Difference between revisions of "1975 USAMO Problems/Problem 3"
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Thus, <math>k=0,1,2,\ldots,n</math> are the roots of <math>Q(x)</math>. | Thus, <math>k=0,1,2,\ldots,n</math> are the roots of <math>Q(x)</math>. | ||
− | Since these are all <math>n+1</math> of the roots of the <math>n+1^{\text{th}}</math> degree polynomial, we can write <math>Q(x)</math> as <cmath>Q(x) = c(x)(x-1)(x-2) \cdots (x-n)</cmath> where <math>c</math> is a constant. | + | Since these are all <math>n+1</math> of the roots of the <math>n+1^{\text{th}}</math> degree polynomial, by the [[Factor Theorem]], we can write <math>Q(x)</math> as <cmath>Q(x) = c(x)(x-1)(x-2) \cdots (x-n)</cmath> where <math>c</math> is a constant. |
Thus, <cmath>(x+1)P(x) - x = c(x)(x-1)(x-2) \cdots (x-n).</cmath> | Thus, <cmath>(x+1)P(x) - x = c(x)(x-1)(x-2) \cdots (x-n).</cmath> |
Latest revision as of 17:29, 8 January 2024
Contents
Problem
If denotes a polynomial of degree
such that
for
, determine
.
Solution 1
Let , and clearly,
has a degree of
.
Then, for ,
.
Thus, are the roots of
.
Since these are all of the roots of the
degree polynomial, by the Factor Theorem, we can write
as
where
is a constant.
Thus,
We plug in to cancel the
and find
:
Finally, plugging in to find
gives:
If is even, this simplifies to
. If
is odd, this simplifies to
.
~Edits by BakedPotato66
Solution 2
It is fairly natural to use Lagrange's Interpolation Formula on this problem:
through usage of the Binomial Theorem.
~lpieleanu (minor editing and reformatting)
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
See Also
1975 USAMO (Problems • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
1 • 2 • 3 • 4 • 5 | ||
All USAMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.