# Quickie #7

• Jan 3rd 2007, 05:34 AM
Soroban
Quickie #7
Show that for all integer values of $x$,

. . $x^9 - 6x^7 + 9x^5 - 4x^3$ is divisible by $8640.$

• Jan 3rd 2007, 03:01 PM
galactus
Soroban, am I on the right track?.

We have to show that $x^{9}-6x^{7}+9x^{5}-4x^{4}$ is divisible by $8640=2^{6}\cdot{3^{3}}\cdot{5}$

Factor: $x^{3}(x-2)(x-1)^{2}(x+1)^{2}(x+2)$

I would say because 5 is a factor of $x(x-1)(x-2)(x+2)$

3 is a factor of $x(x-1)(x-2), \;\ x(x-1)(x+2), \;\ x(x+1)(x+2)$

8 is a factor of $x(x-2)(x-1)(x+1), \;\ x(x-1)(x+1)(x+2)$
• Jan 3rd 2007, 09:41 PM
Soroban
Hello, Galactus!

Of course, you meant: 5 is a factor of $x(x-1)(x+1)(x-2)(x+2)$

The "Quickie" solution goes like this:

We know that the product of $n$ consecutive integers is divisible by $n$
. . and the product of four consecutive integers is divisible by $2^3.$

We have: . $8640 \:=\:2^6\!\cdot\!3^3\!\cdot\!5$

. . and: . $N \;=\;(x-2)\ (x-1)^2\ x^3\ (x+1)^2\ (x+2)$

$\text{Since }N \;=\;\underbrace{[(x-2)(x-1)x]}_{\text{3 consecutive}}\ \underbrace{[(x-1)x(x+1)]}_{\text{3 consecutive}}\ \underbrace{[x(x+1)(x+2)]}_{\text{3 consecutive}}$
. . then $N$ is divisible by $3^3.$

$\text{Since }N \;=\;\underbrace{[(x-2)(x-1)x(x+1)(x+2)]}_{\text{5 consecutive}}\ [(x-1)x^2(x+1)]$
. . then $N$ is divisible by $5.$

$\text{Since }N \;=\;x\ \underbrace{[(x-2)(x-1)x(x+1)]}_{\text{div by }2^3}\ \underbrace{[(x-1)x(x+1)(x+2)]}_{\text{div by }2^3}$
. . then $N$ is divisible by $2^6.$

Therefore, $N$ is divisible by $2^6\!\cdot\!3^3\!\cdot\!5\;=\;8640.$