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Math Whiz Breaks Calculation Record
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."
Just memorize the 13th root of every 100-digit number in existence. Sheesh.
.. 49238826]. That's about 8 million possibilities, and the distribution is far from linear.
:-)
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.
Weeks of coding saves hours of planning.
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.
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.
The 13th root of a 100-digit number is an 8-digit number. Here's how YOU can find TWO of those 8 digits in an instant.
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...
Are described here. Rest of the site is also informative and insane.
http://racine13eme.site.voila.fr/100digang.htm
-pvg