1980 AHSME Problems/Problem 28

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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 (ProblemsAnswer KeyResources)
Preceded by
Problem 27
Followed by
Problem 29
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