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If n>2, then show that n^5-5n^3+4n is divisible by 120.

+2 votes
171 views
asked Aug 17, 2016 in Mathematics by Rahul Roy (7,955 points)

1 Answer

+2 votes
answered Aug 18, 2016 by vikash (21,287 points)
selected Aug 18, 2016 by Rahul Roy
 
Best answer

Solution: We have 120 = 3 * 5 * 8

therefore to show that given expression is divisible by 120 ; it is sufficient to show that 3, 5 and 8 are the factors of n^5-5n^3+4n

let us factorise the given expression:

since (n-2), (n-1), n, (n+1) are 4 consecutive integers.
one of it must be divisible by 4.
and at least two out of 5 consecutive integers must be even.
thus it must be divisible by 8.
now again out of three consecutive integers 1 must be divisible by 3.
and out of 5 consecutive integers 1 must be divisible by 5.
thus the given expression is divisible by 3, 5 and 8
thus it is divisible by 120.

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