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Difference between revisions of "1980 AHSME Problems/Problem 28"

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\text{(E)} \ 65  </math>   
 
\text{(E)} \ 65  </math>   
  
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== Solution ==
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==Solution 1==
 +
The roots of <math>x^2+x+1</math> are <math>-\frac{1}{2} + \frac{\sqrt{3}}{2}i = \cos\frac{2}{3}\pi + i \sin \frac{2}{3}\pi</math> and <math>-\frac{1}{2} - \frac{\sqrt{3}}{2}i = \cos \frac{4}{3}\pi + i \sin\frac{4}{3}\pi</math>, which are the primitive third roots of unity.
 +
 
 +
Let <math>w = -\frac{1}{2} + \frac{\sqrt{3}}{2}i = \cos\frac{2}{3}\pi + i \sin \frac{2}{3}\pi</math>.  Note that <math>-\frac{1}{2} - \frac{\sqrt{3}}{2}i = \cos \frac{4}{3}\pi + i \sin\frac{4}{3}\pi = w^2 = \bar{w}</math>.  So by the conjugate root theorem, <math>x^{2n} + 1 + (1+x)^{2n}</math> is divisible by <math>x^2+x+1</math> if and only if <math>w</math> is a root of <math>x^{2n} + 1 + (1+x)^{2n}</math>.
 +
 
 +
Also, <math>w^{2n} + 1 + (w+1)^{2n} = w^{2n} + 1 + (-w^2)^{2n} = w^{2n} + 1 + w^{4n} = w^{2n} + 1 + w^n = w^{2n} + w^n + 1</math>.
 +
 
 +
If <math>n \equiv 0\ (\textrm{mod}\ 3)</math>, <math>w^{2n} + w^n + 1 \equiv 1 + 1 + 1 = 3</math>.
 +
 +
If <math>n \equiv 1\ (\textrm{mod}\ 3)</math>, <math>w^{2n} + w^n + 1 \equiv w^2 + w + 1 = 0</math>.
 +
 
 +
If <math>n \equiv 2\ (\textrm{mod}\ 3)</math>, <math>w^{2n} + w^n + 1 \equiv w + w^2 + 1 = 0</math>.
 +
 
 +
Therefore, <math>x^{2n} + 1 + (1+x)^{2n}</math> is not divisible by <math>x^2 + x + 1</math> if and only if <math>n</math> is divisible by <math>3</math>.  Of the answer choices, only <math>\boxed{(\textbf{C})\  21}</math> is divisible by <math>3</math>.
 +
 
 +
== Solution 2==
 
Let <math>h(x)=x^2+x+1</math>.
 
Let <math>h(x)=x^2+x+1</math>.
  
−
Then we have
+
Then <math>(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} </math> <math>= \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</math>.
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<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.
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Let <math>g(x) = \sum_{k=0}^{n-1} \binom{n}{k} x^k (h(x))^{n-k-1}</math>.  Then <math>x^{2n} + 1 + (x+1)^{2n} = x^{2n} + 1 + g(x)h(x) + x^n</math>, so <math>x^{2n} + 1 + (x+1)^{2n}</math> is divisible by <math>x^2 + x + 1</math> if and only if <math>x^{2n} +  x^n + 1</math> is divisible by <math>x^2 + x + 1</math>.
 +
 
 +
Let <math>k</math> be any non-negative integer, and let <math>r \in \{0, 1, 2\}</math>.  Then <math>x^{3k+r} - x^r = x^r(x^{3k} - 1) = x^r(x^3 - 1)\left(\sum_{j=0}^{k - 1} x^{3j}\right) </math>
 +
<math>= x^r(x-1)(x^2 + x + 1)\left(\sum_{j=0}^{k - 1} x^{3j}\right)</math>.
 +
 
 +
Therefore, <math>x^{3k+r} - x^r</math> is divisible by <math>x^2 + x + 1</math>.
 +
 
 +
If <math>n</math> is a nonnegative integer, then <math>n</math> can be written as <math>3k + r</math>, where nonnegative integer <math>k</math> and <math>r \in \{0, 1, 2\}</math> are the quotient and remainder when <math>n</math> is divided by <math>3</math>.
 +
 
 +
If <math>r = 2</math>, then <math>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</math>. Since <math>x^{3(2k+1)+1} - x</math> and <math>(x^{3k+2} - x^2)</math> are divisible by <math>x^2 + x + 1</math>, <math>x^{2n} + x^n + 1</math> is divisible by <math>x^2 + x + 1</math>. 
 +
 
 +
If <math>r = 1</math>, then <math>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</math>.  Since <math>x^{3(2k)+2} - x^2</math> and <math>(x^{3k+1} - x)</math> are divisible by <math>x^2 + x + 1</math>, <math>x^{2n} + x^n + 1</math> is divisible by <math>x^2 + x + 1</math>.
 +
 
 +
If <math>r = 0</math>, then <math>x^{2n} + x^n + 1 = x^{6k} + x^{3k} + 1 = (x^{6k} - 1) + (x^{3k} - 1) + 3</math>.  Since <math>x^{6k} - 1</math> and <math>x^{3k} - 1</math> are divisible by <math>x^2 + x + 1</math>, <math>x^{2n} + x^n + 1</math> is not divisible by <math>x^2 + x + 1</math>.
 +
 
 +
So <math>x^{2n} + 1 + (1+x)^{2n}</math> is not divisible by <math>x^2 + x + 1</math> if and only if <math>r = 0</math>, that is, <math>n</math> is divisible by <math>3</math>.  The only answer choice that is divisible by <math>3</math> is <math>\boxed{(\textbf{C})\ 21}</math>.
  
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Notice that
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-Wei, edited by j314andrews
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<cmath>x^2n = x^2n+x^{2n-1}+x^{2n-2}+...x
 
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          -x^{2n-1}-x^{2n-2}-x^{2n-3}
 
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        +...</cmath>
 
  
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<math>x^n = x^n+x^{n-1}+x^{n-2}
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==Solution 3==
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        -x^{n-1}-x^{n-2}-x^{n-3}
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We start by noting that <cmath>x + 1 \equiv -x^2 \mod (x^2+x+1).</cmath>
−
  +....</math>
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Let <math>n = 3k+r</math>, where <math>r \in \{ 0,1,2 \}</math>.
  
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Therefore, the left term from <math>x^2n</math> is <math>x^{(2n-3u)}</math>
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Thus we have <cmath>x^{4n} + x^{2n} + 1 \equiv x^{4r} + x^{2r} + 1 \mod (x^3 -1).</cmath>
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          the left term from <math>x^n</math> is <math>x^{(n-3v)}</math>,
 
  
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If divisible by h(x), we need 2n-3u=1 and n-3v=2 or
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When <math>r = 0</math>, <cmath>x^{4n} + x^{2n} + 1 \equiv 3 \mod (x^3 -1).</cmath>
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                              2n-3u=2 and n-3v=1
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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>.
  
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The solution will be n=1 or 2 mod(3). Therefore n=21 is impossible
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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>.
  
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~~Wei
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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 ==

Latest revision as of 20:03, 17 August 2025

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

The roots of $x^2+x+1$ are $-\frac{1}{2} + \frac{\sqrt{3}}{2}i = \cos\frac{2}{3}\pi + i \sin \frac{2}{3}\pi$ and $-\frac{1}{2} - \frac{\sqrt{3}}{2}i = \cos \frac{4}{3}\pi + i \sin\frac{4}{3}\pi$, which are the primitive third roots of unity.

Let $w = -\frac{1}{2} + \frac{\sqrt{3}}{2}i = \cos\frac{2}{3}\pi + i \sin \frac{2}{3}\pi$. Note that $-\frac{1}{2} - \frac{\sqrt{3}}{2}i = \cos \frac{4}{3}\pi + i \sin\frac{4}{3}\pi = w^2 = \bar{w}$. So by the conjugate root theorem, $x^{2n} + 1 + (1+x)^{2n}$ is divisible by $x^2+x+1$ if and only if $w$ is a root of $x^{2n} + 1 + (1+x)^{2n}$.

Also, $w^{2n} + 1 + (w+1)^{2n} = w^{2n} + 1 + (-w^2)^{2n} = w^{2n} + 1 + w^{4n} = w^{2n} + 1 + w^n = w^{2n} + w^n + 1$.

If $n \equiv 0\ (\textrm{mod}\ 3)$, $w^{2n} + w^n + 1 \equiv 1 + 1 + 1 = 3$.

If $n \equiv 1\ (\textrm{mod}\ 3)$, $w^{2n} + w^n + 1 \equiv w^2 + w + 1 = 0$.

If $n \equiv 2\ (\textrm{mod}\ 3)$, $w^{2n} + w^n + 1 \equiv w + w^2 + 1 = 0$.

Therefore, $x^{2n} + 1 + (1+x)^{2n}$ is not divisible by $x^2 + x + 1$ if and only if $n$ is divisible by $3$. Of the answer choices, only $\boxed{(\textbf{C})\ 21}$ is divisible by $3$.

Solution 2

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 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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