OT: Interesting math problem... - Page 5

Will · #41 September 22nd, 2004, 02:49 PM

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

Narf:

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

narf poit chez BOOM · #42 September 22nd, 2004, 02:54 PM

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

Alneyan · #43 September 22nd, 2004, 04:03 PM

Quote:

Will said:
Oh, and by the way, I'm taking a pre-graduate level theoretical computer science course and a high level number theory course currently.

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.

Jack Simth · #44 September 22nd, 2004, 08:37 PM

Quote:

Kamog said:
If you're allowed to use logarithms:

sqrt (x) = e^(0.5 * ln(x))

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.

Will · #45 September 23rd, 2004, 12:07 AM

Quote:

narf poit chez BOOM said:
Um...Are you hammering me, the medal, yourself or the thread? Or giving me a hammer?

Yes.

narf poit chez BOOM · #46 September 23rd, 2004, 12:22 AM

/me is confused.

Instar · #47 September 23rd, 2004, 01:30 AM

Quote:

Will said:

Quote:

narf poit chez BOOM said:
Um...Are you hammering me, the medal, yourself or the thread? Or giving me a hammer?

Yes.

Quite.

Kamog · #48 September 23rd, 2004, 01:46 AM

Quote:

Jack Simth said:
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.

Hmm, that's a good point.

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.

narf poit chez BOOM · #49 September 23rd, 2004, 01:55 AM

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

Kamog · #50 September 23rd, 2004, 02:12 AM

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.