Powers, and more powers

Numbers · 2005-07-12, 21:36

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

Orgasmic Troll · 2005-07-12, 23:35

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?

Orgasmic Troll · 2005-07-13, 00:03

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?

Numbers · 2005-07-13, 04:42

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.

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

2005-07-13, 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

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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?