Arithmetic problem from Tobaku Haouden Zero.

[daniel_the_smith]

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]

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]

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]

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]

Zero's method:

A lot less sophisticated than some of the answers here . He begins by multiplying 1.414213560 by itself. He uses the other people in the room to do the same calculation for verification. Next, he uses an arithmetic rule regarding multiplication of numbers differing by 1 value. Simply, once you know what 1414213560 * 1414213560 is, to check 1414213561 * 1414213561, you need to only add (1414213560 + 1414213560 + 1) to 1414213560 * 1414213560. In other words: 1414213560 * 1414213560 + 1414213560 + 1414213560 + 1 = 1414213561 * 1414213561

He decides to start with 2 for the 9th digit based on the answer he got for 1414213560 * 1414213560, which is 1.999999993287873600. Using the rule, he arrives at 1.999999998944727844. With 11:53 minutes left on the clock, he applies the same rule once again to arrive at 3 for his 10th digit:



In the game he was allowed 2 tries to determine the 'password' (the 9th and 10th digits). He uses 23 for his first try, only to realize he had to round it up based on:



Hence arriving at 24 as the correct password. If anyone's confused about what I said or want more detail (or just want to read the story), you can read the part in the manga here: http://www.mangafox.com/manga/tobaku_ha ... .2/96.html

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]

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]

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]

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]

Just found this article about Japanese mnemonics for numbers.