Powers, and more powers

[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

[Orgasmic Troll]

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]

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]

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.