Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

1980 AHSME Problems/Problem 28

Revision as of 15:11, 17 August 2025 by J314andrews (talk | contribs) (→‎Solution 1)

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 1

Let $h(x)=x^2+x+1$.

Then $(x+1)^{2n} = (x^2+2x+1)^n = (x+h(x))^n = \sum_{k=0}^{n} \binom{n}{k} x^k (h(x))^{n-k}$ $= \sum_{k=0}^{n-1} \binom{n}{k} x^k (h(x))^{n-k} + x^n = h(x) \left(\sum_{k=0}^{n-1} \binom{n}{k} x^k (h(x))^{n-k-1}\right) + x^n$.

Let $g(x) = \sum_{k=0}^{n-1} \binom{n}{k} x^k (h(x))^{n-k-1}$. Then $x^{2n} + 1 + (x+1)^{2n} = x^{2n} + 1 + g(x)h(x) + x^n$, so $x^{2n} + 1 + (x+1)^{2n}$ is divisible by $x^2 + x + 1$ if and only if $x^{2n} + x^n + 1$ is divisible by $x^2 + x + 1$.

Let $k$ be any non-negative integer, and let $r \in \{0, 1, 2\}$. Then $x^{3k+r} - x^r = x^r(x^{3k} - 1) = x^r(x^3 - 1)\left(\sum_{j=0}^{k - 1} x^{3j}\right)$ $= x^r(x-1)(x^2 + x + 1)\left(\sum_{j=0}^{k - 1} x^{3j}\right)$.

Therefore, $x^{3k+r} - x^r$ is divisible by $x^2 + x + 1$.

If $n$ is a nonnegative integer, then $n$ can be written as $3k + r$, where nonnegative integer $k$ and $r \in \{0, 1, 2\}$ are the quotient and remainder when $n$ is divided by $3$.

If $r = 2$, then $x^{2n} + x^n + 1 = x^{6k+4} + x^{3k+2} + 1 = x^{3(2k+1)+1} + x^{3k+2} + 1 = (x^{3(2k+1)+1} - x) + (x^{3k+2} - x^2) + x^2 + x + 1$. Since $x^{3(2k+1)+1} - x$ and $(x^{3k+2} - x^2)$ are divisible by $x^2 + x + 1$, $x^{2n} + x^n + 1$ is divisible by $x^2 + x + 1$.

If $r = 1$, then $x^{2n} + x^n + 1 = x^{6k+2} + x^{3k+1} + 1 = x^{3(2k)+2} + x^{3k+1} + 1 = (x^{3(2k)+2} - x^2) + (x^{3k+1} - x) + x^2 + x + 1$. Since $x^{3(2k)+2} - x^2$ and $(x^{3k+1} - x)$ are divisible by $x^2 + x + 1$, $x^{2n} + x^n + 1$ is divisible by $x^2 + x + 1$.

If $r = 0$, then $x^{2n} + x^n + 1 = x^{6k} + x^{3k} + 1 = (x^{6k} - 1) + (x^{3k} - 1) + 3$. Since $x^{6k} - 1$ and $x^{3k} - 1$ are divisible by $x^2 + x + 1$, $x^{2n} + x^n + 1$ is not divisible by $x^2 + x + 1$.

So $x^{2n} + 1 + (1+x)^{2n}$ is not divisible by $x^2 + x + 1$ if and only if $r = 0$, that is, $n$ is divisible by $3$. The only answer choice that is divisible by $3$ is $\boxed{(\textbf{C})\ 21}$.

-Wei, edited by j314andrews

Solution 2

Notice that the roots of $w^2+w+1=0$ are also the third roots of unity (excluding $w=1$). This is fairly easy to prove: multiply both sides by $w-1$ and we get \[(w-1)(w^2+w+1) = w^3 - 1 = 0.\] These roots are $w = e^{i \pi /3}$ and $w = e^{2i \pi /3}$.

Now we have \begin{align*} w^{2n} + 1 + (w+1)^{2n} &= w^{2n} + 1 + (-w^2)^{2n} \\ &= w^{4n} + w^{2n} + 1\\ &= 0. \end{align*} Plug in the roots of $w^2+w+1=0$. Note that \[e^{2i \pi /3} + 1 = -\frac{1}{2} + \frac{\sqrt{3}}{2}i + 1 = \frac{1}{2} + \frac{\sqrt{3}}{2}i = e^{i \pi /3}.\] However, this will not work if $n=3m$, so $n$ cannot be equal to $21$. Hence our answer is $\textrm{(C)}$.

Solution 3

We start by noting that \[x + 1 \equiv -x^2 \mod (x^2+x+1).\] Let $n = 3k+r$, where $r \in \{ 0,1,2 \}$.

Thus we have \[x^{4n} + x^{2n} + 1 \equiv x^{4r} + x^{2r} + 1 \mod (x^3 -1).\]

When $r = 0$, \[x^{4n} + x^{2n} + 1 \equiv 3 \mod (x^3 -1).\] When $r = 1$, \[x^{4n} + x^{2n} + 1 \equiv x^2 + x + 1 \mod (x^3 -1),\] which will be divisible by $x^2+x+1$.

When $r = 2$, \[x^{4n} + x^{2n} + 1 \equiv x^2 + x + 1 \mod (x^3 -1),\] which will also be divisible by $x^2+x+1$.

Thus $r \ne 0$, so $n$ cannot be divisible by $3$, and the answer is $\textrm{(C)}$.

See also

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1980_AHSME_Problems/Problem_28&oldid=258003"