A Problem for Geoff Pilling

The number is equivalent to:

$(n-4) + (n-5) \cdot n + (n-2) \cdot n^2 + (n-3) \cdot n^3 +(n-1) \cdot n^4$

Taking $\mod {(n+1)}$ and using the fact that $n = -1\mod {(n+1)}$ :

$(n-4) + (n-5) \cdot -1 + (n-2) \cdot {(-1)}^2 + (n-3) \cdot {(-1)}^3 +(n-1) \cdot {(-1)}^4 \mod {(n+1)}$

$(n-4) - (n-5) + (n-2) - (n-3) +(n-1) \mod {(n+1)}$

$1 + 1 + n - 1 \mod{(n+1)}$

$n+1 \mod{(n+1)}$

$0 \mod{(n+1)}$

The number is therefore divisible.

As stated Alex G also stated in his(or her, to exonerate myself of sexism) solution , the number is equivalent to:

$(n-4) + (n-5).n + (n-2) . n^2 + (n-3) . n^3 + (n-1).n^4$

Now , if $n+1$ were to divide the given number , $n=-1$ must be a solution to the equation :

$(n-4) + (n-5).n + (n-2) . n^2 + (n-3) . n^3 + (n-1).n^4 = 0$

Plugging in $n=-1$ :

$-5 + (-6)(-1) + (-3)(1) + (-4)(-1) + (-2)(1) = 0$

And thus the number is actually divisible by $n+1$ .

Note that the number listed is equivalent to $(n-1)*n^4+(n-3)*n^3+(n-2)*n^2+(n-5)*n+n-4$ Now, we want to know if n+1 divides the given polynomial. To do so, we can use synthetic division. We can do this one of two ways (both of which work.) You can either expand out the polynomial into $n^5-2n^3-n^2-4n-4$ , which either synthetic division or plugging in 1 both prove that n+1 divides the polynomial. The other way (which I initially did) was to keep the coefficients in terms of $n$ (i.e. in the form of the digits shown above). Synthetically dividing that polynomial yields $(n-1)-(n-3)+(n-2)-(n-5)+(n-4)=n+1$ . Since the remainder is $n+1$ when divided by $n+1$ , it is clear that $n+1$ divides the polynomial.

Woo hoo... Cool... Nice problem, Alex... I love bases! :)