My Math Forum Clarification on this problem

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June 12th, 2018, 09:03 AM   #1
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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.
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 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! Thanks from MathsLearner123
 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
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Quote:
 Originally Posted by JeffM1 And the order in which you multiply things is irrelevant.
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 2-timer?

Last edited by Denis; June 12th, 2018 at 02:04 PM.

June 12th, 2018, 04:04 PM   #5
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Quote:
 Originally Posted by Denis 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 2-timer?
24 timer

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