Question

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Answer 1

I've messed with this one a little and have come up with
\[f(x) - x = 0\] so \[f(x) = (x-1)(x-2)..(x-n)+x\] and this will satisfy conditions 1 and 2. I then tried to solve this for f(-1)=2012, but came up with no solution for n. Is there a flaw in my work or assumptions?

Answer 2

I guess my first question is, is my assumption that there can only be 1 possible n to satisfy the 3 conditions correct?

Answer 3

putnam exam problem right?

Answer 4

I'm not sure if it is or not, I did get it from my professor during a putnam exam study period though

Answer 5

I guess it's putnam-like, not sure if something very similar has been on it

Answer 6

you are on the right track, in fact you are most of the way there

Answer 7

\[f(x) -x= a(x-1)(x-2)..(x-n)\] you have to solve for \(a\)

Answer 8

only thing you forgot is that the leading coefficient is not necessarily 1

Answer 9

Ohhh

Answer 10

I guess I glanced over that, my first few attempts at this problem were much more ugly lol

Answer 11

\[2012=f(-1)-1\]
\[=a(n+1)!-1\]or something like that
i think i have the solution somewhere
would you like me to attach it or do you want to keep working?

Answer 12

No need for that I think. I'm jotting it all down now and it's working out nicely

Answer 13

oh i think there is a \((-1)^n\) somewhere as well

Answer 14

I think I should be good from here, thanks satellite.

Answer 15

yw

Answer 16

Yeah I see the (-1)^n coming out