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# Math contest, find the amount of frih\$ you win!

Arnie
Rewrite the following function as simple as possible so that it doesn't use multiplication (factors):

N(x) = a(x-a) * b(x-b) * c(x-c) * ... * y(x-y) * z(x-z)

The first to get the correct answer will get as much frih\$ as the answer is for any set of values for a t/m z you pick, provided that it doesn't exceed my stock...
inventor
N(x)=a(x-a) * b(x-b) * c(x-c) * ... * x(x-x)*y(x-y) * z(x-z)
N(x)=a(x-a) * b(x-b) * c(x-c) * ... * x(0)*y(x-y) * z(x-z)

when you multiply that out you're gona get a 0. you can simplify all you want but in the end the amount of frih you get will be 0?
Arnie
We have a winner... now donating the frih\$

Too bad it was discovered so soon though
imagefree
 Arnie wrote: We have a winner... now donating the frih\$ Too bad it was discovered so soon though

i figured out the solution last day, but didnt reply because there was no return!
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