wow. It’s great task to master powers of 300,000 numbers. Like impossible mission.
Keep in mind, he can do the first 20 powers really FAST: That doesnt mean he can do more, youll find much longer recitations (up to the power of 200!). If you look how much digits that is, you get near a million of digits- the question about HOW to store such a number still is a miracle.
Great to see the discussion coming up again!
To make things more human, in a somewhat weird japanese championships, Rüdiger made a mistake in “putting together” the 17.th power of a number. As far as my knowledge goes, this was the first reported error ever (!) in higher powers.
Learning powers is just plain hard and -possibly- boring. Still working up to the fifth…
Resurrecting this old thread…
While I’m waiting for 3x3 multiplication to get easier I thought I might start to play with logarithms.
The long string addition and subtraction seems like a good memory practice for the major system and a good stretch for holding longer numbers in my head.
Has anyone created practice tools software/spreadsheet for calculating logs mentally. I’m sure I can make something in google docs or excel but if it’s already out there it would be nice to have. I prefer practicing the skill to building the tools for this hobby 
thanks,
Robert
With regard to some of the exponent calculations above it would seem to me a lot easier to use
a^2 + b^2 + 2ab + b^2 … a^2 + b^2 + c^2 + 2ab + 2ac + 2ac … etc for
44^2 444^2…
lots of easy steps this way and you can just keep expanding upwards by 2’s
similarly
for sum of cubes
a^3 + b^3 = (a + b) (a^2 - ab + b^2)
i.e. 84^3
I’m pretty sure Euler has a formula for sums of fourth and fifth powers in his book on elements of algebra. I’ll have to check this evening.
Is there a general algebraic solution to sums of powers for decomposing…
xyz ^ n such that
x^n f(j) + y^n f(j) + z^n f(j) + f(xyz) …
i.e. 123^12
I use either (google) spreadsheets or a programmable calculator for this.
"I'm pretty sure Euler has a formula for sums of fourth and fifth powers"Robert, from the Binomial theorem we can derive formulas for all kinds of powers of (a+b)^n, for n integer
The expansions get complicated after 4th power,but all those formulas definitely exist.
About cubes that you mentioned, (e.g. 84^3) there was a cube root task in MCWC-2012, http://www.recordholders.org/downloads/worldcup/tasks-2012/tasks2012.pdf where we had to extract non integer cubes of six digits, up to 8 significant places
e.g. we should find that cube of 594245 = 84.072736. (8 correct, rounded). One such correct result was marked with 8 points. For all 10 tasks the maximum was 80 points. The gold medal winner Jan van Koningsveld, got 50 points, that’s an average of 5 correct cube digits. I got a 7th place with 33 points. The task was only 10minutes, that’s why in order to save time for calculating decimals,I memorised the first 100 cubes. But I in case I forgot any cube, I also had in mind the formulas in order to calculate, in case needed
a^3 + b^3 = (a + b) (a^2 - ab + b^2)
and (a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3
and (a-b)^3 = a^3 - 3a^2b + 3ab^2 - b^3
I talked with Andreas Berger who came 2nd then and he also told me he had memorised the 100 cubes. I know that’s not calculation, but sometimes it’s useful to have a few more stuff read for recall out of your mental inventory. Besides, it’s only 90 new numbers (assuming that mostmental calculators know the first 10 cubes (1,8,27,64,125, 216, 343, 512, 729
And actually for all non-integer roots the fun starts after the decimals, that’s where the real algorithm starts to unfold.
About the memorised cubes, since the task contain 6-digit roots I payed more attention to the cubes from 46-99, which yield a six-digit cube. It’s not like we memorised 10,000+ powers like Gamm did. The task is simpler. I think with a few mnemonics those cubes can also be memorised, they seem bit like telephone numbers.
46 97336
47 103823
48 110592
49 117649
50 125000
132651
140608
148877
157464
166375
175616
185193
195112
205379
216000
226981
238328
250047
262144
274625
287496
300763
314432
328509
343000
357911
373248
389017
405224
421875
438976
456533
474552
493039
512000
531441
551368
571787
592704
614125
636056
658503
681472
704969
729000
753571
778688
804357
830584
857375
884736
912763
941192
970299
All the above have integer cube root (46-99)
Nodas
To add a newer clip with some interesting answers, I finally come back to this place :-).
This is Rüdiger interviewed by Nelson Dellis.
Rüdiger half admits to memorizing the answers.
When asked how much calculation is going on compared to memorizing he (imo) avoids the answer.
He answers so quick that he (again; imo) cannot do much calculation anyway.
So until I see any evidence to the contrary I believe he has memorized the answers.
Impressive feat though, from 1^1 to 99^99 is about 10,000 numbers!
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