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June 12th, 2018, 09:03 AM  #1 
Member Joined: Aug 2017 From: India Posts: 50 Thanks: 2  Clarification on this problem
I am kind of self learning maths and trying to solve some of the problems online and came across the following question. How to solve it? I really don't understand the question also.

June 12th, 2018, 09:46 AM  #2 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,950 Thanks: 987 
If p=2, q=0, r=1, s=8, then according to the wording of the question, there is only 1 solution....((p^q)^r)^s = 1 If rearranging allowed, then 4! = 24 solutions. Kinda weird, if you ask me! 
June 12th, 2018, 11:52 AM  #3 
Senior Member Joined: May 2016 From: USA Posts: 1,305 Thanks: 548 
Denis needs to stand in the corner AGAIN. Yes, you can permute those 4 integers 4! = 24 ways. HOWEVER, $((p^q)^r)^s = p^{(qrs)}.$ And the order in which you multiply things is irrelevant. $p = 0 \implies qrs = 1 * 2 * 8 = 16 \implies ((p^q)^r)^s = 0^{16} = 0.$ $p \ne 0 \implies qrs = 0 \implies ((p^q)^r)^s = p^0 = 1.$ There are two possible outcomes. 
June 12th, 2018, 02:00 PM  #4 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,950 Thanks: 987  True...BUT you can still multiply 'em irrelevantly! 2 ways to multiply 1 and 2: 1*2 and 2*1: which one is irrelevant? EDIT: does the above make me a 2timer? Last edited by Denis; June 12th, 2018 at 02:04 PM. 
June 12th, 2018, 04:04 PM  #5 
Senior Member Joined: May 2016 From: USA Posts: 1,305 Thanks: 548  

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