 mersenneforum.org Powers, and more powers
 Register FAQ Search Today's Posts Mark Forums Read 2005-07-12, 21:36 #1 Numbers   Jun 2005 Near Beetlegeuse 1100001002 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   2005-07-12, 23:35   #2
Orgasmic Troll
Cranksta Rap Ayatollah

Jul 2003

12018 Posts Quote:
 Originally Posted by Numbers 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
Are you saying ^ is a binary operation, but you've left out the definition?   2005-07-13, 00:03 #3 Orgasmic Troll 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 non-negative 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?   2005-07-13, 04:42 #4 Numbers   Jun 2005 Near Beetlegeuse 22·97 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 2005-07-13 at 04:45  Thread Tools Show Printable Version Email this Page Similar Threads Thread Thread Starter Forum Replies Last Post f1pokerspeed Puzzles 6 2012-11-25 00:05 Uncwilly Lounge 15 2010-03-31 07:13 plandon Math 7 2009-06-30 21:29 roger Math 35 2008-07-17 08:15 nibble4bits Math 31 2007-12-11 12:56

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