Re: OT: Interesting math problem...
 

It's also known as "lucky guess". While it is possible for a puzzle like that one to have one of the letters be zero, it is usually the case that one zero makes all zero. Plus, since we were only going for a right answer to the puzzle, I could have chose V=0 first, found it was not valid, then jumped back into V=9. But since I did V=9 first, and got a good answer, I could halt right there.

Oh, and by the way, I'm taking a pre-graduate level theoretical computer science course and a high level number theory course currently. Might have something to do with the intuition development http://forum.shrapnelgames.com/images/smilies/happy.gif

Narf:

[img]/threads/images/Graemlins/Hammer.gif[/img]

Re: OT: Interesting math problem...
 

Um...Are you hammering me, the medal, yourself or the thread? Or giving me a hammer?

Re: OT: Interesting math problem...
 

You have my sympathy, for dread fills my heart at the very thought of the hardships you are underdoing. I can but express my compassion to you, and wish you well despite your suffering.

For your heroic efforts, I hereby grant you a Medal of Gallanty. Or I would have, had Narf not already done so. Besides, we are in a bit of a shortage of medals these days.

*Ducks for cover, right behind Narf. Hammers happen to fall from the skies these days* All this message was obviously to be taken with a grain of salt, and more than a single grain if more is available.

Re: OT: Interesting math problem...
 

Kamog, I don't mean to be picky, but if you have decimal exponents available to do the e^ portion, you don't need logarithms at all:

sqrt (x) = x^(0.5)

- it's a basic identity.

Re: OT: Interesting math problem...
 

Yes.

Re: OT: Interesting math problem...
 

/me is confused.

Re: OT: Interesting math problem...
 

Quite.

Re: OT: Interesting math problem...
 

Hmm, that's a good point. http://forum.shrapnelgames.com/images/smilies/happy.gif

Yeah, I know that identity. Actually I started off with it to get the equation:
y = x^0.5
ln(y) = ln(x^0.5)
ln(y) = 0.5*ln(x)
y = e^(0.5*ln(x))

I see what you mean, why use logarithms if you have decimal exponents available.

Re: OT: Interesting math problem...
 

So, how do you calculate 64^0.5 without a calculator?

Re: OT: Interesting math problem...
 

Here's a trick to calculate square roots of square numbers without a calculator. Keep subtracting odd numbers like this:

1. 64 - 1 = 63
2. 63 - 3 = 60
3. 60 - 5 = 55
4. 55 - 7 = 48
5. 48- 9 = 39
6. 39 - 11 = 28
7. 28 - 13 = 15
8. 15 - 15 = 0

It took 8 steps to get to 0 so the square root of 64 is 8. I don't know why this works.