Another question:

What is the largest number N for which you can say that n^5-5n^3+4n is divisible by N for every integer n?

Thank you very much.

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- Jan 3rd 2007, 11:33 AMJenny20The largest number N
Another question:

What is the largest number N for which you can say that n^5-5n^3+4n is divisible by N for every integer n?

Thank you very much. - Jan 3rd 2007, 12:52 PMSoroban
Hello, Jenny!

Quote:

What is the largest number $\displaystyle N$ for which you can say that

$\displaystyle n^5-5n^3+4n$ is divisible by $\displaystyle N$ for every integer $\displaystyle n$?

$\displaystyle \text{Factor: }\:n^5 - 5n^3 + 4n \;=\;n(n^4 - 4n^2 + 4)$

. . . . . . . . $\displaystyle = \;n(n^2 - 1)(n^2 - 4)\;=\;n(n-1)(n+1)(n-2)(n+2)$

$\displaystyle \text{We have: }\:n^5 - 5n^3 + 4n \;=\;\underbrace{(n-2)(n-1)(n)(n+1)(n-2)}_{\text{product of 5 consecutive integers}}$

The product of 5 consecutive integers is divisible by $\displaystyle 1,\,2,\,3,\,4,\,5.$

Therefore: .$\displaystyle N \,=\,120$