Slashdot Log In
Math Whiz Breaks Calculation Record
Posted by
samzenpus
on Wed Nov 24, 2004 08:27 PM
from the show-off dept.
from the show-off dept.
keyshawn632 writes "The Associated Press reports that Gert Mittring, 38, needed only 11.8 seconds to calculate the 13th root of a 100-digit number in his head at a math museum in Giessen, a small town, located in western Germany.
It's worth noting though that his feat will not be recognized by The Guinness Book Of World Records because of the difficulty of standardizing such mathematical challenges."
This discussion has been archived.
No new comments can be posted.
The Fine Print: The following comments are owned by whoever posted them. We are not responsible for them in any way.
What? (Score:5, Insightful)
Re:What? (Score:5, Interesting)
Re:What? (Score:5, Funny)
To memorize 22 digits, this guy takes ~4 seconds. So for 100 digits that would take about 18 seconds.
Now I forgot, he did what in 11.8 seconds
That's easy. (Score:5, Funny)
Re:That's easy. (Score:5, Informative)
Let's just think about this for a minute.
100-digit numbers will fall between 10^99 and 10^100. Thirteenth-roots of such numbers will lie between 10^(99 / 13) and 10^(100 / 13), or in the range [41246264
But it's linear enough that the first nine digits of the 100-digit number yield a unique possibility for a root. And the last digit of the root will be the same as the last digit of the 100-digit number, because (N mod 10) always equals (N^13 mod 10). So the problem can be tackled from both ends, with the middle digits of the root being the hardest.
Of course, if the audience members are clued in, they can still beat the mental calculator hands down. Type the first nine digits, take the thirteenth root, and start reading off the digits; round up slightly to make the eighth significant digit match the final digit of the 100-digit number. Done.
A college professor of mine taught us how to square 3-digit numbers in our head in seconds using tricks like this; he was able to multiply arbitrary 5-digit numbers in his head, and often performed this onstage. And for the curious, yes, I do actually have a life outside slashdot.
Sources report... (Score:5, Funny)
Ironic...I'm currently listening to... (Score:5, Funny)
Devi: another brilliant mathematical mind (Score:5, Interesting)
The book itself was an interesting read, and at the time I just ate it up. It has a lot of tricks regarding number theory, mathematical riddles, calendar tricks, and calculation of pi, for example. It teaches how to figure the day of the week for any Gregorian date of any time in a few seconds, a trick which I still remember and use today!
As for the Pi, it contained a few poems and sayings whose letter counts signified the individual digits. I started trying to memorize pi, with my sights set firmly on the world record (as I am not without my own mathematical and mnemonic prowess). However, around grade 9, I decided to abandon my quest in order to get a life. I had memorized 1350 digits at that point.
One such quote held little significance for me at the time, but has since become hilarious. "How I want a drink, alcoholic of course, after the heavy chapters involving quantum mechanics!" Needless to say, my quantum prof found it quite funny. :)
High pi (Score:5, Funny)
This isn't to say that 1350 digits wouldn't be useful! If you ever wake up in an alternate universe (you were warned about operating quantum machinery while drunk!) just look up pi in a math book. The degree of trouble you're in could correlate to the digit at which your memorized value, and the local value of pi, diverge.
If pi only diverges after 1000 or more digits, you're probably alright, except for having to re-memorize pi.
If pi diverges after 100 digits, there may be some minor historical divergences, like, say, President Nixon being impeached, or Bush winning a second term. The mind boggles!
If pi diverges after 30 or 40 digits, look out the window. Do dinosaurs roam the earth? Since you're surrounded by ruthless, math-book-publishing carnivores, consider donating yourself to the primate house of the zoo.
If local pi begins with a number other than 3, you should start to get worried, or maybe implode.
method to calculate the day of the week! (Score:5, Funny)
First, figure out the "year number". This part -- and the month number -- take some practice. Here's the first few to get you started:
1900 - 0
1904 - 5
1908 - 3
1912 - 1
1916 - 6
1920 - 4
1924 - 2
1928 - 0
And it repeats thusly. Note that the "year number" starts at 0 for the beginning of the century, and is decreased by two (modulo seven) every leap year.
In case you're interested in the other 75% of the time, simply add one to the year number for every year you add. Thus, 1901 becomes 1, 1902 becomes 2, etc.
The "month" number requires memorization of another table, which cannot be recalculated as quickly as the year number:
Jan - 0
Feb - 3
Mar - 3
Apr - 6
May - 1
Jun - 4
Jul - 6
Aug - 2
Sep - 5
Oct - 0
Nov - 3
Dec - 5
Add the month number to the year number. If your year is a leap year and your month is January or February, subtract 1.
Next, add the day number. The day number is the day. :P
Now, add or subtract sevens as necessary until you end up with a number between 0 and 6:
0 - Sunday
1 - Monday
2 - Tuesday
3 - Wednesday
4 - Thursday
5 - Friday
6 - Saturday
The result will be the day of the week.
If your desired date does not begin with a "19", you have to add a century number as well. I believe 2000 is a leap year, since every 100 years is not but every 400 years is. Thus, the century number of 2000 is 6 (or, equivalently, -1). 1800 is 5, 1700 is 3, etc. (I am not certain of these.)
As an example, today's year number is 5, the month number is 3, and the day number is 24. After compensating for the century by subtracting 1, we obtain 31. This reduces to 3 (by subtracting 28), which corresponds to Wednesday. Since it is Wednesday, and since I am in a large empty room, I further deduce that the lecture has ended.
Re:method to calculate the day of the week! (Score:5, Funny)
Now where were we? Oh yeah - the important thing was I had an onion on my belt, which was the style at the time. They didn't have white onions because of the war. The only thing you could get was those big yellow ones...
The first mentat? (Score:5, Funny)
I so call bullshit (Score:5, Interesting)
There has to be a trick to it aside from "thinking really fast"
Tom
Re:I so call bullshit (Score:5, Insightful)
Well of course, there is. Probably two or three tricks combined. .
Walking a tightrope is more than just having "good balance," and it's really just a trick, and not necessarily a very useful one, but. .
It is still pretty impressive and you can't do it.
KFG
ahh (Score:5, Funny)
That's the problem when dealing with a highly subjective field like mathematics.
What he will be doing next week... (Score:5, Insightful)
Probably breaking codes for some government or another. Someone with talent with numbers and such will catch the eye of someone out there. Could it be that this was just to show off his talent as a sort of "job interview"? Probably not, but I expect he will get some calls about it anyway.
Gert disqualified and sued! (Score:5, Funny)
The news set off a legal feeding frenzy. SCO sued Mr. Mittring for using the company's super secret 13th root finder source code. Microsoft then added to the man's woes by suing for patent infringement over Microsoft's patents on 100 digit numbers. RIAA then sued him for including "8675309" in the answer -- obviously a stolen clip from "Jenny" by Tommy Tutone.
Oh come on now. (Score:5, Funny)
I can do the 23rd root of a 163 digit number in 5.8 seconds, and I wasn't even trying. I've climbed Mt. Everest in an hour and a half. I can rewrite the Linux kernel in under an hour. I can count up to ten thousand coins in no more than a minute.
And yet, curiously, it takes me almost...-checks watch- five minutes to make a stupid, useless post on /. Strange eh?
To how many significant figures? (Score:5, Interesting)
The 13th root of a 100-digit number will always have 7 digits. If you memorize the first few digits of the 13th powers of numbers between 49 and 58 and you are given a 100-digit number, then you immediately know the first 2 digits of the 13th root. Memorize the initial digits of 13th power of numbers between 491 and 588 and you immediately know the first 3 digits. By memorizing the terminal digits of 13th powers of numbers less than 100, you could similarly immediately get the last 3 digits. That leaves 1 digit to compute, which is a slightly less impressive-sounding feat for 11.8 seconds. It's not a trivial calculation, though, and not at all shabby for 11.8 seconds.
Jonathan
This is not as difficult as it sounds. (Score:5, Informative)
1. The leading digit is ALWAYS 4.
2. The last digit of the 13-th root of N is always the same as the last digit of N.
(The first fact follows because Floor[N[(10^100 - 1)^(1/13)]] = 49238826 and Floor[N[(10^99 - 1)^(1/13)]] = 41246263. The second holds because N^13 is congruent to N modulo 10.)
With minimal practice, you can get the second-highest digit from the magnitude. Beyond that I can only speculate what he's doing. But by taking an alternating sum of the digits, you get its value mod 11, which gives you the value of the root mod 11, which buys you another digit. Now you're halfway there...
Re:And the answer is: (Score:5, Funny)
Re:I can do better (Score:5, Funny)
"Wisconsin"
"W."
Re:And still has 0.00005% of getting laid (Score:5, Funny)
And that would be the only rooting this guy will ever do in his life
Simpler than that (Score:5, Informative)
That leaves you with a mere... 7,193,306 possible roots to memorize.
I don't know how they do it, but I am familiar with modulo-10 math "tricks". For example, did you know that if you add up the individial digits in any number and the result is divisible by 3, then the original number is divisible by 3? For example "621". 6+2+1=9, and so 621 is divisible by 3 (Try it: 621/3=207).
13th root has similar magic: the 13th root of any number will have the same last digit as the number you are trying to take the root of. For example, the 13th root of 2235879388560037062539773567 is 127. Notice that they both end in 7. An integer and its 13th power always ends in the same digit. Try it.
The point is, that little trick itself reduces the problem space by a factor of 10 right there. So I'm assuming they've studied and learned further tricks like these. Ask them for the 11th root of the same number and they'll probably come up completely blank.
Re:I can (seriously) do 43rd root of 100 digit num (Score:5, Informative)
Well, I guess that's not so outrageous depending on the precision you need. All the 43rd roots of 100 digit numbers are greater than 200 and less than 212, so if you only need integer precision you only have 13 choices. And memorizing 12 thresholds is not that hard.