Arithmetic problem from Tobaku Haouden Zero.

hyperpape · **#21**

Oh, and if we want to test ourselves, someone should give us all the first 8 digits of 3 ^ (1/2)...

daniel_the_smith · **#22**

OK, go for it: 1.73205081

Edit: I tried calculating the above by hand with newton's method. I made a mistake somewhere. But using newton's method with a calculator, I got it correct. OK, I will try getting two more places by hand.

hyperpape · **#23**

Owwwww! I did that by hand. I believe 1.73205081 ^ 2 = 3.0000000084211561 (Python reports it as 3.0000000084216563, which fails a basic sanity check).

So we're a victim of rounding. We wanted 1.73205080.

Also, an important question is whether they have graph paper and what their handwriting is like. I did my first try on notebook paper, and it got out of alignment while I was squaring 1.73...

illluck · **#24**

I just tried to use the a^2+2ab approximation for root 3 by hand. It took me 20 minutes to do the calculations (my math sucks) and I made a mistake in there. Using a calculator, though, it does seem like the approximation is good enough.

daniel_the_smith · **#25**

hyperpape wrote:

Owwwww! I did that by hand. I believe 1.73205081 ^ 2 = 3.0000000084211561 (Python reports it as 3.0000000084216563, which fails a basic sanity check).

So we're a victim of rounding. We wanted 1.73205080.

Also, an important question is whether they have graph paper and what their handwriting is like. I did my first try on notebook paper, and it got out of alignment while I was squaring 1.73...

Yeah, I realized it was evil to round it, but: someone already mentioned that the given digits might have been rounded, and ordinarily I would round a number like that if giving it to N digits, so...

I decided I didn't have the patience to try this by hand and instead wrote a computer program to calculate square roots to arbitrary precision. Here's the fist 100k digits of sqrt(3): http://pastebin.com/6aV2Dn68

illluck · **#26**

@hyperpape: My manual calculations give something close to the python result - 3.0000000084216561. I completely messed up the long division - I see no less than 3 errors in calculating 4 digits XD

For someone who doesn't fail as horribly as I do at math, though, the approximation is quite possible to do in 20 minutes (perhaps even less time, I had problems with aligning figures - would have been a good idea to just drew straight lines as grids).

Cassandra · **#27**

I think, hyperpage's suggestion of

b = (2 - a * a) / (2 * a)

provides the fastest method, and it surely can be done in 25 minutes.

@ Araban:
What method has been choosen by the Hero ?

Solomon · **#28**

Zero's method:

daniel_the_smith · **#29**

First 8 digits of square root of five per Google: 2.23606798 (in case someone wants to try Zero's method)

Solomon · **#30**

daniel_the_smith wrote:

First 8 digits of square root of five per Google: 2.23606798 (in case someone wants to try Zero's method)

Sigh...I actually did take the time to multiply this by itself by hand, only to get 5.0000000111812804, which had me confused because it had to be < 5. So I checked the first 8 digits of sqrt(5), only to realize that Google rounded up the 8th digit so the first 8 digits of sqrt(5) is actually 2.23606797__.

Those are 11 minutes I'm never getting back daniel_the_smith

.

daniel_the_smith · **#31**

I'm sorry, that time it wasn't intentional. I guess I should have put a disclaimer that I didn't know if it was rounded or not.

As penance, if I have a chance, I'll see if I can square that by hand in less than 11 minutes...

Edit: Actually, I don't feel too bad because I'm pretty sure I wasted more than 11 minutes writing a square root calculator due to this thread...

hyperpape · **#32**

I can't believe I didn't think of that. I use that shortcut really often.

It's probably not a huge time difference, but you avoid the subtraction, and more importantly, you don't have to worry about whether you're dropping the right digits.

Fedya · **#33**

My first thought was that

2 = (200000/141421) * (282842/200000)

Add those two fractions and divide by 2, and you get

(79999798482/56568400000)

Either do the long division out to ten places, or multiply 56568400000 * 1.41421356 (since you know the first eight places), and then do the division for the last two digits.

As the denominator is eleven digits, you have to be correct to at least 1/56568400000, which is ten places after the decimal.

Bill Spight · **#34**

I think maybe Fedya has the right idea. We can use Newton's method with 1.41421.

A little algebra gives us sqrt(2) - 1.41421 = delta = 1/1.41421 - 1.41421/2 , to 10 digits.

1.41421/2 = 0.707105

And we know that 1/1.41421 starts off as 0.70710, since 1.41421 is less than the square root of 2.

The long division to get the next five digits is not so bad.

So to find delta we have

0.7071085624 -
0.707105
------------
0.0000035624

Which means that digits 9 and 10 are 2 and 4, respectively.

Edit: Actually, we do not have to do that last subtraction, do we.

Edit again: Actually, we do not have to do long division for 5 digits, since we know that 1/1.41421 starts off as 0.70710856, don't we?

We only have to do long division for the final two digits.

Edit again: And to get those two digits, we can divide at that point by 1.414, as that will give us sufficient accuracy. (1.41 might do, but I like a little leeway.

)

Bill Spight · **#35**

Araban wrote:

Zero's method:

What rule is Zero using to get the 2 and 3?

Once he has 1.9999999932878736 he can perform the division 6712/283 to get the next two digits with sufficient accuracy. I. e., (2 - a^2)/2a.

Cassandra · **#36**

Bill Spight wrote:

What rule is Zero using to get the 2 and 3?

Once he has 1.9999999932878736 he can perform the division 6712/283 to get the next two digits with sufficient accuracy. I. e., (2 - a^2)/2a.

Yes, and even the "long" division with the 8 digits-numbers can be done in the remaining 14 minutes. (Devision by 2 should not be the big problem.)

That's relative simple, because we already have most of the n * a (n = 1, 2, ...) from doing the multiplication before.

And it is not necessary to perform the "manual" devision-steps for more that 3 digits. Remember, that "manual" devision here is nothing more than substraction.

Overall, this method has less calculating steps than the method, which has been choosen by the hero. And there remains the fact of his unexplained ability of guessing "2" and "3".

jts · **#37**

Presumably he's multiplying n*3,000,000,000 (and then by 300,000,000) to estimate how many chunks of 1,414,000,000 you need to add to round off to 2 x 10^m... you need about 60,000,000,000 in the first and about 1,200,000,000 in the second.

mitsun · **#38**

This solution is the same as others have posted, but maybe thinking about fractional accuracy makes it a little easier to see. Once the hero calculates that (1.414235600)^2 = 1.9999999933 (note that he does not need further digits), he can observe that this starting point falls short by 67 parts in the last two decimal places. To get those 67 parts back, he has to increase the starting point by 67/2/1.4 (again no more digits are needed) = 23 parts in the last two decimal places. He can spend any remaining time verifying this answer

ChradH · **#39**

Just found this article about Japanese mnemonics for numbers.

Arithmetic problem from Tobaku Haouden Zero.

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