where m = 36q³ + 36q² + 11q + 1 ]
Therefore, It is divisible by 6
_________________
• CASE - 4
n = 6q + 3
n³ - n = ( 6q + 3 )³ - ( 6q + 3 )
= 216q³ + 27 + 324q² + 162q - 6q - 3
= 216q³ + 324q² + 156q + 24
= 6 ( 36q³ + 54q² + 26q + 4 )
= 6m [ Where m = 36q³ + 54q² + 26q + 4 ]
Therefore , It is divisible by 6.
_________________________
• CASE - 5
n = 6q + 4
n³ - n = ( 6q + 4 )³ - ( 6q + 4 )
= 216q³ + 64 + 432q² + 288q - 6q - 4
= 216q³ + 432q² + 282q + 60
= 6 ( 36q³ + 72q² + 47q + 10 )
= 6m [ Where m = 36q³ + 72q² + 47q + 10 ]
Therefore , It is divisible by 6
_____________________
• CASE - 6
n = 6q + 5
n³ - n = ( 6q + 5 )³ - ( 6q + 5 )
= 216q³ + 125 + 540q² + 450q - 6q - 5
= 216q³ + 540q² + 444q + 120
= 6 ( 36q³ + 90q² + 74q + 20 )
= 6m [ Where m = 36q³ + 90q² + 74q + 20 ]
Therefore, It is divisible by 6
__________________
Hence , n³ - n will be divisible by 6 for any positive integer.
Step-by-step explanation:
n³ - n = n (n² - 1) = n (n - 1) (n + 1)
Whenever a number is divided by 3, the remainder obtained is either 0 or 1 or 2.
∴ n = 3p or 3p + 1 or 3p + 2, where p is some integer.
If n = 3p, then n is divisible by 3.
If n = 3p + 1, then n – 1 = 3p + 1 –1 = 3p is divisible by 3.
If n = 3p + 2, then n + 1 = 3p + 2 + 1 = 3p + 3 = 3(p + 1) is divisible by 3.
So, we can say that one of the numbers among n, n – 1 and n + 1 is always divisible by 3.
⇒ n (n – 1) (n + 1) is divisible by 3.
Similarly, whenever a number is divided 2, the remainder obtained is 0 or 1.
∴ n = 2q or 2q + 1, where q is some integer.
If n = 2q, then n is divisible by 2.
If n = 2q + 1, then n – 1 = 2q + 1 – 1 = 2q is divisible by 2 and n + 1 = 2q + 1 + 1 = 2q + 2 = 2 (q + 1) is divisible by 2.
So, we can say that one of the numbers among n, n – 1 and n + 1 is always divisible by 2.
⇒ n (n – 1) (n + 1) is divisible by 2.
Since, n (n – 1) (n + 1) is divisible by 2 and 3.
∴ n (n-1) (n+1) = n³ - n is divisible by 6.( If a number is divisible by both 2 and 3 , then it is divisible by 6)
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