Who looks?

--------------------------
http://www.optimnem.co.uk
Press Math
In just one week, program users can expect the following benefits:
You will be able to calculate any sum up to 99x99 in your head quickly
and confidently (useful for: tax returns, grocery bills, math homework)
--------------------------

Who have an opinion?

Sincerely Yours Oleg Stepanov.

http://www.lk.net/~stepanov/

Dear Alexis Lemaire.

I get your letter to me and answered you two days ago.
But after I get message that your address can not get my message.
On your letter was written same - you have problems. May be we can
discuss here?

    As to me, I suppose... and this is based on my knowledge:
    Lower roots much real than highest - not so many tricks. Real
calculations...
    Very strange compete on different powers.
    I suppose - 9 digits much difficult than 8 digits on answer, if
power same. If we made record on n digits on time less than 60 seconds
we should go to next record with n+1 digits. (I supposed made extracting
root from 113 digits with 9 on answer, but I afraid Alexis made it faster).
    I think limit on 60 seconds is very important to visitors, what
should looks show. Result less 10 seconds not very good for really
understanding what record - what fast talking... More than 2 minutes -
loosing concentrations. But. Of course more that 2 minutes -
demonstrating power of human brain.


    As about your record. Alexis. I suppose answer 44 800 613 little
strange. 00 in middle give me possibility made it on same (or near
same) speed like you. Please, apologize. I did not see this article
when on summer demonstrate how I prepare to broke record. It was very
strange situation. On my program, on beginning, all digits are 0.
I told before attempt - "Now I can not calculate 4-th digit, let
suppose it will be 0 and looks only another 7 digits..." But. On
likelihood that was real "0" and I was surprised. Correspondent
understand nothing and write "Oleg broke record on 30 seconds".
I was VERY much surprised "0"!!! But my surprising much more when I see
"00" on chapter about Wim Klein rooting where was 42 500 764...
May be enough zeroes on really important places? :-))) Or next time I
will made record like 40 000 001 on 1 second... :-)

    If somebody believe that "0" was good for me, I must say - NO.
That 0 give me so many problems... I still try made break, but peoples
will wait from me faster than 30 seconds how it was on that stupid
article... :-(

    BTW. If somebody interest. I suppose give possibility check me to
visitors. I will ask some of four-digit numbers and give last 4 and first
four on supposition number what I get first four or last four digits
on task. After I demonstrate that there are no trick I do full with
computer.

    Apologize if I wrote something not correct...


   P.S. Who suppose we should forbid 0 on rooting tasks?


Sincerely Yours Oleg Stepanov.
http://www.lk.net/~stepanov/

--- In MentalCalculation@y..., "rob1729" <rob1729@y...> wrote:
> Hi Ulrich,
>
> Have you considered using
>
> N mod 19 = (N div 20 + N mod 20) mod 19   ??
>
> Take, for example, N=1853
> N / 20 = 92 r 13, i.e. N div 20 = 92, N mod 20 = 13
>
> 92 + 13 = 105
>
> and 105 mod 19 = 10, or, if you do not know that 5x19 is 95, then
> simply re-apply the formula, i.e.
> 105 / 20 = 5 r 5
>
> 5 + 5 = 10.
>
> Regards,
> Robert Fountain


  Hi Robert

Thanks for the suggestion.

Let me stick to my "years" J = 100*H + E

The question is if   J mod 19 = (J div 20 + J mod 20) mod 19
is to be preferred to   J mod 19 = (10*H mod 19 + E mod 19) mod 19

Looking at it in this way I prefer your formula, because it is more
beautiful. It has more to do with the structure of "Division by 19".
On the other hand my approach has more to do with the structure of
decimal numbers.
Now I try to work with your formula but still I feel reluctant as to
adopt it. Division by 2 looks easy but often leads me to make
mistakes. On the other hand  H is visible and I have always 100*H mod
19 ready made. But this difference looks very small indeed and your
advantage is that you need no preparation. Perhaps a matter of taste?
The price I have to pay, of course, consists in learning actually
100*H mod 19 for nineteen H. This is certainly not cheap. But I used
to consider it a bargain.
Actually I am concerned with calendars only, so that the biggest H
(for me) should be H=20. So I always get the numbers I need very
rapidly and safely.

But what about larger numbers? I never thought about them till to-day.
Look at this:
1.000 mod 19 = 12  (this is the number I have to know for H=10)
and
10.000 mod 19 = 1000 0 mod 19 = 120 mod 19 = 6
or
10.000 mod 19 = 100 00 mod 19 = 500 mod 19 = 6
and
100.000 mod 19 = 1000 00 mod 19 = 1200 mod 19 = 3
or
100.000 mod 19 = 100 000 mod 19 = 500 0 mod 19 = 60 mod 19 = 3
  etc.
  another example
27.594.476.303 mod 19 = 275 94476303 mod 19 = 994 476303 mod 19 = 647
6303 mod 19 = 1630 3 mod 19 = 153 mod 19 = 1
or
27.594.476.303 mod 19 = 275 94 47 63 03 mod 19 = 994 47 63 03 mod 19
= 647 63 03 mod 19 = 163 03 mod 19 =  1103 mod 19 = 1
or
27.594.476.303 mod 19 = 27 594476303 mod 19 = 859 4476303 mod 19 =
444 76303 mod 19 = 776 303 mod 19 = 1630 3 mod 19 = 153 mod 19 = 1

any way does, but I like regular one  ( which puts 27.594.476.303 =
275 94 47 63 03 )  best. Once I write down the number in this way I
can do the calculation mentally.
Please note that I never have to deal with numbers that have more
than 4 digits and that there is no difficulty in remembering results.
Actually this is but normal division made more rapid by using
information about 100*H mod 19.
If now I try your approach, I have to repeat division with 2 instead
of always finding the rest mod 19.
27.594.476.303 div 20 = 1.379.723.815
1.379.723.815 div 20 = 68.986.190
68.986.190 div 20 = 3.449.309
3.449.309 div 20 = 172.465
172.465 div 20 = 8.623
8.623 div 20 = 431
431 div 20 = 16
How could I manage this without writing down results?
kind regards
Ulrich Voigt
www.likanas.de

Hi.
On my page chapter from Smith.
http://www.lk.net/~stepanov/mnemo/smith13e.html
Here how Great Klein calculate.

Also there are article on Internet how extract 137 root from
1000 digits. But I do not remember where.

Just now I get info about extracting 13th from 100 by Lemaire
on 13 seconds. Little strange answer with two "0" in the middle
what made extraction easy on my opinion, but I can be wrong.
Soon I put it to my page. But now it is only on
www.chez.com/rfernand/vi/vi23.pdf

I do not know how mr. Lemaire extract. Looks like he understand it
better than me... Sorry, I have no ideas how made extracting 13th
root less than 30 seconds... Only memorizing all possible answers. :-)

PWRL> Hello,
PWRL> my name is Jan van Koningsveld and I live in Emden, Germany. I always
liked MC but mainly focussed on memorising Pi without using any technic, just
for the fun of it. In July I met Gert
PWRL> Mittring, Alberto Coto and Ralf Laue during the Record Festival in
Flensburg and this gave me enough inspiration to seriously start MC again. I was
able to improve my speed on easy tasks and
PWRL> would now like to concentrate on extracting higher roots like 13th root of
100-digit numbers, 23rd root of 200-digit numbers.
PWRL> I would be very grateful if someone could give me some advice of how to
extract higher roots. Is there just one real way to do this or are there
different possibilities ?

PWRL> Regards,
PWRL> Jan





Sincerely Yours Oleg Stepanov.

http://www.lk.net/~stepanov/

Hello,

my name is Jan van Koningsveld and I live in Emden, Germany. I always liked MC
but mainly focussed on memorising Pi without using any technic, just for the fun
of it. In July I met Gert Mittring, Alberto Coto and Ralf Laue during the Record
Festival in Flensburg and this gave me enough inspiration to seriously start MC
again. I was able to improve my speed on easy tasks and would now like to
concentrate on extracting higher roots like 13th root of 100-digit numbers, 23rd
root of 200-digit numbers.
I would be very grateful if someone could give me some advice of how to extract
higher roots. Is there just one real way to do this or are there different
possibilities ?

Regards,
Jan


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

Hi Ulrich,

Have you considered using

N mod 19 = (N div 20 + N mod 20) mod 19   ??

Take, for example, N=1853
N / 20 = 92 r 13, i.e. N div 20 = 92, N mod 20 = 13

92 + 13 = 105

and 105 mod 19 = 10, or, if you do not know that 5x19 is 95, then
simply re-apply the formula, i.e.
105 / 20 = 5 r 5

5 + 5 = 10.

Regards,
Robert Fountain

Thank you Ralf for your fine answer. In doing calendar calculation I take simple
procedures serious.
So let me think a bit about your proposal.

Our strategy is similar to each other and indeed your method is the more simple
one. And it is effective.

On the other hand, once you are doing this kind of arithmetic very often, you
will - even without mnemonics - learn the crucial numbers  (that is in your case
(19-H)*5  or even  100*H +(19-H)*5  for all those H which are of concern) by
heart. I even dare say that you simply cannot avoid this. After all H will
normally be a number between 15 and 20, so there is not very much to be done.

But then I feel that there is a slight disadvantage with "your" numbers as
compared with  mod ( 100*H ; 19)  because (as your example c shows) you have at
times to make non trivial "decisions" as to what is the best way and (as your
example b shows) the difference between E and 100*H + (19-H)*5 might be > 19
which will never happen with the difference between  mod ( 100*H ; 19) + E and 0
- 19 - 38 - 76 - 95.

The only thing that can happen with "my" numbers is, that mod ( 100*H ; 19) + E
> 100.

Example
year 1897, J = 1897, H = 18, E = 97

mod (  mod ( 1800 ; 19) + 97 ; 19 )  = mod ( 14 + 97 ; 19) = mod ( 111 ; 19) =
111 - 95 = 16 does not look so very comfortable.

In this case I am prepared to take the difference of E with 95  right at the
start, which leads to    mod ( 1897 ; 19)  = 14 + (97 - 95) = 14 + 2 = 16.

My main advantage lies in the constant use of the same sequence 0 - 19 - 38 - 76
- 95.

Mnemonics useful or not?

If you take the range of H between 0 and 20, then it might be a long time till
simple practice will teach you mod ( 100*H ; 19)  for all H.

Sincerely

Ulrich Voigt

www.likanas.de

   ----- Original Message -----
   From: Ralf Laue
   To: MentalCalculation@yahoogroups.com
   Sent: Sunday, October 13, 2002 11:07 PM
   Subject: Re: [Mental Calculation] mnemonics etc. revised version


   Ulrich Voigt wrote:

   > More precisely: To minimize calculation using devices to know useful numbers
by heart.
   > Here is an example which in fact is quite important in calculating the date
of Easter or Pesach.
   > The task is to determine the rest    mod ( J ; 19)   which remains if you
divide an integer  J  by 19.
   > J will be the number of a year, lets say J =  2002, so that  mod (2002;19) 
= 7
   >
   > I write  J = H / E  with H = 20 and  E = 02
   > ...
   > Does anyone know of a more rapid method to produce  mod (J ; 19)?

   Hello,

   Here is a method which may not be faster but easier to remind:

   Of course, we know that 0, 19, 38, 57, 76 and 95 are divisable by 19.
   Knowing these numbers, it is very easy to find the remainder of each
   two-
   digit number.
   Now we can make use of the fact that the years
   100*H + (19-H)*5 have also rest 0 if divided by 19. This means that we
   know one year for each century that is divisable by 19.
   (This fact is ALL you have to memorise).

   Examples:
   a) year 1716, J = 1716, H = 17, E = 16
   (19-H)*5=2*5=10 -> We know that 1710 is divisable by 19 -> 1716 has
   remainder 6

   b) year 698, J = 698, H = 6, E = 98
   (19-H)*5 = (19-6)*5 = 65 -> We know that 665 is divisable by 19.
   We are searching the remainder of 698, and we know that 665 has
   remainder 0.
   For this reason, the remainder of 698 is equal to the remainder of
   (698-665)= 33 -> The remainder for 698 is 12.

   c) year 2177, J = 2177, H = 21, E = 77
   (19-H)*5 = (-2)*5 =-10 -> 2100-10 =2090 is divisable by 19.
   This can be used to find that the remainder of 2177 is equal to the
   remainder
   of 2090.
   However, maybe it is even easier to find (using H=22) that
   (19-H)*5 = (-3)*5 = -15 and 2200-15=2185 is divisable by 19.
   If we know this, it is trivial to find the remainder for 2177.

   Best Wishes,
   Ralf Laue

   --
   _____________________________________________________________________
   Please excuse me for a delay in replying to your e-mail.
   I have a lot of e-mail to answer day by day.
   ----------------------------------------------------------------------
   Ralf Laue       e-mail: info@...
   P. O. Box 80    Visit my Homepage about unusual world records:
   04181 Leipzig   http://www.recordholders.org/en/
   Germany
   ----------------------------------------------------------------------


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--- In MentalCalculation@y..., "pjlmem" <pjlmem@y...> wrote:

>      I am working hard to improve my performance.
>       If anyone has any windows program to generate random dates to
> find out the day for it, kindly provide me the same to train myself.
> memorably yours,

Have a look at http://groups.yahoo.com/group/MentalCalculation/links.
There are two links to such programs.
Also, there is a Windows program, announced in a posting to another
mailing list:
http://groups.yahoo.com/group/wwbc/message/12554

You should ask the author of this program to send you a copy -
or even better to put it on the Links section of our group at
http://groups.yahoo.com/group/MentalCalculation/links.

Ralf Laue

Ulrich Voigt wrote:

> More precisely: To minimize calculation using devices to know useful numbers
by heart.
> Here is an example which in fact is quite important in calculating the date of
Easter or Pesach.
> The task is to determine the rest    mod ( J ; 19)   which remains if you
divide an integer  J  by 19.
> J will be the number of a year, lets say J =  2002, so that  mod (2002;19)  =
7
>
> I write  J = H / E  with H = 20 and  E = 02
> ...
> Does anyone know of a more rapid method to produce  mod (J ; 19)?

Hello,

Here is a method which may not be faster but easier to remind:

Of course, we know that 0, 19, 38, 57, 76 and 95 are divisable by 19.
Knowing these numbers, it is very easy to find the remainder of each
two-
digit number.
Now we can make use of the fact that the years
100*H + (19-H)*5 have also rest 0 if divided by 19. This means that we
know one year for each century that is divisable by 19.
(This fact is ALL you have to memorise).

Examples:
a) year 1716, J = 1716, H = 17, E = 16
(19-H)*5=2*5=10 -> We know that 1710 is divisable by 19 -> 1716 has
remainder 6

b) year 698, J = 698, H = 6, E = 98
(19-H)*5 = (19-6)*5 = 65 -> We know that 665 is divisable by 19.
We are searching the remainder of 698, and we know that 665 has
remainder 0.
For this reason, the remainder of 698 is equal to the remainder of
(698-665)= 33 -> The remainder for 698 is 12.

c) year 2177, J = 2177, H = 21, E = 77
(19-H)*5 = (-2)*5 =-10 -> 2100-10 =2090 is divisable by 19.
This can be used to find that the remainder of 2177 is equal to the
remainder
of 2090.
However, maybe it is even easier to find (using H=22) that
(19-H)*5 = (-3)*5 = -15 and 2200-15=2185 is divisable by 19.
If we know this, it is trivial to find the remainder for 2177.

Best Wishes,
Ralf Laue

--
_____________________________________________________________________
Please excuse me for a delay in replying to your e-mail.
I have a lot of e-mail to answer day by day.
----------------------------------------------------------------------
Ralf Laue       e-mail: info@...
P. O. Box 80    Visit my Homepage about unusual world records:
04181 Leipzig   http://www.recordholders.org/en/
Germany
----------------------------------------------------------------------

Greetings,

Thank you Ralf for creating a group dedicated to MC. I am sure the
discussions that lie ahead will be most interesting.

Regards,
Robert Fountain
England, UK

--- In MentalCalculation@y..., JohnLouis Louis <pjlmem@y...> wrote:
>
>  dear bruno,
>          regarding memorising binary numbers, convert every 3
binary number into a random number ETC

---------
Thanks for your help:
   I will be doing a variation of what u suggest.  Instead of 3 binary
digits I will be using 4.  WHY?  Because it can benefit some computer
programming as well.  The native tongue of computers is binary
arranged in multiples of 4-bit binaries. (this is for those not
familiar with computers)AS IN 1101 0100 0110 etc.  It can be useful
at times to convert binary to decimal or hexadecimal.  This chart
shows the relationship:

HEX   DECIMAL     BINARY
0     0 = 0+0+0+0 0000
1     1 = 0+0+0+1 0001
2     2 = 0+0+2+0 0010
3     3 = 0+0+2+1 0011
4     4 = 0+2+0+0 0100
5     5 = 0+4+0+1 0101
6     6 = 0+4+2+0 0110
7     7 = 0+4+2+1 0111
8     8 = 8+0+0+0 1000
9     9 = 8+0+0+1 1001
A    10 = 8+0+2+0 1010
B    11 = 8+0+0+1 1011
C    12 = 8+4+0+0 1100
D    13 = 8+4+0+1 1101
E    14 = 8+4+2+0 1110
F    15 = 8+4+2+1 1111

There are only 16 possibilities.  So "F" in hexadecimal is"15" in
decimal and "1111" in binary.
    I therefore "kill 3 birds with one stone" (English expression)so
to speak....memorizing binaries then has a practical use as well.

    The rest I will do as you have suggested, name/action/journey.
Regards,
Bruno

UV> And what about you. Do you have am mnemonic method to deal with
UV> such matters?

I am author of book "Mnemonic. Truth and false." :-)))
But I do not suppose mnemonic good for Mental Calculations.
And I do not interest calendar calculating.


Sincerely Yours Oleg Stepanov.
http://www.lk.net/~stepanov/

Dear Oleg Stepanov!
>> Also, do you like use "-"? I mean 18 are same like -1, but calculate more
easy...<<

Yes, of course I use  -1 instead of  +18,  -2 instead of +17 etc, this is often
very convenient. For instance I know mod (1500 ; 19) = 18 but mostly take this
to mean -1.

  >>Why do you not memorize the reminder for all numbers from 1500 to 2010??? <<

I do not believe in this idea because there would be too many years to be
memorized. As they overlap the centuries you cannot easily use simple mnemonic
words.

>>Do you know a method of memorizing groups?<<

Yes. Indeed this is my favourite method. In my book Esels Welt. Mnemotechnik
zwischen Simonides und Hary Lorayne I dwelt at length about the theory and
history of this method which I call "Klumpenmethode" (= method by clusters). But
in this case I feel that this method would be really inadequate.

And what about you. Do you have am mnemonic method to deal with such matters?

  Sincerely Ulrich Voigt

   ----- Original Message -----
   From: Oleg Stepanov
   To: Ulrich Voigt
   Sent: Saturday, October 12, 2002 11:18 PM
   Subject: Re: [Mental Calculation] mnemonics and mental calculation


   UV> The task is to determine the rest  J 19  which remains if you divide an
integer  J  by 19.
   UV> Does anyone know of a more rapid method to produce  J 19 ?

   Dear mr. Voigt. Why do not only memorize reminder for all numbers form
   1500 to 2010??? It can be faster.

   Do you know method of memorizing groups? I means all numbers what
   have reminder 0 on 19 on one place... all 5 - on another place,
   all 10 on another.... And add 1-5 will be easy. Also, do you like use
   "-"? I mean 18 are same like -1, but calculate more easy...

      Isn't it?


   Sincerely Yours Oleg Stepanov.

   http://www.lk.net/~stepanov/



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

----- Original Message -----
   From: Oleg Stepanov
   To: Ulrich Voigt
   Sent: Saturday, October 12, 2002 11:18 PM
   Subject: Re: [Mental Calculation] mnemonics and mental calculation


   UV> The task is to determine the rest  J 19  which remains if you divide an
integer  J  by 19.
   UV> Does anyone know of a more rapid method to produce  J 19 ?

   Dear mr. Voigt. Why do not only memorize reminder for all numbers form
   1500 to 2010??? It can be faster.

   Do you know method of memorizing groups? I means all numbers what
   have reminder 0 on 19 on one place... all 5 - on another place,
   all 10 on another.... And add 1-5 will be easy. Also, do you like use
   "-"? I mean 18 are same like -1, but calculate more easy...

      Isn't it?


   Sincerely Yours Oleg Stepanov.

   http://www.lk.net/~stepanov/



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   To unsubscribe from this group, send an email to:
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[Non-text portions of this message have been removed]

UV> The task is to determine the rest  J 19  which remains if you divide an
integer  J  by 19.
UV> Does anyone know of a more rapid method to produce  J 19 ?

Dear mr. Voigt. Why do not only memorize reminder for all numbers form
1500 to 2010??? It can be faster.

Do you know method of memorizing groups? I means all numbers what
have reminder 0 on 19 on one place... all 5 - on another place,
all 10 on another.... And add 1-5 will be easy. Also, do you like use
"-"? I mean 18 are same like -1, but calculate more easy...

    Isn't it?


Sincerely Yours Oleg Stepanov.

http://www.lk.net/~stepanov/

The expressions I wrote are now looking so very strange, so let me put here  a
revised version:

I am a teacher (mathematics-history-philosophy) at the Matthias Claudius
Gymnasium in Hamburg.
Besides I do some work as a mnemonist. Actually I try to construct
calendar-mnemonics.

My idea: To mix mental calculation with mnemonics.

More precisely: To minimize calculation using devices to know useful numbers by
heart.

Here is an example which in fact is quite important in calculating the date of
Easter or Pesach.

The task is to determine the rest    mod ( J ; 19)   which remains if you divide
an integer  J  by 19.

J will be the number of a year, lets say J =  2002, so that  mod (2002;19)  = 7

I write  J = H / E  with H = 20 and  E = 02

and I make use of

mod ( J ; 19 )  = mod ( mod (100 H ; 19)  + mod ( E ; 19 ) ; 19 )

Now I learn by heart

   a.. mod (100 H ; 19)   for all H of historical concern.
   b.. the sequence  0 - 19 - 38 - 57 - 76 - 95
This is the mnemonic part which needs careful preparation and some exercise.

But once I got it, it proved very fast.

After this I get   mod (2002 ; 19) =  mod (5 + 2 ; 19)  = 7

Another example:  J = 1688

mod (1688 ; 19) =  mod (4 + 12 ; 19) = 16

mod (E ; 19)  = mod (88 ; 19) =  88 - 76 = 12   is a very simple matter indeed,
so simple that it hardly deserves the name of "calculation".

Does anyone know of a more rapid method to produce  mod (J ; 19)?

Ulrich Voigt      Bornstr. 6      20146 Hamburg     Germany

webside  www.likanas.de





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

JL> Suppose your person for 13 is george bush & the action
JL> of the person for 54 is shooting, then you ridiclously
JL> associate as if george bush is shooting & locate this
JL> picture in the first spot of your journey.
JL> this is how every one does in the world memory championship.
Please, I can be wrong. Probably not "every one". Another peoples
have another systems. Karsten Gunter memorize by 3 (9 binary) digits.

    Apologize if this is not important.

Sincerely Yours Oleg Stepanov.

http://www.lk.net/~stepanov/

dear bruno,
          regarding memorising binary numbers, convert every 3 binary number into
a random number, for ex 000 = 0, 100 = 1, 110 = 2, 111 = 3, 001 = 4, 010 = 5,
011 = 6 & 101 = 7. These are the only 8 possiblities. So,  you can convert every
6 binary digits into a 2 random digit number.
          If you have a standard person & action for each number, then the job
becomes very simple.that is, ridiculously associate the person with the action
of the second person and locate this action in a spot of your journey.
example 100 111 010 001 . If you convert this to random numbers you will get 13
54. Suppose your person for 13 is george bush & the action of the person for 54
is shooting, then you ridiclously associate as if george bush is shooting &
locate this picture in the first spot of your journey.
this is how every one does in the world memory championship.I am the one
requested  the memory software.
memorably yours,
p.john louis (india)
   klinger5757 <btemp@...> wrote: Greetings all:
   I was about to embark on a program of learning for MC when I came
upon this group.  Sorry I don't have much to offer in terms of
experience or advise--my journey must start the way all beginners do.
Hopefully though I can avoid some pitfalls based on your experience.
One area I am somewhat uncertain about is with regards to binary
digit memorization.   101001101.  If any would be willing to share
how they do this, it would be appreciated.  I have been considering
converting 4 binaries to a number/letter at a time as in assembly
lang.  Any other suggestions?
    Also I have noted numerous requests for a software program
mimicking the World MC Championships.  I will, next month, be
teaching myself a new computer language (Python)and thought to apply
it to that request.  If all goes well, I should have something up by
the end of the year.  It won't be pretty, but it should get the job
done.

Regards,
Bruno


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I am a teacher (mathematics-history-philosophy) at the Matthias Claudius
Gymnasium in Hamburg.
Besides I do some work as a mnemonist. Actually I try to construct
calendar-mnemonics.

My idea: To mix mental calculation with mnemonics.

More precisely: To minimize calculation using devices to know useful numbers by
heart.

Here is an example which in fact is quite important in calculating the date of
Easter or Pesach.

The task is to determine the rest  J 19  which remains if you divide an integer 
J  by 19.

J will be the number of a year, lets say J =  2002, so that  2002 19  = 7

I write  J = H / E  with  H = 20  and  E = 02

and I make use of    J 19  =  ( (100 H) 19  + E 19 ) 19

Now I learn by heart

   a.. (100 H ) 19   for all H of historical concern.
   b.. the sequence  0 - 19 - 38 - 57 - 76 - 95
This is the mnemonic part which needs careful preparation and some exercise.

But once I got it, it proved very fast.

After this I get   2002 19 = ( 2000 19  + 02 19 ) 19  =  ( 5 + 2 ) 19  = 7

Another example:  J = 1688

1688 19 = ( 1600 19  + 88 19 ) 19  =  ( 4 + 12 ) 19  = 16

E 19  = 88 19 =  88 - 76 = 12   is a very simple matter indeed, so simple that
it hardly deserves the name of "calculation".

Does anyone know of a more rapid method to produce  J 19 ?

Ulrich Voigt      Bornstr. 6      20146 Hamburg     Germany

webside  www.likanas.de





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Greetings all:
   I was about to embark on a program of learning for MC when I came
upon this group.  Sorry I don't have much to offer in terms of
experience or advise--my journey must start the way all beginners do.
Hopefully though I can avoid some pitfalls based on your experience.
One area I am somewhat uncertain about is with regards to binary
digit memorization.   101001101.  If any would be willing to share
how they do this, it would be appreciated.  I have been considering
converting 4 binaries to a number/letter at a time as in assembly
lang.  Any other suggestions?
    Also I have noted numerous requests for a software program
mimicking the World MC Championships.  I will, next month, be
teaching myself a new computer language (Python)and thought to apply
it to that request.  If all goes well, I should have something up by
the end of the year.  It won't be pretty, but it should get the job
done.

Regards,
Bruno

Hi. My name is Michel Asselin, in am in my forties and have had a
fascination for mnemmonics and mental calculation ever since my teens, when
I came across booklets profiling very basic and simple methods to achieve
both. It has stuck with me ever since. I even bought math software to teach
myself calculus, specifically to enhance my understanding of a Newton
equation which is said to facilitate the extraction of roots mentally (i.e.,
roots of any numbers).

I thank Ralf for inviting me, and hope that I may contribute to the group.

la> Hello, hi friends
Hi.

la> I am a french mental calculator able to break records like 13th
la> root of 100-digit number, 23rd root of 200-digit number, etc...
la> I would like to create a new world record, called: the largest
la> high root extracted in 5 minutes.
I am interesting only highest number of digits on answer.
But. So kind to find you here. Long time I want to find somebody from
France. Of course, I know a lot of french, but they not interest
MC. Would be you so kind. Do you have information about Mourice
Dagbert? Do you interest history of subject? I mean we find
book Binet and want translate it to different languages. I am
apologize. Do you interest help us open information about Great
Mental Calculators of your country?


la> The competitor wins a point for each possibility (only near 200
la> 000 points for the 73rd root of a 500-digit number) and may choose
la> the root in a list of roots, like 13th, 23rd, 53rd, 73rd, 83rd...
la> Roots really harder or easier than standard tasks are
la> forbidden.(this includes 7th, 11st, 12nd, 16th, 33th, 67th, etc... )
la> I think this record is much more difficult than the other ones and
la> more interesting, more straightforward, ...
Please, introduce what you mean. I want broke extracting 13th root,
write article how to do it, and stop to do it. After I suppose
concentrate only on 2 and 3 powers... I mean - increase number of
digits at answer.


la> Another question:
la> true or false?
la> R.Gamm is able to calculate or recall a 2-digit number raised to a power<20?
What problem? Mille. Osaka made it too. Not many peoples believe she
calculate, but memorize it not difficult... Or you mean really source
where written about check him? BTW. I am so apologize. Do you really
suppose Rudiger Gamm are Mental Calculator? This is very difficult
question, but probably not all peoples what call themself MC are real.
Sorry, if I am wrong.

la> A.Lemaire




Sincerely Yours Oleg Stepanov.

http://www.lk.net/~stepanov/

Hello, hi friends

I am a french mental calculator able to break records like 13th root of
100-digit number, 23rd root of 200-digit number, etc...
I would like to create a new world record, called: the largest high root
extracted in 5 minutes.

The competitor wins a point for each possibility (only near 200 000 points for
the 73rd root of a 500-digit number) and may choose the root in a list of roots,
like 13th, 23rd, 53rd, 73rd, 83rd...
Roots really harder or easier than standard tasks are forbidden.(this includes
7th, 11st, 12nd, 16th, 33th, 67th, etc... )
I think this record is much more difficult than the other ones and more
interesting, more straightforward, ...

Another question:
true or false?
R.Gamm is able to calculate or recall a 2-digit number raised to a power<20?

Sincerely yours, cheers

A.Lemaire
------------------------------------------

Faites un voeu et puis Voila ! www.voila.fr

Hi,

I like mental and very fast calculations, though i dont have very
good records like you guys have but i can certainly improve on it and
of course i hope to learn from your experiences.

Cheers,

Darshan Purandare
Orlando, FL
USA

Wow, this is fascinating. Really hope to learn something from this
group.
Working in San Jose, CA
--Rishi

--- In MentalCalculation@y..., "Rex Boggs" <rboggs@b...> wrote:
> BlankI have just joined the group.  My interest is teaching mental
> calculation in school.
>
> Cheers
>
> Rex
> --
> Rex Boggs
> Glenmore SHS
> PO Box 5822, RMC
> Rockhampton   Q    4702
> Australia

BlankI have just joined the group.  My interest is teaching mental
calculation in school.

Cheers

Rex
--
Rex Boggs
Glenmore SHS
PO Box 5822, RMC
Rockhampton   Q    4702
Australia

p> hi friends,
Hi.

p> I am john louis, an indian. I am happy to join this group.To say
p> about myself, I am master of chemistry & teaching in a school for
p> the last 14 years.I am very much interested in mental calculation
p> also.
Would you be so kind to help me? We have some questions connected
with your country. Just now I interest Hary Prasad. His record
on dates on Guinness page like movie, but there written, that
it is video where he extract root from 6 digits. Do you think
possible connect with him? Or find more correct time for
extracting?


Sincerely Yours Oleg Stepanov.

http://www.lk.net/~stepanov/

hi friends,
      I am john louis, an indian. I am happy to join this group.To say
about myself, I am master of chemistry & teaching in a school for
the last 14 years.I am very much interested in mental calculation
also.
      I can calculate day for any date at an average  speed of 3 sec
per date.I am working hard to improve upon it.
      Recently, I took part in world memory championship-2002 held at
london & finished with 19th place.
1.I memorised a deck of card in 132.5 sec & recalled them perfectly.
2.I memorised 8 decks of cards in 60 min & recalled them perfectly.
3.I memorised 840 binary digits in 30 min & recalled them perfectly.
4.I memorised 680 digits random numbers in 60 min & recalled them
perfectly.
5.I memorised 85 random english words in 15 min & recalled them
perfectly.etc.
      I am working hard to improve my performance.
       If anyone has any windows program to generate random dates to
find out the day for it, kindly provide me the same to train myself.
memorably yours,
p.john louis

This is a test.

    My name Oleg Stepanov.
    I lake mathematic and mental calculations.
    On my country Mental Calculations are very close subject,
and my country long time was closed. I want open information
about us and to us.
    Little problem - information on Russia not same like on
another world - many books open on Internet, and there are a lot of
libraries what compete who first open good book. I understand it
is not normal and will try be under western laws.
    My page - http://www.lk.net/~stepanov/mnemo/mnemoare.html
I search information and open it with translating.

    Now I write a book on Russian "Mental Calculators. Truth and
myth." And for showing why I can write this book I want break
record of Greatest calculator Wim Klein on extracting 13-th
root from 100 digits.


Sincerely Yours Oleg Stepanov.
http://www.lk.net/~stepanov/

Hello,

This is the very first posting to the new "Mental Calculation"
mailing list.

After I have created my web page about world records for memory
and mental calculation at
http://www.recordholders.org/en/list/memory.html, I have got a
lot of e-mail from other people asking details about these
achievements. Some of them suggested to create something like
an Association of Mental Calculators, calculating championships
etc. I realised that it would be a good idea to bring the best
calculators I know in contact with each other as a first step.

This was the reason for creating this discussion list.

In this first posting, I also want to introduce myself, and I
would be glad if other first time posters also could do so.
I come from Leipzig, Germany, and I have studied mathematics
(sic!) Now I am working as a syste administrator in a small
Internet company.
Some time ago, I broke some records for calendar computing and
was mentioned in the Guinness Book of Records for this feature.
(Of course everyone of you know that this is an easy task - but
you have to be very fast to get the record!)
I am not very good in other calculations, but I hope to improve
in the next years.
This year, I helped to organize a record breakers' festival in
Flensburg. We had the opportunity to invite Alberto Coto and
Gert Mittring, two of the best calculators at present. I would
be glad if we could organize such meetings in the future, and
I wish that this list helps to bring the calculators in contact
with each other.

Best Wishes,
Ralf