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Difference between revisions of "2023 AIME II Problems/Problem 8"
(→‎Solution 1)
 
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\left( \omega^{3\left( 7 - k \right)} + \omega^{\left( 7 - k \right)} + 1 \right) \\
 
\left( \omega^{3\left( 7 - k \right)} + \omega^{\left( 7 - k \right)} + 1 \right) \\
 
& = 3 \cdot 2^3 \\
 
& = 3 \cdot 2^3 \\
−
& = \boxed{\textbf{(024) }}.
+
& = \boxed{\textbf{024}}.
 
\end{align*}
 
\end{align*}
 
</cmath>
 
</cmath>
  
 
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
 
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
 +
 +
<math>k</math> cannot be a multiple of <math>7</math>, otherwise the first equation would equal <math>9</math> instead of <math>2</math> ~inaccessibles
  
 
==Solution 2 (Moduli)==
 
==Solution 2 (Moduli)==
Line 45: Line 47:
 
<cmath>|z_0||z_1||z_2||z_3||z_4||z_5||z_6|.</cmath>
 
<cmath>|z_0||z_1||z_2||z_3||z_4||z_5||z_6|.</cmath>
  
−
<math>z_0 = 0</math>, and we may observe that <math>z_x</math> and <math>z_{7-x}</math> are conjugates for any <math>x</math>, meaning that their magnitudes are the same. Thus, our product is
+
<math>z_0 = 3</math>, and we may observe that <math>z_x</math> and <math>z_{7-x}</math> are conjugates for any <math>x</math>, meaning that their magnitudes are the same. Thus, our product is
  
 
<cmath>3|z_1|^2|z_2|^2|z_3|^2</cmath>
 
<cmath>3|z_1|^2|z_2|^2|z_3|^2</cmath>
Line 85: Line 87:
 
~mathboy100
 
~mathboy100
  
−
 
+
==Solution 3 (Inspecting the exponents of powers of <math>\omega</math>)==
−
==Solution 3==
 
 
We write out the product in terms of <math>\omega</math>:
 
We write out the product in terms of <math>\omega</math>:
 
<cmath>\prod_{k=0}^6 \left(\omega^{3k} + \omega^k + 1\right)=3(\omega^3+\omega+1)(\omega^6+\omega^2+1)(\omega^9+\omega^3+1)(\omega^{12}+\omega^4+1)(\omega^{15}+\omega^5+1)(\omega^{18}+\omega^6+1).</cmath>
 
<cmath>\prod_{k=0}^6 \left(\omega^{3k} + \omega^k + 1\right)=3(\omega^3+\omega+1)(\omega^6+\omega^2+1)(\omega^9+\omega^3+1)(\omega^{12}+\omega^4+1)(\omega^{15}+\omega^5+1)(\omega^{18}+\omega^6+1).</cmath>
Line 133: Line 134:
 
To obtain the correct exponents, we seperately add each element of the lower row to one element of the top row.  
 
To obtain the correct exponents, we seperately add each element of the lower row to one element of the top row.  
  
−
Therefore (after reducing <math>\mod 14</math> again), we get the following sums:  
+
Therefore (after reducing <math>\mod 14</math> again), we get the following sets:  
  
 
<cmath>\textbf{(A)}\ \{0, 4, 6, 10, 0, 2, 8, 12, 0\}</cmath>
 
<cmath>\textbf{(A)}\ \{0, 4, 6, 10, 0, 2, 8, 12, 0\}</cmath>
Line 139: Line 140:
 
<cmath>\textbf{(C)}\ \{0, 12, 4, 2, 0, 8, 10, 8, 0\}.</cmath>
 
<cmath>\textbf{(C)}\ \{0, 12, 4, 2, 0, 8, 10, 8, 0\}.</cmath>
  
−
Raising <math>\omega</math> to the power of each of these, then multiplying over <math>\textbf{(A)}, \textbf{(B)},</math> and <math>\textbf{(C)}</math> yields  
+
Raising <math>\omega</math> to the power of each element in every set then multiplying over <math>\textbf{(A)}, \textbf{(B)},</math> and <math>\textbf{(C)}</math> yields  
 +
 
 +
<cmath>\frac{1}{3} \prod_{k=0}^6 \left(\omega^{3k} + \omega^k + 1\right)=\left(\sum_{a\in \textbf{(A)}} \omega^a\right)\left(\sum_{b\in \textbf{(B)}} \omega^b\right)\left(\sum_{c\in \textbf{(C)}} \omega^c\right)</cmath>
  
−
<cmath>\frac{1}{3} \prod_{k=0}^6 \left(\omega^{3k} + \omega^k + 1\right)=\left(\omega^0+\omega^4+\omega^6+\omega^{10}+\omega^0+\omega^2+\omega^8+\omega^{12}+\omega^0\right)^3</cmath>
+
<cmath>=\left(\sum_{a\in \textbf{(A)}} \omega^a\right)^3</cmath>
 +
 
 +
<cmath>=\left(\omega^0+\omega^4+\omega^6+\omega^{10}+\omega^0+\omega^2+\omega^8+\omega^{12}+\omega^0\right)^3</cmath>
  
 
<cmath>=\left(3+\omega^2+\omega^4+\omega^6+\omega^8+\omega^{10}+\omega^{12}\right)^3,</cmath>
 
<cmath>=\left(3+\omega^2+\omega^4+\omega^6+\omega^8+\omega^{10}+\omega^{12}\right)^3,</cmath>
Line 163: Line 168:
  
 
-Benedict T (countmath1)
 
-Benedict T (countmath1)
 +
 +
 +
==Solution 4==
 +
 +
The product can be factored into <math>-(r-1)(s-1)(t-1)(r-w)(s-w)(t-w)(r-w^2)(s-w^2)(t-w^2)....(r-w^6)(s-w^6)(t-w^6)</math>,
 +
 +
 +
where <math>r,s,t</math> are the roots of the polynomial <math>x^3+x+1=0</math>.
 +
 +
 +
This is then <math>-(r^7-1)(s^7-1)(t^7-1)</math> because <math>(r^7-1)</math> and <math>(r-1)(r-w)(r-w^2)...(r-w^6)</math> share the same roots.
 +
 +
 +
To find <math>-(r^7-1)(s^7-1)(t^7-1)</math>,
 +
 +
 +
Notice that <math>(r^7-1)=(r-1)(r^6+r^5+r^4+r^3+r^2+r+1)</math>. Since r satisfies <math>x^3+x+1=0</math>, <math>r^6+r^4+r^3=0</math>
 +
 +
 +
Substituting, you are left with <math>r^5+r^2+r+1</math>. This is <math>r^2(r^3+1)+r+1</math>, and after repeatedly substituting <math>r^3+x+1=0</math> you are left with <math>-2r^3</math>.
 +
 +
 +
So now the problem is reduced to finding <math>-(r-1)(s-1)(t-1)(-2r^3)(-2s^3)(-2t^3)=8(rst)^3(r-1)(s-1)(t-1)</math>, and vietas gives you the result of <math>\boxed{24}</math> -resources
 +
 +
==Solution 5==
 +
The product can be written as <math>(\omega^0+\omega^0+1)(\omega^3+\omega+1)(\omega^6+\omega^2+1)(\omega^9+\omega^3+1)\\(\omega^{12}+\omega^4+1)(\omega^{15}+\omega^5+1)(\omega^{18}+\omega^6+1).</math>
 +
 +
The key here then is noticing that cis<math>2\pi=1.</math>
 +
 +
From this we realize <math>\omega^{7n+a}=\omega^a</math> where <math>a</math> is a residue<math>\pmod 7.</math>
 +
 +
Our expression then simplifies to <math>(3)(\omega^3+\omega+1)(\omega^6+\omega^2+1)(\omega^2+\omega^3+1)\\(\omega^5+\omega^4+1)(\omega+\omega^5+1)(\omega^4+\omega^6+1).</math>
 +
 +
The key then becomes multiplying the 2nd and 7th, 3rd and 6th, 4th and 5th terms because they would result in a couple of <math>\omega^7</math> which will cancel out.
 +
 +
Multiplying them you obtain <math>\omega^6+\omega^5+\omega^4+\omega^3+\omega^2+\omega+3</math>. Recognizing this as <math>\frac{\omega^7-1}{\omega-1}+2.</math> For terms <math>2*7, 3*6, 4*5.</math>
 +
 +
<math>\omega^7-1=0</math> (cis<math>2\pi=1.</math>) We are left with <math>(3)(2)(2)(2)=\boxed{024}.</math>
 +
 +
~BigBrain_2009
 +
 +
==Video Solution by The Power of Logic==
 +
https://youtu.be/o6w9t43GpJs?si=aoe-uM3m5AIwpz_H
  
 
== See also ==
 
== See also ==

Latest revision as of 11:09, 25 December 2024

Problem

Let $\omega = \cos\frac{2\pi}{7} + i \cdot \sin\frac{2\pi}{7},$ where $i = \sqrt{-1}.$ Find the value of the product\[\prod_{k=0}^6 \left(\omega^{3k} + \omega^k + 1\right).\]

Solution 1

For any $k\in Z$, we have, \begin{align*} & \left( \omega^{3k} + \omega^k + 1 \right) \left( \omega^{3\left( 7 - k \right)} + \omega^{\left( 7 - k \right)} + 1 \right) \\ & = \omega^{3 \cdot 7} + \omega^{2k + 7} + \omega^{3k} + \omega^{-2k + 3 \cdot 7} + \omega^7 + \omega^k + \omega^{3\left( 7 - k \right)} + \omega^{\left( 7 - k \right)} + 1 \\ & = 1 + \omega^{2k} + \omega^{3k} + \omega^{-2k} + 1 + \omega^k + \omega^{-3k} + \omega^{-k} + 1 \\ & = 2 + \omega^{-3k} \sum_{j=0}^6 \omega^{j k} \\ & = 2 + \omega^{-3k} \frac{1 - \omega^{7 k}}{1 - \omega^k} \\ & = 2 . \end{align*} The second and the fifth equalities follow from the property that $\omega^7 = 1$.

Therefore, \begin{align*} \Pi_{k=0}^6 \left( \omega^{3k} + \omega^k + 1 \right) & = \left( \omega^{3 \cdot 0} + \omega^0 + 1 \right) \Pi_{k=1}^3 \left( \omega^{3k} + \omega^k + 1 \right) \left( \omega^{3\left( 7 - k \right)} + \omega^{\left( 7 - k \right)} + 1 \right) \\ & = 3 \cdot 2^3 \\ & = \boxed{\textbf{024}}. \end{align*}

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

$k$ cannot be a multiple of $7$, otherwise the first equation would equal $9$ instead of $2$ ~inaccessibles

Solution 2 (Moduli)

Because the answer must be a positive integer, it is just equal to the modulus of the product. Define $z_n = \left(\textrm{cis }\frac{2n\pi}{7}\right)^3 + \textrm{cis }\frac{2n\pi}{7} + 1$.

Then, our product is equal to

\[|z_0||z_1||z_2||z_3||z_4||z_5||z_6|.\]

$z_0 = 3$, and we may observe that $z_x$ and $z_{7-x}$ are conjugates for any $x$, meaning that their magnitudes are the same. Thus, our product is

\[3|z_1|^2|z_2|^2|z_3|^2\] \[= 3\left((\cos \frac{6\pi}{7} + \cos \frac{2\pi}{7} + 1)^2 + (\sin \frac{6\pi}{7} + \sin \frac{2\pi}{7})^2\right) \left((\cos \frac{12\pi}{7} + \cos \frac{4\pi}{7} + 1)^2 + (\sin \frac{12\pi}{7} + \sin \frac{4\pi}{7})^2\right) \left((\cos \frac{4\pi}{7} + \cos \frac{6\pi}{7} + 1)^2 + (\sin \frac{4\pi}{7} + \sin \frac{6\pi}{7})^2\right)\]

Let us simplify the first term. Expanding, we obtain

\[\cos^2 \frac{6\pi}{7} + \cos^2 \frac{2\pi}{7} + 1 + 2\cos \frac{6\pi}{7}\cos \frac{2\pi}{7} + 2\cos \frac{6\pi}{7} + 2\cos \frac{2\pi}{7} + \sin^2 \frac{6\pi}{7} + \sin^2 \frac{2\pi}{7} + 2\sin \frac{6\pi}{7}\sin \frac{2\pi}{7}.\]

Rearranging and cancelling, we obtain

\[3 + 2\cos \frac{6\pi}{7} + 2\cos \frac{2\pi}{7} + 2\cos \frac{6\pi}{7}\cos \frac{2\pi}{7} + 2\sin \frac{6\pi}{7}\sin \frac{2\pi}{7}.\]

By the cosine subtraction formula, we have $2\cos \frac{6\pi}{7}\cos \frac{2\pi}{7} + 2\sin \frac{6\pi}{7}\sin \frac{2\pi}{7} = \cos \frac{6\pi - 2\pi}{7} = \cos \frac{4\pi}{7}$.

Thus, the first term is equivalent to

\[3 + 2(\cos \frac{2\pi}{7} + \cos \frac{4\pi}{7} + \cos \frac{6\pi}{7}).\]

Similarly, the second and third terms are, respectively,

\[3 + 2(\cos \frac{4\pi}{7} + \cos \frac{8\pi}{7} + \cos \frac{12\pi}{7}),\textrm{ and}\] \[3 + 2(\cos \frac{6\pi}{7} + \cos \frac{12\pi}{7} + \cos \frac{4\pi}{7}).\]

Next, we have $\cos \frac{2\pi}{7} + \cos \frac{4\pi}{7} + \cos \frac{6\pi}{7} = -\frac{1}{2}$. This is because

\[\cos \frac{2\pi}{7} + \cos \frac{4\pi}{7} + \cos \frac{6\pi}{7} = \frac{1}{2}(\textrm{cis }\frac{2\pi}{7} + \textrm{cis }\frac{4\pi}{7} + \textrm{cis }\frac{6\pi}{7} + \textrm{cis }\frac{8\pi}{7} + \textrm{cis }\frac{10\pi}{7} + \textrm{cis }\frac{12\pi}{7})\]

\[= \frac{1}{2}(-1)\] \[= -\frac{1}{2}.\]

Therefore, the first term is simply $2$. We have $\cos x = \cos 2\pi - x$, so therefore the second and third terms can both also be simplified to $3 + 2(\cos \frac{2\pi}{7} + \cos \frac{4\pi}{7} + \cos \frac{6\pi}{7}) = 2$. Thus, our answer is simply

\[3 \cdot 2 \cdot 2 \cdot 2\] \[= \boxed{\mathbf{024}}.\]

~mathboy100

Solution 3 (Inspecting the exponents of powers of $\omega$)

We write out the product in terms of $\omega$: \[\prod_{k=0}^6 \left(\omega^{3k} + \omega^k + 1\right)=3(\omega^3+\omega+1)(\omega^6+\omega^2+1)(\omega^9+\omega^3+1)(\omega^{12}+\omega^4+1)(\omega^{15}+\omega^5+1)(\omega^{18}+\omega^6+1).\]

Grouping the terms in the following way exploits the fact that $\omega^{7k}=1$ for an integer $k$, when multiplying out two adjacent products from left to right:

\[\frac{1}{3} \prod_{k=0}^6 \left(\omega^{3k} + \omega^k + 1\right)=(\omega^3+\omega+1)(\omega^{18}+\omega^6+1)(\omega^6+\omega^2+1)(\omega^{15}+\omega^5+1)(\omega^9+\omega^3+1)(\omega^{12}+\omega^4+1).\]


When multiplying two numbers with like bases, we add the exponents. We can now rewrite the exponents of each product (two at a time, where $1$ is treated as the identity) as a series of arrays:

\[\textbf{(A)}\begin{bmatrix} 3&1 &0 \\ 18&6&0\\ \end{bmatrix}\]

\[\textbf{(B)}\begin{bmatrix} 6&2 &0 \\ 15&5&0\\ \end{bmatrix}\]

\[\textbf{(C)}\begin{bmatrix} 9&3 &0 \\ 12&4&0\\ \end{bmatrix}.\]


Note that $\omega=e^{\frac{2\pi i}{7}}$. When raising $\omega$ to a power, the numerator of the fraction is $2$ times whatever power $\omega$ is raised to, multiplied by $\pi i$. Since the period of $\omega$ is $2\pi,$ we multiply each array by $2$ then reduce each entry $\mod{14},$ as each entry in an array represents an exponent which $\omega$ is raised to.


\[\textbf{(A)}\begin{bmatrix} 6&2 &0 \\ 8&12&0\\ \end{bmatrix}\]

\[\textbf{(B)}\begin{bmatrix} 12&4 &0 \\ 2&10&0\\ \end{bmatrix}\]

\[\textbf{(C)}\begin{bmatrix} 4&6 &0 \\ 10&8&0\\ \end{bmatrix}.\]

To obtain the correct exponents, we seperately add each element of the lower row to one element of the top row.

Therefore (after reducing $\mod 14$ again), we get the following sets:

\[\textbf{(A)}\ \{0, 4, 6, 10, 0, 2, 8, 12, 0\}\] \[\textbf{(B)}\ \{0, 8, 12, 6, 0, 4, 2, 10, 0\}\] \[\textbf{(C)}\ \{0, 12, 4, 2, 0, 8, 10, 8, 0\}.\]

Raising $\omega$ to the power of each element in every set then multiplying over $\textbf{(A)}, \textbf{(B)},$ and $\textbf{(C)}$ yields

\[\frac{1}{3} \prod_{k=0}^6 \left(\omega^{3k} + \omega^k + 1\right)=\left(\sum_{a\in \textbf{(A)}} \omega^a\right)\left(\sum_{b\in \textbf{(B)}} \omega^b\right)\left(\sum_{c\in \textbf{(C)}} \omega^c\right)\]

\[=\left(\sum_{a\in \textbf{(A)}} \omega^a\right)^3\]

\[=\left(\omega^0+\omega^4+\omega^6+\omega^{10}+\omega^0+\omega^2+\omega^8+\omega^{12}+\omega^0\right)^3\]

\[=\left(3+\omega^2+\omega^4+\omega^6+\omega^8+\omega^{10}+\omega^{12}\right)^3,\] as these sets are all identical.

Summing as a geometric series,

\[\frac{1}{3} \prod_{k=0}^6 \left(\omega^{3k} + \omega^k + 1\right)=\left(3+\frac{\omega^2(\omega^{12}-1)}{\omega^2-1}\right)^3\]

\[=\left(3+\frac{\omega^{14}-\omega^2}{\omega^2-1}\right)^3\]

\[=\left(3+\frac{1-\omega^2}{\omega^2-1}\right)^3\]

\[=(3-1)^3=8.\]

Therefore,

\[\frac{1}{3} \prod_{k=0}^6 \left(\omega^{3k} + \omega^k + 1\right)=8,\] and \[\prod_{k=0}^6 \left(\omega^{3k} + \omega^k + 1\right)=3\cdot8=\boxed{\textbf{(024)}}.\]

-Benedict T (countmath1)


Solution 4

The product can be factored into $-(r-1)(s-1)(t-1)(r-w)(s-w)(t-w)(r-w^2)(s-w^2)(t-w^2)....(r-w^6)(s-w^6)(t-w^6)$,


where $r,s,t$ are the roots of the polynomial $x^3+x+1=0$.


This is then $-(r^7-1)(s^7-1)(t^7-1)$ because $(r^7-1)$ and $(r-1)(r-w)(r-w^2)...(r-w^6)$ share the same roots.


To find $-(r^7-1)(s^7-1)(t^7-1)$,


Notice that $(r^7-1)=(r-1)(r^6+r^5+r^4+r^3+r^2+r+1)$. Since r satisfies $x^3+x+1=0$, $r^6+r^4+r^3=0$


Substituting, you are left with $r^5+r^2+r+1$. This is $r^2(r^3+1)+r+1$, and after repeatedly substituting $r^3+x+1=0$ you are left with $-2r^3$.


So now the problem is reduced to finding $-(r-1)(s-1)(t-1)(-2r^3)(-2s^3)(-2t^3)=8(rst)^3(r-1)(s-1)(t-1)$, and vietas gives you the result of $\boxed{24}$ -resources

Solution 5

The product can be written as $(\omega^0+\omega^0+1)(\omega^3+\omega+1)(\omega^6+\omega^2+1)(\omega^9+\omega^3+1)\\(\omega^{12}+\omega^4+1)(\omega^{15}+\omega^5+1)(\omega^{18}+\omega^6+1).$

The key here then is noticing that cis$2\pi=1.$

From this we realize $\omega^{7n+a}=\omega^a$ where $a$ is a residue$\pmod 7.$

Our expression then simplifies to $(3)(\omega^3+\omega+1)(\omega^6+\omega^2+1)(\omega^2+\omega^3+1)\\(\omega^5+\omega^4+1)(\omega+\omega^5+1)(\omega^4+\omega^6+1).$

The key then becomes multiplying the 2nd and 7th, 3rd and 6th, 4th and 5th terms because they would result in a couple of $\omega^7$ which will cancel out.

Multiplying them you obtain $\omega^6+\omega^5+\omega^4+\omega^3+\omega^2+\omega+3$. Recognizing this as $\frac{\omega^7-1}{\omega-1}+2.$ For terms $2*7, 3*6, 4*5.$

$\omega^7-1=0$ (cis$2\pi=1.$) We are left with $(3)(2)(2)(2)=\boxed{024}.$

~BigBrain_2009

Video Solution by The Power of Logic

https://youtu.be/o6w9t43GpJs?si=aoe-uM3m5AIwpz_H

See also

2023 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 7 		Followed by
Problem 9
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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