My Math Forum I found new solution for this problem
 User Name Remember Me? Password

 Algebra Pre-Algebra and Basic Algebra Math Forum

 May 6th, 2019, 04:14 PM #11 Newbie   Joined: Oct 2017 From: Egypt Posts: 15 Thanks: 0 Okay, but what if the given was only three brackets? 59934030_653367038444064_1692238877223288832_n.jpg Last edited by yoyo145; May 6th, 2019 at 04:42 PM.
 May 6th, 2019, 04:54 PM #12 Global Moderator   Joined: Dec 2006 Posts: 21,028 Thanks: 2259 If the number of factors is odd, the right-hand side will need to involve ω.
May 6th, 2019, 04:56 PM   #13
Newbie

Joined: Oct 2017
From: Egypt

Posts: 15
Thanks: 0

Quote:
 Originally Posted by skipjack If the number of factors is odd, the right-hand side will need to involve ω.
The number of factors is even: 2n

Last edited by skipjack; May 7th, 2019 at 02:28 AM.

 May 7th, 2019, 02:27 AM #14 Global Moderator   Joined: Dec 2006 Posts: 21,028 Thanks: 2259 I took "only three brackets" to mean "only three factors", which won't arise (because 2n is even).
May 7th, 2019, 02:37 AM   #15
Newbie

Joined: Oct 2017
From: Egypt

Posts: 15
Thanks: 0

Quote:
 Originally Posted by skipjack I took "only three brackets" to mean "only three factors", which won't arise (because 2n is even).
can you solve the new problem and explain what do you mean ?
prove that:
59934030_653367038444064_1692238877223288832_n.jpg
=1

Last edited by yoyo145; May 7th, 2019 at 02:55 AM.

 May 7th, 2019, 03:01 AM #16 Global Moderator   Joined: Dec 2006 Posts: 21,028 Thanks: 2259 What new problem? For problem (8), consider the factors in successive pairs, as already described. Can you try it for yourself and post your work?
May 7th, 2019, 03:20 AM   #17
Newbie

Joined: Oct 2017
From: Egypt

Posts: 15
Thanks: 0

Quote:
 Originally Posted by skipjack What new problem? For problem (8), consider the factors in successive pairs, as already described. Can you try it for yourself and post your work?
Okay, but really I don't understand why 2n factors means successive pairs. When I read it, I think the question mean to 8 factor or to 10 factors.

Last edited by skipjack; May 7th, 2019 at 06:47 AM.

 May 7th, 2019, 06:57 AM #18 Global Moderator   Joined: Dec 2006 Posts: 21,028 Thanks: 2259 The question doesn't specify successive pairs. I suggested successive pairs because that's a convenient way to find the product.
 May 7th, 2019, 07:13 AM #19 Newbie   Joined: Oct 2017 From: Egypt Posts: 15 Thanks: 0 So your answer is not general because if you get three factors we will not figure out the right answer. Please check my proof. Last edited by skipjack; May 7th, 2019 at 07:18 AM.
 May 7th, 2019, 07:17 AM #20 Global Moderator   Joined: Dec 2006 Posts: 21,028 Thanks: 2259 There can't be just three factors because it is stated that there are 2n factors.

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post Hardik New Users 0 October 18th, 2015 10:06 AM Einstein's_Riddle_2 Applied Math 4 June 18th, 2015 05:38 AM Ganesh Ujwal Physics 3 January 5th, 2015 01:58 AM Obsessed_Math Calculus 4 February 9th, 2012 03:18 PM TreeTruffle Algebra 2 March 27th, 2010 01:22 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top

Copyright © 2019 My Math Forum. All rights reserved.