1980 AHSME Problems/Problem 28

Problem

The polynomial $x^{2n}+1+(x+1)^{2n}$ is not divisible by $x^2+x+1$ if $n$ equals

$\text{(A)} \ 17 \qquad \text{(B)} \ 20 \qquad \text{(C)} \ 21 \qquad \text{(D)} \ 64 \qquad \text{(E)} \ 65$

Solution

Let $h(x)=x^2+x+1$ .

Then we have $(x+1)^2n = (x^2+2x+1)^n = (h(x)+x)^n = g(x) \cdot h(x) + x^n,$ where $g(x)$ is $h^{n-1}(x) + nh^{n-2}(x) \cdot x + ... + x^{n-1}$ (after expanding $(h(x)+x)^n$ according to the Binomial Theorem.

Notice that $x^2n = x^2n+x^{2n-1}+x^{2n-2}+...x -x^{2n-1}-x^{2n-2}-x^{2n-3} +...$

$x^n = x^n+x^{n-1}+x^{n-2} -x^{n-1}-x^{n-2}-x^{n-3} +....$

Therefore, the left term from $x^2n$ is $x^{(2n-3u)}$

          the left term from  is ,

If divisible by h(x), we need 2n-3u=1 and n-3v=2 or

                             2n-3u=2 and n-3v=1

The solution will be n=1 or 2 mod(3). Therefore n=21 is impossible

~~Wei

1980 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 27	Followed by Problem 29
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

1980 AHSME Problems/Problem 28

Problem

Solution

See also