Modulo120

May 2009
596
31
ALGERIA
Prove that : \(\displaystyle \forall n\in N:9n^{5}-5n^{3}-4n\equiv 0(mod120\))
 
Nov 2009
485
184
Note that \(\displaystyle 120=8\cdot 3\cdot 5\) and \(\displaystyle 9n^5-5n^3-4n=(n-1)n(n+1)(4+9n^2)\). It suffices to show that \(\displaystyle 8, 3, \text{ and } 5\) divide \(\displaystyle (n-1)n(n+1)(4+9n^2)\) for any \(\displaystyle n\).

Can you take it from here?
 
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simplependulum

MHF Hall of Honor
Jan 2009
715
427
\(\displaystyle 9n^5 - 5n^3 - 4n = 9(n^5 - 5n^3 + 4n) - 5n^3 + 45n^3 - 4n - 36n \)


\(\displaystyle = 9n(n^2-4)(n^2-1) + 40n^3 - 40n \)

\(\displaystyle = 9(n-2)(n-1)(n)(n+1)(n+2) + 40n(n^2-1)\)

We know the product of five consective integers is divisible by \(\displaystyle 5! = 120 \) so we only consider \(\displaystyle 40n(n^2-1) \) .

We need to show that \(\displaystyle n(n^2-1)\) is the multiple of \(\displaystyle 3 \)

\(\displaystyle n(n^2-1) = n(n-1)(n+1) = (n-1)(n)(n+1) \) which is the multiple of \(\displaystyle 3 \) .

Therefore , we prove it .
 
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