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

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

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 ,

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

1980 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 27	Followed by Problem 29
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