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• Dear all I ve spent some weeks now looking at 13th root of 100 digits. My main motivation for this was the way this category of mental calculation was put on a
Message 1 of 17 , Aug 5, 2003
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Dear all

I've spent some weeks now looking at 13th root of 100 digits. My main motivation for this was the way this category of mental calculation was put on a pedestal as the ultimate category. I have yet to really start practicing in earnest but already believe I have derived a method for 13th root of odd powers that does not require an incredibly high IQ as has been suggested, nor does it require memorisation of log tables, or great ability at factoring numbers a la Kleins method. I have not yet finalised the method for the last 4 digits of even powers. Also, odd powers ending in 5 are a bit tricky still - I will work on these next. However, my method for odd powers (except 5) is very rapid indeed.

My method uses mnemonics for digits 1 to 4 (based on the first 4 digits of the power) and exceedingly simple calculations and for digits 5 to 8...... I have to remember less than 1000 numbers using mnemonics and only simple division by three or simple subtraction for the mental calculations. Surely this is easier than learning up to 100 x 100 table yet 13th root of 100 digits was talked about as if it was some kind of holy grail.

13.5 seconds is Alexis record I believe - well, with digit 8 being immediately recalled, and digits 1 to 4 being recalled within a couple of seconds through my mnemonic system, this would then leave approx 10 seconds to make three further small calculations......hmm, seems imminently possible! A nice power (ending in 1 for example) could be done in 6 or 7 seconds I believe......

So, what are the parameters for setting world records? How many attempts in what timeframe? Where does this take place??? Are you able to pass on some attempts to wait for a 'nicer' calculation?

I believe I will be <20 seconds within 1 month - I shall keep you all informed of my progress.

Paul

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• ... attempts in what timeframe? Where does this take place??? Are you able to pass on some attempts to wait for a nicer calculation? The suggested rules for
Message 1 of 17 , Aug 5, 2003
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--- In MentalCalculation@yahoogroups.com, Paul Reed <t_paulr@y...> wrote:

> So, what are the parameters for setting world records? How many
attempts in what timeframe? Where does this take place??? Are you able
to pass on some attempts to wait for a 'nicer' calculation?

The suggested rules for the record list at
http://www.recordholders.org/en/list/memory.html
can be found at
http://www.recordholders.org/en/rules/calculating.html.

The number of attempts per day / per year must be limited in order
to avoid good luck-"records".

Ralf
• Dear Calculator friends, What is the special reason that the 13th root - of an huge number- seems to be the magic standard? Does one of you have an idea? Eg I
Message 1 of 17 , Jun 15, 2008
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Dear Calculator friends,

What is the special reason that the 13th root - of an huge number- seems to be the magic standard? Does one of you have an idea?

Eg I never see something of 5th root, a11 th or a 15th or a 17th. Why is that?

And then I renounce of even roots. I know: a 4th root gives heaps of possibilities and an 8th still much more! So I can imagine this is almost impossible.

But why only 13th??

Regards,

Willem Bouman

[Non-text portions of this message have been removed]
• Hi Willem, I think one reason for this is because prime roots are more difficult than composite roots. For example if we have the 15th root then we can take
Message 1 of 17 , Jun 15, 2008
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Hi Willem,
I think one reason for this is because prime roots are more difficult than composite roots.
For example if we have the 15th root then we can take the cube root, then the 5th root. Though obviously one must remember some long numbers while doing this!
So we've got the 7th, 11th, 13th and 17th roots to contend with.
I know the number from which we need to find the root has 100 digits. The 13th root of a 100digit number will have 8 digits. The 11th root will have 10 digits. The 17th root will have just 6 digits - maybe this isn't quite enough to make the problem difficult.
But certainly the 7th and 11th roots would be tricky enough!
Will you be getting exciting about the football on the Sunday in Leipzig? England have done me a big favour by not qualifying (I am disappointed really!), but the Dutch are looking very strong indeed - they have to be my tip to win Euro 2008!
Andy

----- Original Message ----
From: A.W.A.P. Bouman <awap.bouman@...>
To: mental calculation <MentalCalculation@yahoogroups.com>
Sent: Sunday, 15 June, 2008 10:40:11 AM
Subject: [Mental Calculation] roots

Dear Calculator friends,

What is the special reason that the 13th root - of an huge number- seems to be the magic standard? Does one of you have an idea?

Eg I never see something of 5th root, a11 th or a 15th or a 17th. Why is that?

And then I renounce of even roots. I know: a 4th root gives heaps of possibilities and an 8th still much more! So I can imagine this is almost impossible.

But why only 13th??

Regards,

Willem Bouman

[Non-text portions of this message have been removed]

__________________________________________________________
Sent from Yahoo! Mail.
A Smarter Email http://uk.docs.yahoo.com/nowyoucan.html

[Non-text portions of this message have been removed]
• The nth root challenge requires n to be prime because the fourth root for example could involve taking the square root twice and so may be less challenging
Message 1 of 17 , Jun 15, 2008
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The nth root challenge requires n to be prime because the fourth root
for example could involve taking the square root twice and so may be
less challenging than taking cube roots. This is likely only part of
the answer. This also applies to integer exponentiation. Raising a
number like 57 to the power of 8 is easier than raising it to the power
of 7 because one can square it 3 times. E.g.

57^2 = 3249 (Known before hand)

3249^2 = (3200^2) + (here comes the difficult part) 2*3200*49 (This is
6400 * 49 (hey both start with squares! a shortcut) = 560^2) + (49^2)
=10556001

As you can see the only problem now is squaring that number which I
would choose over 57*57*57*57*57*57*57 anyday.

From Jsh
• It was 23 and 73 too... http://stepanov.lk.net/mnemo/smith13e.html AWAPB Dear Calculator friends, AWAPB What is the special reason that the 13th root - of an
Message 1 of 17 , Jun 15, 2008
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It was 23 and 73 too...
http://stepanov.lk.net/mnemo/smith13e.html

AWAPB> Dear Calculator friends,

AWAPB> What is the special reason that the 13th root - of an
AWAPB> huge number- seems to be the magic standard? Does one of you
AWAPB> have an idea?

AWAPB> Eg I never see something of 5th root, a11 th or a
AWAPB> 15th or a 17th. Why is that?

AWAPB> And then I renounce of even roots. I know: a 4th root
AWAPB> gives heaps of possibilities and an 8th still much more! So I
AWAPB> can imagine this is almost impossible.

AWAPB> But why only 13th??

AWAPB> Regards,

AWAPB> Willem Bouman

Sincerely Yours, Oleg Stepanov.
http://stepanov.lk.net/
• Dear Mr. Stepanov, Thank you very much for your interesting article, which I printed immediately. I ll tell you something very interesting. When I met Wim
Message 1 of 17 , Jun 15, 2008
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Dear Mr. Stepanov,

Thank you very much for your interesting article, which I printed immediately.

I'll tell you something very interesting. When I met Wim Klein in December 1959 he examined me and asked me which problems I could lose. And among others I mentioned cubis roots. "impossible" he reacted. Even roots can be done, odd ones cannot. "Make me please an excersize of an integer cubic root of 15 digits". Klein did, and asked of course how I did this. In a flash he understood and later on he worked my system more and more out.

He gave me the advise not do calendar calculation, it is a banality, he said. And now I pay the price: now I have to do in the competition, and although I make some progress, I'll never be a star in it.

In August 1986 I wanted to renew the contact, but unfortunately he was murdered.

Regards,

Willem Bouman

----- Oorspronkelijk bericht -----
Van: Oleg Stepanov
Aan: A.W.A.P. Bouman
Verzonden: zondag 15 juni 2008 14:48
Onderwerp: Re: [Mental Calculation] roots

It was 23 and 73 too...
http://stepanov.lk.net/mnemo/smith13e.html

AWAPB> Dear Calculator friends,

AWAPB> What is the special reason that the 13th root - of an
AWAPB> huge number- seems to be the magic standard? Does one of you
AWAPB> have an idea?

AWAPB> Eg I never see something of 5th root, a11 th or a
AWAPB> 15th or a 17th. Why is that?

AWAPB> And then I renounce of even roots. I know: a 4th root
AWAPB> gives heaps of possibilities and an 8th still much more! So I
AWAPB> can imagine this is almost impossible.

AWAPB> But why only 13th??

AWAPB> Regards,

AWAPB> Willem Bouman

Sincerely Yours, Oleg Stepanov.
http://stepanov.lk.net/

[Non-text portions of this message have been removed]
• Dear Mr. Flynn, With all respect: I do not agree with you. An integer cubic root is far less challenging: when you work on the right way in combination with
Message 1 of 17 , Jun 15, 2008
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Dear Mr. Flynn,

With all respect: I do not agree with you. An integer cubic root is far less challenging: when you work on the right way in combination with modulo calculation, there is always 1 possibility. 4 th roots can surely be more tricky because of the number of possibilities.

I agree with your argument in multiplication, my question nevertheless is the opposite: the roots.

Regards,

Willem Bouman

----- Oorspronkelijk bericht -----
Van: jsh.flynn
Aan: MentalCalculation@yahoogroups.com
Verzonden: zondag 15 juni 2008 14:23
Onderwerp: [Mental Calculation] Re: roots

The nth root challenge requires n to be prime because the fourth root
for example could involve taking the square root twice and so may be
less challenging than taking cube roots. This is likely only part of
the answer. This also applies to integer exponentiation. Raising a
number like 57 to the power of 8 is easier than raising it to the power
of 7 because one can square it 3 times. E.g.

57^2 = 3249 (Known before hand)

3249^2 = (3200^2) + (here comes the difficult part) 2*3200*49 (This is
6400 * 49 (hey both start with squares! a shortcut) = 560^2) + (49^2)
=10556001

As you can see the only problem now is squaring that number which I
would choose over 57*57*57*57*57*57*57 anyday.

From Jsh

[Non-text portions of this message have been removed]
• Ha, Andy, Thank you for your reaction. you are right when you speak about the number of digits of the answer. If I remember wll it was Wim Klein who said: The
Message 1 of 17 , Jun 15, 2008
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Ha, Andy,

Thank you for your reaction. you are right when you speak about the number of digits of the answer. If I remember wll it was Wim Klein who said: "The trickyness of a root is not in the exponent, but in the number of digits of the answer". From this point of view a 17 th root of 100 digits would be easier than a 13 th root of 100 digit. With an answer of 8 digits a 17 th power couild have between 120 and 136 digits, as you know.

29 June there will be a Dutch performance in Leipzig: a certain Mr. Willem Bouman will give then a presentation with exersizes followed by an explanation. You can ask Ralf Laue if there is a place for you, the spoken language will be German.

Till then!

Willem Bouman

----- Oorspronkelijk bericht -----
Van: Andy Robertshaw
Aan: MentalCalculation@yahoogroups.com
Verzonden: zondag 15 juni 2008 13:22
Onderwerp: Re: [Mental Calculation] roots

Hi Willem,
I think one reason for this is because prime roots are more difficult than composite roots.
For example if we have the 15th root then we can take the cube root, then the 5th root. Though obviously one must remember some long numbers while doing this!
So we've got the 7th, 11th, 13th and 17th roots to contend with.
I know the number from which we need to find the root has 100 digits. The 13th root of a 100digit number will have 8 digits. The 11th root will have 10 digits. The 17th root will have just 6 digits - maybe this isn't quite enough to make the problem difficult.
But certainly the 7th and 11th roots would be tricky enough!
Will you be getting exciting about the football on the Sunday in Leipzig? England have done me a big favour by not qualifying (I am disappointed really!), but the Dutch are looking very strong indeed - they have to be my tip to win Euro 2008!
Andy

----- Original Message ----
From: A.W.A.P. Bouman <awap.bouman@...>
To: mental calculation <MentalCalculation@yahoogroups.com>
Sent: Sunday, 15 June, 2008 10:40:11 AM
Subject: [Mental Calculation] roots

Dear Calculator friends,

What is the special reason that the 13th root - of an huge number- seems to be the magic standard? Does one of you have an idea?

Eg I never see something of 5th root, a11 th or a 15th or a 17th. Why is that?

And then I renounce of even roots. I know: a 4th root gives heaps of possibilities and an 8th still much more! So I can imagine this is almost impossible.

But why only 13th??

Regards,

Willem Bouman

[Non-text portions of this message have been removed]

__________________________________________________________
Sent from Yahoo! Mail.
A Smarter Email http://uk.docs.yahoo.com/nowyoucan.html

[Non-text portions of this message have been removed]

[Non-text portions of this message have been removed]
• AWAPB Dear Mr. Stepanov, AWAPB Thank you very much for your interesting article, which I printed immediately. This is not article. This is chapter from book.
Message 1 of 17 , Jun 15, 2008
View Source
AWAPB> Dear Mr. Stepanov,

AWAPB> Thank you very much for your interesting article, which I printed immediately.
This is not article. This is chapter from book.

Sincerely Yours, Oleg Stepanov.
http://stepanov.lk.net/
• Dear Mr. Stepanov, Thanks again for your interesting reaction and the link it contents. My problem: it is more than 50 years ago that I left the secundary
Message 1 of 17 , Jun 15, 2008
View Source
Dear Mr. Stepanov,

My problem: it is more than 50 years ago that I left the secundary school, I have to study the log questions and mantissas thoroughly.

My solution: not so long ago I found a formula for finding end figures of roots.
BN = basic number, in this case 11, I speak about the second problem in your link, the 13 th rot of 75185.....85831.
N= exponent, or power, in this case 13.
Jump: the number which with the end figures change. 11^13 end on 3931, the problem ends on 3931.
My formula - on which I am a little proud- is

Jump = BN^(n-1) × power × 100. In this case:

11^12×13×100= 721×13×100= ....7300.

Now we subtract 5831-3931= 1900. The question is yet how many ...7300's go in ...1900. It is 3, so the answer ends on 0311.

Regards,

Willem Bouman

----- Oorspronkelijk bericht -----
Van: Oleg Stepanov
Aan: A.W.A.P. Bouman
Verzonden: zondag 15 juni 2008 14:48
Onderwerp: Re: [Mental Calculation] roots

It was 23 and 73 too...
http://stepanov.lk.net/mnemo/smith13e.html

AWAPB> Dear Calculator friends,

AWAPB> What is the special reason that the 13th root - of an
AWAPB> huge number- seems to be the magic standard? Does one of you
AWAPB> have an idea?

AWAPB> Eg I never see something of 5th root, a11 th or a
AWAPB> 15th or a 17th. Why is that?

AWAPB> And then I renounce of even roots. I know: a 4th root
AWAPB> gives heaps of possibilities and an 8th still much more! So I
AWAPB> can imagine this is almost impossible.

AWAPB> But why only 13th??

AWAPB> Regards,

AWAPB> Willem Bouman

Sincerely Yours, Oleg Stepanov.
http://stepanov.lk.net/

[Non-text portions of this message have been removed]
• Dear Mr. Bouman. I do not know is it interesting. But here one more chapter from that book: http://stepanov.lk.net/mnemo/sm16.html AWAPB Dear Mr. Stepanov,
Message 1 of 17 , Jun 16, 2008
View Source
Dear Mr. Bouman.
I do not know is it interesting. But here one more chapter from
that book:
http://stepanov.lk.net/mnemo/sm16.html

AWAPB> Dear Mr. Stepanov,

AWAPB> Thanks again for your interesting reaction and the link it contents.

AWAPB> My problem: it is more than 50 years ago that I left
AWAPB> the secundary school, I have to study the log questions and
AWAPB> mantissas thoroughly.

AWAPB> My solution: not so long ago I found a formula for
AWAPB> finding end figures of roots.
AWAPB> BN = basic number, in this case 11, I speak about the
AWAPB> second problem in your link, the 13 th rot of 75185.....85831.
AWAPB> N= exponent, or power, in this case 13.
AWAPB> Jump: the number which with the end figures change.
AWAPB> 11^13 end on 3931, the problem ends on 3931.
AWAPB> My formula - on which I am a little proud- is

AWAPB> Jump = BN^(n-1) × power × 100. In this case:

AWAPB> 11^12×13×100= 721×13×100= ....7300.

AWAPB> Now we subtract 5831-3931= 1900. The question is yet
AWAPB> how many ...7300's go in ...1900. It is 3, so the answer ends
AWAPB> on 0311.

AWAPB> Regards,

AWAPB> Willem Bouman

Sincerely Yours, Oleg Stepanov.
http://stepanov.lk.net/
• Dear Mr. Stepanov, Thanks again. Willem Bouman ... Van: Oleg Stepanov Aan: A.W.A.P. Bouman Verzonden: maandag 16 juni 2008 10:20 Onderwerp: Re: [Mental
Message 1 of 17 , Jun 16, 2008
View Source
Dear Mr. Stepanov,

Thanks again.

Willem Bouman

----- Oorspronkelijk bericht -----
Van: Oleg Stepanov
Aan: A.W.A.P. Bouman
Verzonden: maandag 16 juni 2008 10:20
Onderwerp: Re: [Mental Calculation] roots

Dear Mr. Bouman.
I do not know is it interesting. But here one more chapter from
that book:
http://stepanov.lk.net/mnemo/sm16.html

AWAPB> Dear Mr. Stepanov,

AWAPB> Thanks again for your interesting reaction and the link it contents.

AWAPB> My problem: it is more than 50 years ago that I left
AWAPB> the secundary school, I have to study the log questions and
AWAPB> mantissas thoroughly.

AWAPB> My solution: not so long ago I found a formula for
AWAPB> finding end figures of roots.
AWAPB> BN = basic number, in this case 11, I speak about the
AWAPB> second problem in your link, the 13 th rot of 75185.....85831.
AWAPB> N= exponent, or power, in this case 13.
AWAPB> Jump: the number which with the end figures change.
AWAPB> 11^13 end on 3931, the problem ends on 3931.
AWAPB> My formula - on which I am a little proud- is

AWAPB> Jump = BN^(n-1) × power × 100. In this case:

AWAPB> 11^12×13×100= 721×13×100= ....7300.

AWAPB> Now we subtract 5831-3931= 1900. The question is yet
AWAPB> how many ...7300's go in ...1900. It is 3, so the answer ends
AWAPB> on 0311.

AWAPB> Regards,

AWAPB> Willem Bouman

Sincerely Yours, Oleg Stepanov.
http://stepanov.lk.net/

[Non-text portions of this message have been removed]
• Dear mathalthletes,   I had notice a great issue regarding 8 digit, my point of view   it had been agreed   8 digit x 8 digit mulltiplication even you can
Message 1 of 17 , Jun 22, 2008
View Source
Dear mathalthletes,

I had notice a great issue regarding 8 digit, my point of view

8 digit x 8 digit mulltiplication even you can do more than 8.

for square root, 8 digit answer

it may had been thought a root of 100 digit number
therefore they had choosen the 13 root, because it have 8 digit answer and
As ralf explanined once, there a little tricks regarding certain numbers in 13th root, so only look for the few first both end number and you can give a quick the answer, therefore
Alex and the big GERT during a show when more than 100 attempt had been displayed,
in one of their attempts they had had done that magic of less than 3 second answer for 13th root. which led to don t accept anymore one attempt shot for any mental calculation operation since 2004.
At that time it was for my inconvinience because i was preparing to break the 44.7 s gert metring record for square root. I remember while practicing once i had done like a square in 12.7 second in july 2004, for 20 % of number i can go for less than than 44.7 s and in others i go for between 80 to 90 s.

See you guys next weeks and have a great sunday, I hope that germany will be in final of euro 2008 so we will have a great gathering sunday night.

Final world, we need ALL to thank RALF for all his efforts for organizing the world championship for mental calcualtion, Big THANKS RALF and see you next week.

regards
Issam

--- On Sun, 6/15/08, Andy Robertshaw <robertshaw_andy@...> wrote:

From: Andy Robertshaw <robertshaw_andy@...>
Subject: Re: [Mental Calculation] roots
To: MentalCalculation@yahoogroups.com
Date: Sunday, June 15, 2008, 2:22 PM

Hi Willem,
I think one reason for this is because prime roots are more difficult than composite roots.
For example if we have the 15th root then we can take the cube root, then the 5th root. Though obviously one must remember some long numbers while doing this!
So we've got the 7th, 11th, 13th and 17th roots to contend with.
I know the number from which we need to find the root has 100 digits. The 13th root of a 100digit number will have 8 digits. The 11th root will have 10 digits. The 17th root will have just 6 digits - maybe this isn't quite enough to make the problem difficult.
But certainly the 7th and 11th roots would be tricky enough!
Will you be getting exciting about the football on the Sunday in Leipzig? England have done me a big favour by not qualifying (I am disappointed really!), but the Dutch are looking very strong indeed - they have to be my tip to win Euro 2008!
Andy

----- Original Message ----
From: A.W.A.P. Bouman <awap.bouman@ casema.nl>
To: mental calculation <MentalCalculation@ yahoogroups. com>
Sent: Sunday, 15 June, 2008 10:40:11 AM
Subject: [Mental Calculation] roots

Dear Calculator friends,

What is the special reason that the 13th root - of an huge number- seems to be the magic standard? Does one of you have an idea?

Eg I never see something of 5th root, a11 th or a 15th or a 17th. Why is that?

And then I renounce of even roots. I know: a 4th root gives heaps of possibilities and an 8th still much more! So I can imagine this is almost impossible.

But why only 13th??

Regards,

Willem Bouman

[Non-text portions of this message have been removed]

____________ _________ _________ _________ _________ _________ _
Sent from Yahoo! Mail.
A Smarter Email http://uk.docs. yahoo.com/ nowyoucan. html

[Non-text portions of this message have been removed]

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