20050712, 21:36  #1 
Jun 2005
Near Beetlegeuse
2^{2}×97 Posts 
Powers, and more powers
I don't think you can solve this by maths alone.
IF 6^4 = 6^2 * (6^2 – 4^3 – 3^2) And 2^7 – 6^2 = 4^3 + 3^4 + 4^3 Then (4^3 + 2^3) / (4^3 – 5^2 – 3^2) = ? You may assume the following: 1) The signs +  = / and * all mean what they normally mean, (plus, minus, equals, divide and multiply) 2) Brackets mean what they normally mean 3) a^b is a term in the domain of the problem which does not (as presented) include negative numbers. Good luck 
20050712, 23:35  #2  
Cranksta Rap Ayatollah
Jul 2003
641_{10} Posts 
Quote:


20050713, 00:03  #3 
Cranksta Rap Ayatollah
Jul 2003
641 Posts 
I'm inclined to think this problem is gibberish.
I'm assuming that you mean ^ is a binary operation on the nonnegative integers. Giving no other restrictions besides the equations given, I can choose a myriad number of operations that satisfy the restrictions given and give me an infinite amount of choices for what the last expression can equal. Note that 2^3 is independent of all other terms in the problem. Therefore, I can define 2^3 to be anything I want and still have a valid operation. What exactly did you have in mind with this? 
20050713, 04:42  #4 
Jun 2005
Near Beetlegeuse
110000100_{2} Posts 
TravisT,
Fair comment. I tried asking questions about puzzles but no one seemed too inclined to answer (or maybe I just asked the wrong questions) so I just dived in and had a go at devising one for myself. I cannot think of a way to define the term 2^3 except by giving you another clue. Does this help? (11^2 – 3^7) – (6^3 – 9^2) = 2^3 BTW, what I meant was that a^b cannot represent a negative value. Last fiddled with by Numbers on 20050713 at 04:45 
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