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June 12th, 2018, 08:03 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, 08:46 AM   #2
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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!
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June 12th, 2018, 10:52 AM   #3
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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.
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June 12th, 2018, 01:00 PM   #4
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Quote:
Originally Posted by JeffM1 View Post
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 01:04 PM.
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June 12th, 2018, 03:04 PM   #5
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Quote:
Originally Posted by Denis View Post
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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