bashing out 3^7 terms

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i_equal_tan_90

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i_equal_tan_90

#1 h Feb 16, 2023, 5:56 PM

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Let $\omega=\cos\frac{2\pi}{7}+i\cdot\sin\frac{2\pi}{7}$ , where $i=\sqrt{-1}$ . Find
$\prod_{k=0}^{6}(\omega^{3k}+\omega^k+1).$

This post has been edited 1 time. Last edited by i_equal_tan_90, Feb 16, 2023, 7:39 PM
Reason: original wording

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Rubikscube3.1415

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Rubikscube3.1415

#2 h Feb 16, 2023, 5:58 PM • 1 Y

Y by mathiscool12

Anyone else get 024? (a bit worried about this one)

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Mathdreams

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Mathdreams

#3 h Feb 16, 2023, 5:58 PM • 1 Y

Y by Spakian

Let me calculate while doing english

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RetroTurtle

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RetroTurtle

#4 h Feb 16, 2023, 5:59 PM

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Rubikscube3.1415 wrote:

Anyone else get 024? (a bit worried about this one)

Yes someone confirmed 024. I didn’t solve this unfortunately

This post has been edited 1 time. Last edited by RetroTurtle, Feb 16, 2023, 5:59 PM

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DerpoFanBoy

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DerpoFanBoy

#5 h Feb 16, 2023, 5:59 PM

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yep checked in wolfram alpha afterwards it is 024

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Inconsistent

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Inconsistent

#6 h Feb 16, 2023, 6:13 PM • 2 Y

Y by Dansman2838, yshk

24. Just pair w and w inverse in the product and get $2 \cdot 2 \cdot 2 \cdot 3$ .

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ihatemath123

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ihatemath123

#7 h Feb 16, 2023, 6:21 PM

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This problem really scared me when I first read it.

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djmathman

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djmathman

#8 h Feb 16, 2023, 6:26 PM • 13 Y

Y by CT17, mannshah1211, Inconsistent, brainfertilzer, khina, rayfish, Dansman2838, EpicBird08, MathFan335, Amir Hossein, Danielzh, huluwa, Sedro

Alternatively, if you want to carry out a more general solution...

Let $P$ denote the product in question, and let $Q(x) = x^3 + x + 1$ . Denote by $\alpha_1$ , $\alpha_2$ , $\alpha_3$ the roots of $Q$ . Then
$P = \prod_{k=0}^6(\omega^k-\alpha_1)(\omega^k-\alpha_2)(\omega^k-\alpha_3) = -(\alpha_1^7-1)(\alpha_2^7-1)(\alpha_3^7-1).$ Now let $z$ be any root of $x^3+x+1$ for ease of typesetting. Then $z^3 = -(z+1)$ so $z^6 = z^2 + 2z+ 1$ , which implies
$z^7 = z^3 + 2z^2 + z = 2z^2 - 1.$ It follows that
$P = -(2\alpha_1^2 - 2)(2\alpha_2^2 - 2)(2\alpha_3^2 - 2) = 8Q(1)Q(-1) = \boxed{24}.$

This post has been edited 3 times. Last edited by djmathman, Oct 1, 2023, 6:10 PM
Reason: fixed a minus sign; see post #32 for a more detailed explanation

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JacobGuo

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JacobGuo

#9 h Feb 16, 2023, 6:42 PM • 2 Y

Y by Inconsistent, honey_lemon

interesting

Edit: It should also be noted that if $P(x) = 1 + x + x^3$ , then $P(x^k)$ is the conjugate of $P(x^{7-k})$ , so the answer is indeed real.

This post has been edited 1 time. Last edited by JacobGuo, Feb 16, 2023, 6:55 PM

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mathleticguyyy

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mathleticguyyy

#10 h Feb 16, 2023, 6:46 PM • 2 Y

Y by centslordm, hdanger

The answer must be congruent to 2187 mod 7 by symmetry. This makes estimating the magnitude of each term fairly viable

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brainfertilzer

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brainfertilzer

#11 h Feb 16, 2023, 7:37 PM

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This feels “””inspired””” by #8 on 2015 bmt analysis round: https://bmt.berkeley.edu/wp-content/uploads/2022/03/AnaS2015.pdf. The solution for the bmt problem applies perfectly to this (and is the same as the one in post #8)

This post has been edited 1 time. Last edited by brainfertilzer, Feb 16, 2023, 7:37 PM

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jeff10

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jeff10

#12 h Feb 16, 2023, 8:26 PM • 10 Y

Y by Inconsistent, CoolCarsOnTheRun, IAmTheHazard, centslordm, ike.chen, mathleticguyyy, rayfish, yshk, Math4Life7, CyclicISLscelesTrapezoid

Direct

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Reason: grammar

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ASnooby

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ASnooby

#13 h Feb 17, 2023, 2:44 AM

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Yay I got this.
Bad solution bash

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mathboy100

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mathboy100

#14 h Feb 17, 2023, 5:24 AM

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Moduli sol, long but not that bashy:

sol

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am07

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am07

#15 h Feb 17, 2023, 2:56 PM

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trig bash

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lifeismathematics

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lifeismathematics

#16 h Feb 20, 2023, 6:36 PM • 1 Y

Y by aidensharp

let's bash it :gleam:

we have the product as follows :

$3(\omega^{3}+\omega+1)(\omega^{6}+\omega^2+1)(\omega^2+\omega^3+1)(\omega^{5}+\omega^{4}+1)(\omega+\omega^{5}+1)(\omega^{4}+\omega^{6}+1)$

now club any two of them, do the product of them to get it is nothing but $3\cdot 2\cdot 2\cdot 2=\boxed{24}$

This post has been edited 1 time. Last edited by lifeismathematics, Feb 20, 2023, 6:43 PM

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Arrowhead575

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Arrowhead575

#17 h Feb 20, 2023, 6:42 PM

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@above, u can make it easier to bash by letting w^k= w^k-7 for k>3

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Tizzy

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Tizzy

#18 h Feb 21, 2023, 4:11 AM

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I solved this question on the test by grouping some terms together and then applying trigonometric identities which simplified it to 3x2x2x2= 024

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resources

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resources

#19 h Feb 21, 2023, 1:17 PM

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Who else misses the old aime trig style problems?
Glad that they brought it bAck

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waffles_123

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waffles_123

#21 h Feb 21, 2023, 3:19 PM

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mathboy100 wrote:

Moduli sol, long but not that bashy:

sol

i like this solution

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gracemoon124

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gracemoon124

#22 h Mar 31, 2023, 2:07 AM

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solution

This post has been edited 1 time. Last edited by gracemoon124, Mar 31, 2023, 2:13 AM
Reason: lay-tecks

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ryanc226

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ryanc226

#23 h Mar 31, 2023, 2:38 AM

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did anyone get 024? i am cofused.

This post has been edited 5 times. Last edited by ryanc226, Apr 1, 2023, 3:02 AM

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gracemoon124

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gracemoon124

#24 h Mar 31, 2023, 5:34 PM

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@above how exactly do you just let an expression in complex numbers be a variable $x$ ?

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kn07

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kn07

#26 h Jun 19, 2023, 11:42 PM

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brainfertilzer wrote:

This feels “””inspired””” by #8 on 2015 bmt analysis round: https://bmt.berkeley.edu/wp-content/uploads/2022/03/AnaS2015.pdf. The solution for the bmt problem applies perfectly to this (and is the same as the one in post #8)

also looks like kmo first round 2007 p7

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ex-center

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ex-center

#27 h Sep 3, 2023, 5:16 PM

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$\prod_{k=0}^{6}(\omega^{3k}+\omega^k+1)=(\omega^{0}+\omega^{0}+\omega^{0})\prod_{k=1}^{3}(\omega^{3k}+\omega^{k}+1)(\omega^{3(7-k)}+\omega^{7-k}+1)$ $3\prod_{k=1}^{3}(2+\omega^{0k}+\omega^{k}+\omega^{2k}+\omega^{3k}+\omega^{4k}+\omega^{5k}+\omega^{6k})=3\prod_{k=1}^{3}2$ $3 \cdot 2^{3}=24$

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shendrew7

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shendrew7

#28 h Oct 1, 2023, 12:14 AM

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Suppose $r_1$ , $r_2$ , and $r_3$ are the roots of $x^3+x+1$ . First we note $r_i^7 = r_i(r^3)^2 = r_1(-r_1-1)^2 = r_i^3 + 2r_i^2 + r_i = 2r_i - 1.$ Then we compute our product to be

$\begin{align*} \prod_{j=0}^{6}(\omega^{3j}+\omega^j+1) &= \prod_{j=0}^6(\omega^j-r_1)(\omega^j-r_2)(\omega^j-r_3) \\ &= \prod_{i=1}^3 \prod_{j=0}^6 -(r_i - \omega^j) \\ &= \prod_{i=1}^3 \left((-1)^7 (a_i^7-1)\right) \\ &= -(2r_1^2 - 2)(2r_2^2 - 2)(2r_3^2 - 2) \\ &= \boxed{24} \end{align*}$

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am07

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am07

#29 h Oct 1, 2023, 1:31 PM

Y by

lifeismathematics wrote:

let's bash it :gleam:

we have the product as follows :

$3(\omega^{3}+\omega+1)(\omega^{6}+\omega^2+1)(\omega^2+\omega^3+1)(\omega^{5}+\omega^{4}+1)(\omega+\omega^{5}+1)(\omega^{4}+\omega^{6}+1)$

now club any two of them, do the product of them to get it is nothing but $3\cdot 2\cdot 2\cdot 2=\boxed{24}$

i solved the same way

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PaperMath

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PaperMath

#30 h Oct 1, 2023, 2:59 PM

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Is there a way to solve this problem without trig identities?

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PaperMath

958 posts

PaperMath

#31 h Oct 1, 2023, 3:10 PM • 1 Y

Y by djmathman

djmathman wrote:

Alternatively, if you want to carry out a more general solution...

Let $P$ denote the product in question, and let $Q(x) = x^3 + x + 1$ . Denote by $\alpha_1$ , $\alpha_2$ , $\alpha_3$ the roots of $Q$ . Then
$P = \prod_{k=0}^6(\omega^k-\alpha_1)(\omega^k-\alpha_2)(\omega^k-\alpha_3) = -(\alpha_1^7-1)(\alpha_2^7-1)(\alpha_3^7-1).$ Now let $z$ be any root of $x^3+x+1$ for ease of typesetting. Then $z^3 = -(z+1)$ so $z^6 = z^2 + 2z+ 1$ , which implies
$z^7 = z^3 + 2z^2 + z = 2z^2 - 1.$ It follows that
$P = -(2\alpha_1^2 - 2)(2\alpha_2^2 - 2)(2\alpha_3^2 - 2) = -8Q(1)Q(-1) = \boxed{24}.$

Wait how is $P = \prod_{k=0}^6(\omega^k-\alpha_1)(\omega^k-\alpha_2)(\omega^k-\alpha_3) = -(\alpha_1^7-1)(\alpha_2^7-1)(\alpha_3^7-1)?$

Also for your last part - $P = -(2\alpha_1^2 - 2)(2\alpha_2^2 - 2)(2\alpha_3^2 - 2) = -8Q(1)Q(-1) = \boxed{24}$ , should’t it be $-24$ since $Q(1)$ is $3$ , and $Q(-1)$ is $1$ ?

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djmathman

7938 posts

djmathman

#32 h Oct 1, 2023, 6:09 PM

Y by

Ah, I got my signs mixed up on that second part! The first part is correct, though.

One of the annoying things about the Fundamental Theorem of Algebra is that all terms must be of the form $x-r$ , where $r$ is a root of the given polynomial. In the first expression, the $\omega^k$ terms are positive when they should be negative, so I need to negate them to make the signs work out. That is,
$\begin{align*} \prod_{k=0}^6(\omega^k-\alpha_1)(\omega^k-\alpha_2)(\omega^k-\alpha_3) &= (-1)^{21}\prod_{k=0}^6(\alpha_1-\omega^k)(\alpha_2-\omega^k)(\alpha_3-\omega^k)\\ &= -(\alpha_1^7 - 1)(\alpha_2^7 - 1)(\alpha_3^7 - 1). \end{align*}$ The second part is similar, though I missed a negative sign when typing this up. The correct argument should be
$\begin{align*} P &= -(2\alpha_1^2 - 2)(2\alpha_2^2 - 2)(2\alpha_3^2 - 2)\\ &= -8(\alpha_1-1)(\alpha_1+1)(\alpha_2-1)(\alpha_2+1)(\alpha_3-1)(\alpha_3+1)\\ &= 8\left[(\alpha_1-1)(\alpha_2-1)(\alpha_3-1)\right]\left[(-1-\alpha_1)(-1-\alpha_2)(-1-\alpha_3)\right]\\ &= 8Q(1)Q(-1). \end{align*}$ There's probably a better way to organize the negative signs in this computation, but eh, that's the price you pay when moving between different polynomials.

This post has been edited 1 time. Last edited by djmathman, Oct 1, 2023, 6:09 PM

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OlympusHero

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OlympusHero

#33 h Dec 27, 2023, 6:23 PM • 1 Y

Y by ilikemath247365

Notice that $(\omega^{3k}+\omega^k+1)(\omega^{3(7-k)}+\omega^{7-k}+1) = (\omega^{3k}+\omega^k+1)(\omega^{-3k}+\omega^{-k}+1) = 1 + \omega^{2k}+\omega^{3k}+\omega^{-2k}+1+\omega^{k}+\omega^{-3k}+\omega^{-k}+1 = 3 + \omega^{k}+\omega^{2k}+\omega^{3k}+\omega^{-k}+\omega^{-2k}+\omega^{-3k}$ . But this is equal to $3 + (\omega^k+\omega^{2k}+\omega^{3k}+\omega^{4k}+\omega^{5k}+\omega^{6k}) = 3-1 = 2$ . To finish, note that the first term is just $3$ , and we can pair off the remaining terms, giving $3 \cdot 2^3 = 24$ .

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Mr.Sharkman

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Mr.Sharkman

#35 h Apr 4, 2024, 12:52 AM

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Solution

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KnowingAnt

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KnowingAnt

#36 h Jul 10, 2024, 3:31 PM

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$3\prod_{k = 1}^{3} \left(\omega^{3k} + \omega^k + 1\right)\left(\omega^{-3k} + \omega^{-k} + 1\right) = 3\prod_{k = 1}^{3}2 = 24$
........

This post has been edited 1 time. Last edited by KnowingAnt, Jul 10, 2024, 3:32 PM

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FliX0onbo

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FliX0onbo

#37 h Jul 10, 2024, 5:31 PM

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My solution

This post has been edited 3 times. Last edited by FliX0onbo, Jul 10, 2024, 5:33 PM

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joshualiu315

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joshualiu315

#38 h Aug 17, 2024, 6:24 PM

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If $k=0$ , the value equals $3$ , so we need to find

$3 \prod_{k=1}^{6} \left(\omega^{3k} + \omega^k + 1\right)$
Then, note that

$\begin{align*} &\phantom{=} (x^{3k}+x^k+1)(x^{3(7-k)}+x^{7-k}+1) \\ &= x^{21}+x^{2k+7}+x^{3k}+x^{21-2k}+x^7+x^k + x^{21-3k}+x^{7-k}+1 \\ &= 3+x^{-3k}+x^{-2k}+x^{-k}+x^k+x^{2k}+x^{3k} \\ &= \sum_{i=0}^6 x^{ik} + 2 = 2. \end{align*}$
The answer is consequently $3 \cdot 2^3 = \boxed{24}$ .

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lpieleanu

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lpieleanu

#39 h Aug 18, 2024, 7:10 AM

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Solution

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eg4334

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eg4334

#40 h Nov 30, 2024, 8:06 PM

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The idea is to pair up terms with $k$ summing to $7$ : $(w^{3k}+w^k + 1)(w^{21-3k} + w^{7-k} + 1) = (w^{3k}+w^k+1)(\frac{1}{w^{3k}} + \frac{1}{w^k} + 1) = 3 + (w+w^2+\dots+w^6) = 2$ The $k=0$ case obvioulsy gives $3$ , so we have $3 \cdot 2^3 = \boxed{24}$

This post has been edited 1 time. Last edited by eg4334, Nov 30, 2024, 8:06 PM

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MathRook7817

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MathRook7817

#41 h Dec 1, 2024, 7:35 PM

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can u use roots of unity for this one?

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anduran

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anduran

#42 h Dec 1, 2024, 7:48 PM

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MathRook7817 wrote:

can u use roots of unity for this one?

Mostly (or all) solutions rely on the fact that $\omega^7=1,$ or that $\omega$ is a $7$ -th root of unity.

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SpeedCuber7

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SpeedCuber7

#43 h Dec 1, 2024, 7:53 PM

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i could swear $\omega$ was commonly used for the 3rd root but i guess its any complex unity root huh

what do i know

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Mr.Sharkman

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Mr.Sharkman

#44 h Dec 9, 2024, 11:35 PM

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Bro its the og 2024 AIME II/8

Solution

This post has been edited 2 times. Last edited by Mr.Sharkman, Dec 18, 2024, 12:16 AM

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Mr.Sharkman

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Mr.Sharkman

#45 h Dec 22, 2024, 10:40 PM

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SpeedCuber7 wrote:

$\omega$ was commonly used for the 3rd root

yeah this is true; like no one uses $\zeta$ for a 3rd root of unity, but $\omega$ is definitely not only a third root of unity

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ilikemath247365

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ilikemath247365

#46 h Dec 23, 2024, 3:01 AM

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OlympusHero wrote:

Notice that $(\omega^{3k}+\omega^k+1)(\omega^{3(7-k)}+\omega^{7-k}+1) = (\omega^{3k}+\omega^k+1)(\omega^{-3k}+\omega^{-k}+1) = 1 + \omega^{2k}+\omega^{3k}+\omega^{-2k}+1+\omega^{k}+\omega^{-3k}+\omega^{-k}+1 = 3 + \omega^{k}+\omega^{2k}+\omega^{3k}+\omega^{-k}+\omega^{-2k}+\omega^{-3k}$ . But this is equal to $3 + (\omega^k+\omega^{2k}+\omega^{3k}+\omega^{4k}+\omega^{5k}+\omega^{6k}) = 3-1 = 2$ . To finish, note that the first term is just $3$ , and we can pair off the remaining terms, giving $3 \cdot 2^3 = 24$ .

This is one of the simplest and most effective solutions to this problem.

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akliu

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akliu

#47 h Apr 7, 2025, 12:11 PM

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Note that for $k = 0$ , the value in the product is $(1+1+1)=3$ . Rewrite the last $6$ terms of our product as:

$\begin{align*} \prod_{k = 1}^6(\omega^{3k} + \omega^k + 1) &= \prod_{k = 1}^3(\omega^{3k} + \omega^k + 1)(\omega^{-3k} + \omega^{-k} + 1) \\ &= \prod_{k=1}^3((1 + \omega^{-2k} + \omega^{-3k}) + (\omega^{2k} + 1 + \omega^{-k}) + (\omega^{3k} + \omega^k + 1)) \\ &= \prod_{k=1}^3(2 + (\omega^{-3k} + \omega^{-2k} + \omega^{-k} + 1 + \omega^{k} + \omega^{2k} + \omega^{3k})) \\ &= \prod_{k=1}^3 (2 + 0) \text{ (since for k=1, 2, 3 we have primitive roots of unity)} \\ &= 8 \end{align*}$
Meaning that our answer is $3 \cdot 8 = \boxed{24}$

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