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1980 AHSME Problems/Problem 28: Difference between revisions

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\text{(E)} \ 65  </math>   
\text{(E)} \ 65  </math>   


== Solution ==
== Solution 1==
Let <math>h(x)=x^2+x+1</math>.
Let <math>h(x)=x^2+x+1</math>.


Then we have
Then we have
<cmath>(x+1)^2n = (x^2+2x+1)^n = (h(x)+x)^n = g(x) \cdot h(x) + x^n,</cmath>
<cmath>(x+1)^2n = (x^2+2x+1)^n = (h(x)+x)^n = g(x) \cdot h(x) + x^n,</cmath>
where <math>g(x)</math> is <math>h^{n-1}(x) + nh^{n-2}(x) \cdot x + ... + x^{n-1}</math> (after expanding <math>(h(x)+x)^n</math> according to the Binomial Theorem.
where <math>g(x)</math> is <math>h^{n-1}(x) + nh^{n-2}(x) \cdot x + ... + x^{n-1}</math> (after expanding <math>(h(x)+x)^n</math> according to the Binomial Theorem).


Notice that
Notice that
Line 34: Line 34:


~~Wei
~~Wei
==Solution 2==
Notice that the roots of <math>w^2+w+1=0</math> are also the third roots of unity (excluding <math>w=1</math>). This is fairly easy to prove: multiply both sides by <math>w-1</math> and we get <cmath>(w-1)(w^2+w+1) = w^3 - 1 = 0.</cmath> These roots are <math>w = e^{i \pi /3}</math> and <math>w = e^{2i \pi /3}</math>.
Now we have
<cmath>\begin{align*}
w^{2n} + 1 + (w+1)^{2n} &= w^{2n} + 1 + (-w^2)^{2n} \\
&= w^{4n} + w^{2n} + 1\\
&= 0.
\end{align*}</cmath>
Plug in the roots of <math>w^2+w+1=0</math>. Note that
<cmath>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}.</cmath>
However, this will not work if <math>n=3m</math>, so <math>n</math> cannot be equal to <math>21</math>. Hence our answer is <math>\textrm{(C)}</math>.
==Solution 3==
We start by noting that <cmath>x + 1 \equiv -x^2 \mod (x^2+x+1).</cmath>
Let <math>n = 3k+r</math>, where <math>r \in \{ 0,1,2 \}</math>.
Thus we have <cmath>x^{4n} + x^{2n} + 1 \equiv x^{4r} + x^{2r} + 1 \mod (x^3 -1).</cmath>
When <math>r = 0</math>, <cmath>x^{4n} + x^{2n} + 1 \equiv 3 \mod (x^3 -1).</cmath>
When <math>r = 1</math>, <cmath>x^{4n} + x^{2n} + 1 \equiv x^2 + x + 1 \mod (x^3 -1),</cmath> which will be divisible by <math>x^2+x+1</math>.
When <math>r = 2</math>, <cmath>x^{4n} + x^{2n} + 1 \equiv x^2 + x + 1 \mod (x^3 -1),</cmath> which will also be divisible by <math>x^2+x+1</math>.
Thus <math>r \ne 0</math>, so <math>n</math> cannot be divisible by <math>3</math>, and the answer is <math>\textrm{(C)}</math>.


== See also ==
== See also ==

Revision as of 19:04, 30 October 2021

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 we have \[(x+1)^2n = (x^2+2x+1)^n = (h(x)+x)^n = g(x) \cdot h(x) + x^n,\] where $g(x)$ is $h^{n-1}(x) + nh^{n-2}(x) \cdot x + ... + x^{n-1}$ (after expanding $(h(x)+x)^n$ according to the Binomial Theorem).

Notice that \[x^2n = x^2n+x^{2n-1}+x^{2n-2}+...x -x^{2n-1}-x^{2n-2}-x^{2n-3} +...\]

$x^n = x^n+x^{n-1}+x^{n-2} -x^{n-1}-x^{n-2}-x^{n-3} +....$

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 or 2 mod(3). Therefore n=21 is impossible

~~Wei

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

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