Question

Q.

Answer 1

My thoughts:

1. The polynomial is of the form \(a(x-2)(x -5)\cdots(x-3n+1)\).

2. Since \(P(0) = 2\), we have the last coefficient = 2.

Answer 2

There are \(3n+1\) terms.

Answer 3

yes you are actually correct as far as I can see

Answer 4

How to solve the question, then?

Answer 5

I'm not sure how to proceed.

Answer 6

OS is buggy.

Answer 7

Oh my God, it shows that I'm typing all the time, when I'm not.

Answer 8

@Zarkon @amistre64 @ganeshie8 @FoolForMath @FoolAroundMath @mathslover @vishweshshrimali5 @mathmale

Answer 9

@AravindG @whpalmer4 @wio

Answer 10

I'm gonna go to sleep. Please feel free to type in the solution if possible. I've tried what I could. :|

Answer 11

Ey, gimme a second. The sum of roots is\[\large \sum_{\Large i =1 }^{\large 3n}3i - 1 = 3\left(\dfrac{3n(3n+1)}{2} \right)- \dfrac{n(n+1)}{2}\]

Answer 12

Like that helps. >_<

Good night, beautiful people.

Answer 13

im not sure i even understand the question ....

\[P(x) =\sum_{i=0}^{k}c_i~x^i\]
\[P(x) =c_0~x^0+c_1~x^1+c_2~x^2+c_3~x^3+c_4~x^4+...\]
or written another way:
\[P(x)=\sum~c_i\frac{(x-x_0)(x-x_1)(x-x_2)(x-x_3)...(x-x_q)}{x-x_i}\]

Answer 14

maybe this reduces to:
\[P(x)=\sum~c_i\frac{(x-x_{3n})(x-x_{3n-2})(x-x_{3n-1})}{x-x_i}\]

Answer 15

or better yet:\[P(x)=\sum~c_i\frac{x(x-2)(x-1)}{x-x_i}\]
such that x_i is some mod3 value?

Answer 16

x- (i mod 3)

i mod 3 ... 3(i/3 - int(i/3))
i - 3 int(i/3)

just some thoughts is all, nothing that useful of course

Answer 17

I actually got this from an MSE question. The OP is in the same grade as I am, so I concur that there must be actually a way to get there with no big issues.

Answer 18

Wow, this is what you purple people do. Mirin

Answer 19

Well, I can tell you that if the polynomial has degree \(3n\), then it won't be of the form \[a(x-2)(x -5)\cdots(x-3n+1)\]as that only has \(n\) roots.

Answer 20

School time. Bye bye.

Answer 21

Oh!

Answer 22

That's a good start... anyway, gotta go.

Answer 23

Well, I have a polynomial that satisfies everything, but obtaining it used non-trivial methods, so I feel like that probably wasn't the intended method.

Answer 24

*

Answer 25

5th question
http://www.math-olympiad.com/13th-usa-mathematical-olympiad-1984.htm#5

this solutions looks bit simple... check it out :)

Answer 26

Doesn't look that simple. :P

Thanks, though!

Answer 27

\[

\begin{array}{|c|c|}
\hline
\text{x}&\text{P(x)}&\text{1st diff}&\text{2nd diff}&\text{3rd diff}&\text{...}&\text{n th diff}\\
\hline
\text{0}&\text{P(0)}&\\
\hline
\text{1}&\text{P(1)}&\\
\hline
\text{2}&\text{P(2)}&\\
\hline
\text{3}&\text{P(3)}&\\
\hline
\text{4}&\text{P(4)}&\\
\hline
\text{..}&\text{..}&\\
\hline
\text{n}&\text{P(n)}&\\
\hline

\end{array}
\]

Answer 28

Filling this table may make it easy to interpret the method given in that link..

Answer 29

Can we do an ... like ... Indian-ish method?

Answer 30

Notice that if P(x) is linear, then the "2nd diff" column will have all "0"s

Answer 31

finite differences is not Indian-ish method ha ?

Answer 32

I'm searching for that kind of solution. I really hope there is one. I Googled this and nothing too special came up.

Answer 33

It is?

Answer 34

If the polynomial degree is "n", then the "n+1"th difference will be 0 :

P(x) = ax + b : 2nd differnces are 0
P(x) = ax^2+bx+c : 3rd differences are 0
...

Answer 35

http://education.ti.com/en/timathnspired/~/media/Images/Activities/US/Math/Algebra%20II/Finite%20Differences/A2_Finite_Differences_sm.jpg

^^thats how second differences look for a polynomial of degree "1"-
the 2nd differences are 0

Answer 36

for quadratic :
http://thatsparabolic.files.wordpress.com/2010/11/table-and-finite.png

Answer 37

If you're curious, this is the actual polynomial that satisfies the conditions.
\[\begin{aligned}
2-\frac{95499}{770}x &+\frac{22990311}{61600} x^2-\frac{10409769}{22400}x^3+\frac{14204367}{44800}x^4-\frac{2383359}{17920}x^5
\\ &+\frac{6520011}{179200}x^6-\frac{598923}{89600}x^7+\frac{148959}{179200}x^8-\frac{1237}{17920}x^9\\
&+\frac{657}{179200}x^{10}-\frac{111}{985600}x^{11}+\frac{3}{1971200}x^{12}
\end{aligned}\]