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 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

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