1980 AHSME Problems/Problem 28

Revision as of 20:04, 28 March 2020 by Wwei.yu (talk | contribs) (Solution)

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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