# Difference between revisions of "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

Assume $h(x)=x^2+x+1$ $(x+1)^2n = (h(x)+x)^n = g(x)*h(x) + x^n$

$x^2n = x^2n+x^(2n-1)+x^(2n-2) -x^(2n-1)-x^(2n-2)-x^(2n-3) +...$

$x^n = x^n+x^(n-1)+x^(n-2) -x^(n-1)-x^(n-2)-x^(n-3) +....$

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

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


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=1/2 mod(3). Therefore n=21 is impossible

~~Wei

## See also

 1980 AHSME (Problems • Answer Key • Resources) Preceded byProblem 27 Followed byProblem 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.

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