Hello,

I just want to tell you that the application form and the general
information sheet about the Mental Calculation World Cup 2006 is now
available in Spanish language as well. It can be downloaded from
http://www.recordholders.org/en/events/worldcup/2006/.
(where can find an English, German, Russian and French version as well.)

Thank you to Yusnier Viera Romero for doing the translation.

Best Wishes,
Ralf
--
_____________________________________________________________________
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
----------------------------------------------------------------------

http://www.doubledivision.org/


Mike

thnaks
----- Original Message -----
From: "Ralf Laue" <info@...>
To: <MentalCalculation@yahoogroups.com>
Sent: Sunday, October 30, 2005 1:24 AM
Subject: [Mental Calculation] Mental Calculation World Cup Forms in Spanish


> Hello,
>
> I just want to tell you that the application form and the general
> information sheet about the Mental Calculation World Cup 2006 is now
> available in Spanish language as well. It can be downloaded from
> http://www.recordholders.org/en/events/worldcup/2006/.
> (where can find an English, German, Russian and French version as well.)
>
> Thank you to Yusnier Viera Romero for doing the translation.
>
> Best Wishes,
> Ralf
> --
> _____________________________________________________________________
> 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
> ----------------------------------------------------------------------
>
>
>
>
>
>
> Yahoo! Groups Links
>
>
>
>
>
>
>
>

I know that it has been EONS since I have posted a message; however, I
would like to know the best ways to calculate the 2nd, 3rd, 4th, etc.
roots of a number. Please respond.

Joshua Greene

As av ery good decimalizer, I think you should be at your ease with the ways
employed by A.C Aitken


joshulator84 <joshulator84@...> a écrit :
   I know that it has been EONS since I have posted a message; however, I
would like to know the best ways to calculate the 2nd, 3rd, 4th, etc.
roots of a number. Please respond.

Joshua Greene








Yahoo! Groups Links











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Hi all,

I'm new here...and I KNOW this is off topic.  But I'd like to ask
you all your opinions/theories on this

I'm reteaching myself calculus, and that requires a knowledge of
functions.  One of the "famous" "is this a function?" question
is "If set (S) is all real numbers and T is the square root of the
numbers in S, is that a function?

The book answer is no.  I submit that that is true, as the square
root of x (lets say 64) has a negative and a positive solution (-8
and 8 in this case).

Yet my TI-83 (yes...i'm an old man...30) gives me a -64 when i
square -8 and an 64 when i pull an -8*-8

Is this an anomaly or am i missing a "new" mathematical pricipal? (I
think the calculator is trying to force things that aren't "true" to
make calcs easier)

Thanks all!!

Hi Joshua

   Square roots are covered in many, many posts to the group, but the deeper
roots are less commonly reported. Many people who are practiced at extracting
these 'deep roots' (as I call them) tend to memorise logarithmic tables and
calculate them in this manner, but my memory is not good enough for this and so
I tend to rely on the purely mathematical method of calculus.

   The nature of calculus (which can be taken from any one of a great number of
maths text books) is easy to understand, and the method by which it can be
applied - and here comes the 'plug' - is laid out in my own book. I'm not sure
if I've posted this section to the group, but I have sent quite a few sections.

   Please feel free to e-mail me if you would like me to explain in greater
detail just what I do and how I do it; though I do, of course, know I'm not the
only one around who can help; there are many capable experts within this group.

   All the best,

   George Lane

joshulator84 <joshulator84@...> wrote:
   I know that it has been EONS since I have posted a message; however, I
would like to know the best ways to calculate the 2nd, 3rd, 4th, etc.
roots of a number. Please respond.

Joshua Greene








Yahoo! Groups Links











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

--- In MentalCalculation@yahoogroups.com, "shinobisoulxxx2"
<shinobisoulxxx2@y...> wrote:
>
> Yet my TI-83 (yes...i'm an old man...30) gives me a -64 when i
> square -8 and an 64 when i pull an -8*-8
>
> Is this an anomaly or am i missing a "new" mathematical pricipal? (I
> think the calculator is trying to force things that aren't "true" to
> make calcs easier)
>
> Thanks all!!
>

I suspect that your calculator is doing -(8^2) instead of (-8)^2.

Rob F

George and Ron Doerfler have already been very generous with their
posts on this subject. See, for instance, messages 1144 and 1154. -
RobF

--- In MentalCalculation@yahoogroups.com, George Lane
<george972453@y...> wrote:
>
> Hi Joshua
>
>   Square roots are covered in many, many posts to the group, but
the deeper roots are less commonly reported. Many people who are
practiced at extracting these 'deep roots' (as I call them) tend to
memorise logarithmic tables and calculate them in this manner, but
my memory is not good enough for this and so I tend to rely on the
purely mathematical method of calculus.
>
>   The nature of calculus (which can be taken from any one of a
great number of maths text books) is easy to understand, and the
method by which it can be applied - and here comes the 'plug' - is
laid out in my own book. I'm not sure if I've posted this section to
the group, but I have sent quite a few sections.

Hello,

I wish all the best for 2006 to all of you. I hope that many of us will
meet at the Mental Calculation World Cup in November in Giessen (see
http://www.recordholders.org/en/events/worldcup/2006/index.html). It
would be nice to send back the entry form if you have the intention to
take part.

It is a pleasure for me to announce that new world records have been
verified for all three calendar computing categories. (see
http://www.recordholders.org/en/records/dates.html) These records will
be published in the "Book of Alternative Records"
(http://www.alternativerecords.co.uk) as well.
Yusnier Viera Romero (Cuba) bettered the record for calculating the day
of the week for dates of one century twice on 31 October 2005 at the
University of Havana. He first achieved 23.2 sec and 19.8 sec in another
attempt. On the same occasion, he calculated 42 days of the week in one
minute for days chosen from the years 1600-2100.
Only a few days later, on 6 November, Matthias Kesselschläger got back
his world records with a time of 17.9 seconds (20 days from one century)
and 45 dates from 1600-2100. Also, he set a record for the 365 days of
2005 in 214 seconds only.

I am looking forward to see the contest between Yusnier and Matthias at
the World Cup, both told me that they want to take part.

Best Wishes,
Ralf

Hello everyone,


I have just joined this group. I have pretty poor mental calculation
skills, can you guys recommend me any books which will suit a guy of
my skill so that i can become proficient in this gr8 art.

thanks in advance

--- In MentalCalculation@yahoogroups.com, "asadrizvi7"
<asadrizvi7@g...> wrote:

> I have just joined this group. I have pretty poor mental calculation
> skills, can you guys recommend me any books which will suit a guy of
> my skill so that i can become proficient in this gr8 art.

Just have a look at
http://www.recordholders.org/en/book/memory.html#Calculation

Hello Ralf. Best wishes to you too.

I wonder what kind of preparation is required in order to achieve
scores of 40+ in the 1 minute calendar???

Lewis Carroll would certainly have been impressed with the new
records.
http://www.gutenberg.org/files/11662/11662-h/11662-h.htm#10

Robert F



--- In MentalCalculation@yahoogroups.com, Ralf Laue <info@r...>
wrote:
>
> Hello,
>
> I wish all the best for 2006 to all of you. I hope that many of us
will
> meet at the Mental Calculation World Cup in November in Giessen
(see
> http://www.recordholders.org/en/events/worldcup/2006/index.html).
It
> would be nice to send back the entry form if you have the
intention to
> take part.
>
> It is a pleasure for me to announce that new world records have
been
> verified for all three calendar computing categories. (see
> http://www.recordholders.org/en/records/dates.html) These records
will
> be published in the "Book of Alternative Records"
> (http://www.alternativerecords.co.uk) as well.
> Yusnier Viera Romero (Cuba) bettered the record for calculating
the day
> of the week for dates of one century twice on 31 October 2005 at
the
> University of Havana. He first achieved 23.2 sec and 19.8 sec in
another
> attempt. On the same occasion, he calculated 42 days of the week
in one
> minute for days chosen from the years 1600-2100.
> Only a few days later, on 6 November, Matthias Kesselschläger got
back
> his world records with a time of 17.9 seconds (20 days from one
century)
> and 45 dates from 1600-2100. Also, he set a record for the 365
days of
> 2005 in 214 seconds only.

Hello Ralf,

   thanks for this information.
   Congratulations to Matthias and Yusnier!
   Do you think you can go above 60 per minute ? (1600-2100)

   Regards, Jean-Eric.

Ralf Laue <info@...> a écrit :
   Hello,

I wish all the best for 2006 to all of you. I hope that many of us will
meet at the Mental Calculation World Cup in November in Giessen (see
http://www.recordholders.org/en/events/worldcup/2006/index.html). It
would be nice to send back the entry form if you have the intention to
take part.

It is a pleasure for me to announce that new world records have been
verified for all three calendar computing categories. (see
http://www.recordholders.org/en/records/dates.html) These records will
be published in the "Book of Alternative Records"
(http://www.alternativerecords.co.uk) as well.
Yusnier Viera Romero (Cuba) bettered the record for calculating the day
of the week for dates of one century twice on 31 October 2005 at the
University of Havana. He first achieved 23.2 sec and 19.8 sec in another
attempt. On the same occasion, he calculated 42 days of the week in one
minute for days chosen from the years 1600-2100.
Only a few days later, on 6 November, Matthias Kesselschläger got back
his world records with a time of 17.9 seconds (20 days from one century)
and 45 dates from 1600-2100. Also, he set a record for the 365 days of
2005 in 214 seconds only.

I am looking forward to see the contest between Yusnier and Matthias at
the World Cup, both told me that they want to take part.

Best Wishes,
Ralf








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

Hello, I am Yusnier Viera Romero. Congratulations to Matthias for his
records.
> Hello,
>
> I wish all the best for 2006 to all of you. I hope that many of us will
>  meet at the Mental Calculation World Cup in November in Giessen (see
> http://www.recordholders.org/en/events/worldcup/2006/index.html). It
> would be nice to send back the entry form if you have the intention to
> take part.
>
> It is a pleasure for me to announce that new world records have been
> verified for all three calendar computing categories. (see
> http://www.recordholders.org/en/records/dates.html) These records will
> be published in the "Book of Alternative Records"
> (http://www.alternativerecords.co.uk) as well.
> Yusnier Viera Romero (Cuba) bettered the record for calculating the day
>  of the week for dates of one century twice on 31 October 2005 at the
> University of Havana. He first achieved 23.2 sec and 19.8 sec in
> another  attempt. On the same occasion, he calculated 42 days of the
> week in one  minute for days chosen from the years 1600-2100.
> Only a few days later, on 6 November, Matthias Kesselschläger got back
> his world records with a time of 17.9 seconds (20 days from one
> century)  and 45 dates from 1600-2100. Also, he set a record for the
> 365 days of  2005 in 214 seconds only.
>
> I am looking forward to see the contest between Yusnier and Matthias at
>  the World Cup, both told me that they want to take part.
>
> Best Wishes,
> Ralf
>
>
>
>
>
> Yahoo! Groups Links
>
>
>
>

--- In MentalCalculation@yahoogroups.com, "rob221b" <rob221b@y...> wrote:
> I wonder what kind of preparation is required in order to achieve
> scores of 40+ in the 1 minute calendar???

My experience with the "one year" category was that doing a lot of
training can bring you to a point where you do not actually calculate
anymore - you know the answer instantly.
It was helpful to recognize patterns, for example the dates 04-04,
06-06, 08-08, 10-10 and 12-12 are all the same for a given year.
The same is true for 03-03, 05-05 and 07-07, for 01-11, 11-01 and
22-02 (unless the year is a leap year) etc. No more calculation is
needed for such dates.
--- In MentalCalculation@yahoogroups.com, jean <jean2ms@y...> wrote:
>   Do you think you can go above 60 per minute ? (1600-2100)

To be honest, 60 dates from 1600-2100 in one minute seems to be almost
impossible for me. It is interesting that at this speed the problem is
the concentration, not the calculating, at least if the answer has to
be spoken.

Ralf

--- In MentalCalculation@yahoogroups.com, "ralf_laue" <info@r...>
wrote:
> To be honest, 60 dates from 1600-2100 in one minute seems to be
almost
> impossible for me. It is interesting that at this speed the
problem is
> the concentration, not the calculating, at least if the answer has
to
> be spoken.
>
> Ralf
>

I noticed that when Matthias was attempting the 1 minute record,
speaking aloud the days of the week, he did not pause between his
answers, but rather spoke more slowly so as to maintain a steady
rhythm, like this:

"Montaaaag, Freitaaaag, Sontaaaag"

instead of like this:

"Montag, ... , Freitag, ... , Sontag"

Presumably he speaks the answer to one problem while reading and
calculating the next one. I suspect the calculations involve little
more than adding together two single-digit numbers, an offset having
been memorised for all 501 years in the range 1600-2001 and all 366
dates in a year (the year 1900, say). A rate of one answer per
second does not look so unattainable if this is the case.

Robert F

> been memorised for all 501 years in the range 1600-2001 and all 366

I meant 1600-2100, not 1600-2001, of course.
Perhaps Matthais or Yusnier can reveal something of their approach, so
I do not need to speculate (such action might breach MC secrecy
rules???) -Robert

After experimenting a little speaking the days insytead of writing
them, I discovered that a score of 40+ can be achieved using the
standard four-offset method, after all. RobF

--- In MentalCalculation@yahoogroups.com, "rob221b" <rob221b@y...>
wrote:
>
> --- In MentalCalculation@yahoogroups.com, "ralf_laue" <info@r...>
> wrote:
> > To be honest, 60 dates from 1600-2100 in one minute seems to be
> almost

Are we talking about "spoken records" or "written records" ?
   If it is the first choice, what is the best "written record" in relation to
the next World Cup please ? Congratulations for the records !

   Regards,

   Jean-Eric.

rob221b <rob221b@...> a écrit :
   --- In MentalCalculation@yahoogroups.com, "ralf_laue" <info@r...>
wrote:
> To be honest, 60 dates from 1600-2100 in one minute seems to be
almost
> impossible for me. It is interesting that at this speed the
problem is
> the concentration, not the calculating, at least if the answer has
to
> be spoken.
>
> Ralf
>

I noticed that when Matthias was attempting the 1 minute record,
speaking aloud the days of the week, he did not pause between his
answers, but rather spoke more slowly so as to maintain a steady
rhythm, like this:

"Montaaaag, Freitaaaag, Sontaaaag"

instead of like this:

"Montag, ... , Freitag, ... , Sontag"

Presumably he speaks the answer to one problem while reading and
calculating the next one. I suspect the calculations involve little
more than adding together two single-digit numbers, an offset having
been memorised for all 501 years in the range 1600-2001 and all 366
dates in a year (the year 1900, say). A rate of one answer per
second does not look so unattainable if this is the case.

Robert F










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

Hi Robert,

what exactly is the standard four-offset method ?

Jan


   After experimenting a little speaking the days insytead of writing
   them, I discovered that a score of 40+ can be achieved using the
   standard four-offset method, after all. RobF



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

I should read these things BEFORE I send them. There is a glaring
error in this segment..

============================================
We test first the offset for the year 27, which is 5. We can then
reel off all the years (the next must be 6 years later, then 11
years, and so on according to 5 6 11 6 5 6 11).
Hence, the years are:
1922, 1927, 1933, 1944, 1950, 1955, 1961, 1972.
Notice the pattern 5, 6, 11, 6, 5, 6, 11.
============================================

27 DOES have offset 5, but we are looking for years with offset 6.
Doh! Since 28 has offset 0, i.e. 6 has been skipped, then we know
the next year to have offset 6 is 11 years after year 27. So the
CORRECT sequence of years between 1920 and 1975 for which the 18th
of Aug fell on a Friday, is

1922, 1933, 1939, 1944, 1950, 1961, 1967, 1972.

Notice the pattern 11, 6, 5, 6, 11, 6, 5.

Robert

HI!!

   A friend have a question for the group:

   Why 31-12-2999 is a twesday ?

   Could somebody tell me about this question quiqly , please ??

   Thanks.

   Alberto Coto

Jan van Koningsveld <jan.van.koningsveld@...> escribió:
   Hi Robert,

what exactly is the standard four-offset method ?

Jan


   After experimenting a little speaking the days insytead of writing
   them, I discovered that a score of 40+ can be achieved using the
   standard four-offset method, after all. RobF



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








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

Hi Jan,

By "standard four-offset method" I mean the method of assigning
offsets to the day of the month, the month, the century and the year,
and adding them up to get the offset for the day of the week. You know
all about this of course, but in case anyone is not familiar, here's a
summary of the principles underlying the method used by most calendar
calculators:

Base date: Sunday 0 January 1900
1900 is convenient for the base century/year because most people
living were born in the 20th century, and so the "On which day of the
week was I born?" questions come mainly from them.
There is no 0th day of the month, but that doesn't matter (really the
base date is the 31st Dec 1899, but better to have zero point
comprising zeros).

The 0th is the base day of the month, and has offset zero.
If the 0th of a particular month is a Tuesday, then the 24th day of
the same month must be a Friday, because 24 mod 7 = 3, and so the 24th
falls on a day of the week that is 3 days later than the 0th. The 24th
is therefore assigned an offset 3 relative to the base day of the
month (the 0th).

January, the base month has offset zero.
January has 31 days, so February is assigned offset 3 (31 mod 7 = 3),
i.e. date in Feb is 3 days later than the same date in January.
Example: Today is Saturday 21 Jan, so 21 Feb this year will be a
Tuesday.
February has 28 days (for now ignoring leap years), so a date in March
will fall on the same day of the week as it does in February (28 mod 7
= 0). Therefore March has offset zero relative to February and an
offset 3 relative to January. The 21 Mar this year will be a Tuesday,
same as February.
March has 31 days and 31 mod 7 = 3, so dates in April fall on day of
the week that is 3 days later than the same date in March. So April
has offset 3 relative to March, and therefore an offset 6 relative to
January. So 21 April this year will be a Friday.
Continuing this, offsets relative to January can be defined for every
month:

Jan 0
Feb 3
Mar 3
Apr 6
May 1
Jun 4
Jul 6
Aug 2
Sep 5
Oct 0
Nov 3
Dec 5

It is not necessary to commit this table to memory if you know the
number of days in a month, but it speeds up the calc if it IS
memorised. Possibly at first just remember (besides Jan 0) two or
three offsets and generate the rest (I remember May 1, because "May
day" and Oct 0 because Oct begins with 0). Example, if know offset for
October is zero and know September has 30 days, then offset for
September must be 5, since 30 mod 7 + 5 = 0.

The base year of any century is 00, which is assigned offset 0.
365 mod 7 =1, so if a certain date one year falls on a Friday, then
the same date will fall on a Saturday the following year. If it were
not for leap years, then the year offsets would simply increase by 1
each year: 00 would have offset 0, 01 would have offset 1, 02 would
have offset 2. 03 offset 3, 04 offset 4, and so on up to 99, which has
offset 1 (=99 mod 7). Leap years complicate matters slightly and the
simple offset is increased by 1 every February the 29th. Rather than
have two offsets for leap years (one applicable up to Feb 29th and one
after Feb 29th) it is simpler to have one offset for the whole leap
year, and since Jan-Feb is a much shorter period than Mar-Dec, then it
is better to use the offset applicable to the latter period and treat
Jan/Feb in Leap years as a special case (deducting one from the final
offset – more on this later).

The offsets for the years are therefore:

00 0
01 1
02 2
03 3
04 5
05 6
06 0
07 1
08 3
09 4
.
.
.
98 3
99 4

I should have emphasised earlier that Offset is identical to (Offset
mod 7), because the days have a 7 day cycle -a week, and that all
arithmetic can be performed modulo 7.

It is of course very easy to calculate the year offsets as explained
above instead of memorizing them -for year Y, evaluate (Y mod 7 + Y
div 4) mod 7, or (Y + Y div 4) mod 7 or an equivalent expression, but
the fastest calculations are those that involve the least calculating.
Also can exploit the fact that Y+28 has the same year offset as Y (28
is the smallest multiple of 7 and 4).

The base century is the 20th century (or the 1900s), which has offset
0.
A date in the 21st century falls on day of the week that is one day
before the same date in the 20th century. To see why, look at the
above table of year offsets. The year 99 has an offset 4. If continue
the table, the next year – year 100 would have offset 6, so the year
2000 has offset 6 relative to the base year, 1900. This means the 21st
century has offset 6 relative to the 20th century. In the same way,
offsets can be assigned to other centuries.
If a year cc00 is such that cc is not divisible by 4, then it is not a
leap year. So, 1700,1800,1900, 2100, 2200, 2300 are not leap years,
but 1600, 2000 and 2400 are leap years. The offsets for the centuries
are therefore:

20 0
21 6
22 4
23 2
24 0
25 6
and the pattern continues.

The pattern extends in the other direction also, so the 19th century
has offset 2, the 18th century has offset 4, etc. The calendar clearly
repeats itself every 400 years.


With all 4 offsets now defined, an example is in order:

16 April 1758
Day of the month 16 has offset 2
Month April has offset 6
Century 18 (i.e. the 1700s) has offset 4
Year 58 has offset 2

Total offset 0, i.e. the day of the week is zero days after the day of
the week of our base date of 0 January 1900 – a Sunday.


An example to illustrate special case for Jan/Feb in Leap Year..

11 February 13852
11 has offset 4
Month Feb has offset 3
Century 139 (the 13800s) has offset 2 (use fact that calendar repeats
every 400 years, so just subtract multiple of 4 from 38 to get near
familiar territory – here 38 – 20 = 18, so offset is same as that for
the 1800s, i.e. 2)
Year 52 has offset 2

Total offset 4, so Thursday. BUT 52 is a leap year and month is
February, so, as explained before, should be day earlier, i.e.
Wednesday. Wednesday is correct.


That's the theory. Here's the practice..

If the date is proposed verbally (not written), then the calculation
can begin before the full date has been given.
Example:
22 July 1960
When hear "22" immediately have in mind 1 (offset for 22)
When hear "July", calculate in no time at all 0 (either do 1+6 or 1-1,
i.e. offset for July is either 6 or –1, they are equal modulo 7).
When hear "19", do nothing because that's the base century, so total
offset will just be whatever the year offset is (month is not Jan/Feb,
so no leap year correction to make whatever the year turns out to be).
When hear "60", can say immediate "Friday" (year 60 has offset 5).

If the date is proposed in written form, then it might be better not
to cast out sevens at all (the final offset will never exceed 49 and
with practice the association `offset-day', e.g. 49-Sunday, will
become automatic). I don't know this for sure, since I automatically
cast out sevens – cannot change a habit of 20 years!


There are a number of advantages in having the year offsets memorised
rather than calculate them. They can be used as a look-up table when
answering questions such as, in which years between 1920 and 1975 did
the 18th of August fall on a Friday?

Rain man? It is easy, especially if you know the magic sequence 6, 5,
6, 11, 6, 5, 6, 11, ...

18 August has offset 6 (4+2).
If the Year offset is Y, then 6 + Y = 5 (Friday).
So Y is 6.

Now we search for the first 6 in our memorised table of year offsets
starting with the year 20 (i.e. the year 1920).
Year 20 has offset 4
Year 21 has offset 5
Year 22 has offset 6
So the first year after 1920 for which 18 Aug was Friday, is 1922.
Now, years, including leap years, with identical calendars follow the
sequence 6, 11, 6, 5, 6, 11, 6, 5, …11. Example: 1900, 1906, 1917,
1923, 1928, 1934,..., so the next year after 1920 for which the 18th
Aug fell on a Friday is either 1927, 1928 or 1933. We test first the
offset for the year 27, which is 5. We can then reel off all the years
(the next must be 6 years later, then 11 years, and so on according to
5 6 11 6 5 6 11).
Hence, the years are:
1922, 1927, 1933, 1944, 1950, 1955, 1961, 1972.
Notice the pattern 5, 6, 11, 6, 5, 6, 11.

If the first offset had been a 6 instead of 5, then the second year in
the sequence must be identified from the look-up table, since 6 can be
followed by either 5 or 11.


And that's the standard four-offset method.

Robert



--- In MentalCalculation@yahoogroups.com, "Jan van Koningsveld"
<jan.van.koningsveld@e...> wrote:
>
> Hi Robert,
>
> what exactly is the standard four-offset method ?
>
> Jan
>
>
>   After experimenting a little speaking the days insytead of writing
>   them, I discovered that a score of 40+ can be achieved using the
>   standard four-offset method, after all. RobF
>
>
>
> [Non-text portions of this message have been removed]
>

hi rob,

   WOW, thank you from every body in this group for the information. but
unfortunately, i had discover that alone. i should had waited till i read all
these details from you.
   and not waisted my time to explore all those combinations.

   using your method to compute Alberto Question:
   Why 31-12-2999 is a tuesday?

   it is true to be a tuesday just read Robert explanation below

   best regards
   Issam KHNEISSER


rob221b <rob221b@...> wrote:
   Hi Jan,

By "standard four-offset method" I mean the method of assigning
offsets to the day of the month, the month, the century and the year,
and adding them up to get the offset for the day of the week. You know
all about this of course, but in case anyone is not familiar, here's a
summary of the principles underlying the method used by most calendar
calculators:

Base date: Sunday 0 January 1900
1900 is convenient for the base century/year because most people
living were born in the 20th century, and so the "On which day of the
week was I born?" questions come mainly from them.
There is no 0th day of the month, but that doesn't matter (really the
base date is the 31st Dec 1899, but better to have zero point
comprising zeros).

The 0th is the base day of the month, and has offset zero.
If the 0th of a particular month is a Tuesday, then the 24th day of
the same month must be a Friday, because 24 mod 7 = 3, and so the 24th
falls on a day of the week that is 3 days later than the 0th. The 24th
is therefore assigned an offset 3 relative to the base day of the
month (the 0th).

January, the base month has offset zero.
January has 31 days, so February is assigned offset 3 (31 mod 7 = 3),
i.e. date in Feb is 3 days later than the same date in January.
Example: Today is Saturday 21 Jan, so 21 Feb this year will be a
Tuesday.
February has 28 days (for now ignoring leap years), so a date in March
will fall on the same day of the week as it does in February (28 mod 7
= 0). Therefore March has offset zero relative to February and an
offset 3 relative to January. The 21 Mar this year will be a Tuesday,
same as February.
March has 31 days and 31 mod 7 = 3, so dates in April fall on day of
the week that is 3 days later than the same date in March. So April
has offset 3 relative to March, and therefore an offset 6 relative to
January. So 21 April this year will be a Friday.
Continuing this, offsets relative to January can be defined for every
month:

Jan 0
Feb 3
Mar 3
Apr 6
May 1
Jun 4
Jul 6
Aug 2
Sep 5
Oct 0
Nov 3
Dec 5

It is not necessary to commit this table to memory if you know the
number of days in a month, but it speeds up the calc if it IS
memorised. Possibly at first just remember (besides Jan 0) two or
three offsets and generate the rest (I remember May 1, because "May
day" and Oct 0 because Oct begins with 0). Example, if know offset for
October is zero and know September has 30 days, then offset for
September must be 5, since 30 mod 7 + 5 = 0.

The base year of any century is 00, which is assigned offset 0.
365 mod 7 =1, so if a certain date one year falls on a Friday, then
the same date will fall on a Saturday the following year. If it were
not for leap years, then the year offsets would simply increase by 1
each year: 00 would have offset 0, 01 would have offset 1, 02 would
have offset 2. 03 offset 3, 04 offset 4, and so on up to 99, which has
offset 1 (=99 mod 7). Leap years complicate matters slightly and the
simple offset is increased by 1 every February the 29th. Rather than
have two offsets for leap years (one applicable up to Feb 29th and one
after Feb 29th) it is simpler to have one offset for the whole leap
year, and since Jan-Feb is a much shorter period than Mar-Dec, then it
is better to use the offset applicable to the latter period and treat
Jan/Feb in Leap years as a special case (deducting one from the final
offset – more on this later).

The offsets for the years are therefore:

00 0
01 1
02 2
03 3
04 5
05 6
06 0
07 1
08 3
09 4
.
.
.
98 3
99 4

I should have emphasised earlier that Offset is identical to (Offset
mod 7), because the days have a 7 day cycle -a week, and that all
arithmetic can be performed modulo 7.

It is of course very easy to calculate the year offsets as explained
above instead of memorizing them -for year Y, evaluate (Y mod 7 + Y
div 4) mod 7, or (Y + Y div 4) mod 7 or an equivalent expression, but
the fastest calculations are those that involve the least calculating.
Also can exploit the fact that Y+28 has the same year offset as Y (28
is the smallest multiple of 7 and 4).

The base century is the 20th century (or the 1900s), which has offset
0.
A date in the 21st century falls on day of the week that is one day
before the same date in the 20th century. To see why, look at the
above table of year offsets. The year 99 has an offset 4. If continue
the table, the next year – year 100 would have offset 6, so the year
2000 has offset 6 relative to the base year, 1900. This means the 21st
century has offset 6 relative to the 20th century. In the same way,
offsets can be assigned to other centuries.
If a year cc00 is such that cc is not divisible by 4, then it is not a
leap year. So, 1700,1800,1900, 2100, 2200, 2300 are not leap years,
but 1600, 2000 and 2400 are leap years. The offsets for the centuries
are therefore:

20 0
21 6
22 4
23 2
24 0
25 6
and the pattern continues.

The pattern extends in the other direction also, so the 19th century
has offset 2, the 18th century has offset 4, etc. The calendar clearly
repeats itself every 400 years.


With all 4 offsets now defined, an example is in order:

16 April 1758
Day of the month 16 has offset 2
Month April has offset 6
Century 18 (i.e. the 1700s) has offset 4
Year 58 has offset 2

Total offset 0, i.e. the day of the week is zero days after the day of
the week of our base date of 0 January 1900 – a Sunday.


An example to illustrate special case for Jan/Feb in Leap Year..

11 February 13852
11 has offset 4
Month Feb has offset 3
Century 139 (the 13800s) has offset 2 (use fact that calendar repeats
every 400 years, so just subtract multiple of 4 from 38 to get near
familiar territory – here 38 – 20 = 18, so offset is same as that for
the 1800s, i.e. 2)
Year 52 has offset 2

Total offset 4, so Thursday. BUT 52 is a leap year and month is
February, so, as explained before, should be day earlier, i.e.
Wednesday. Wednesday is correct.


That's the theory. Here's the practice..

If the date is proposed verbally (not written), then the calculation
can begin before the full date has been given.
Example:
22 July 1960
When hear "22" immediately have in mind 1 (offset for 22)
When hear "July", calculate in no time at all 0 (either do 1+6 or 1-1,
i.e. offset for July is either 6 or –1, they are equal modulo 7).
When hear "19", do nothing because that's the base century, so total
offset will just be whatever the year offset is (month is not Jan/Feb,
so no leap year correction to make whatever the year turns out to be).
When hear "60", can say immediate "Friday" (year 60 has offset 5).

If the date is proposed in written form, then it might be better not
to cast out sevens at all (the final offset will never exceed 49 and
with practice the association `offset-day', e.g. 49-Sunday, will
become automatic). I don't know this for sure, since I automatically
cast out sevens – cannot change a habit of 20 years!


There are a number of advantages in having the year offsets memorised
rather than calculate them. They can be used as a look-up table when
answering questions such as, in which years between 1920 and 1975 did
the 18th of August fall on a Friday?

Rain man? It is easy, especially if you know the magic sequence 6, 5,
6, 11, 6, 5, 6, 11, ...

18 August has offset 6 (4+2).
If the Year offset is Y, then 6 + Y = 5 (Friday).
So Y is 6.

Now we search for the first 6 in our memorised table of year offsets
starting with the year 20 (i.e. the year 1920).
Year 20 has offset 4
Year 21 has offset 5
Year 22 has offset 6
So the first year after 1920 for which 18 Aug was Friday, is 1922.
Now, years, including leap years, with identical calendars follow the
sequence 6, 11, 6, 5, 6, 11, 6, 5, …11. Example: 1900, 1906, 1917,
1923, 1928, 1934,..., so the next year after 1920 for which the 18th
Aug fell on a Friday is either 1927, 1928 or 1933. We test first the
offset for the year 27, which is 5. We can then reel off all the years
(the next must be 6 years later, then 11 years, and so on according to
5 6 11 6 5 6 11).
Hence, the years are:
1922, 1927, 1933, 1944, 1950, 1955, 1961, 1972.
Notice the pattern 5, 6, 11, 6, 5, 6, 11.

If the first offset had been a 6 instead of 5, then the second year in
the sequence must be identified from the look-up table, since 6 can be
followed by either 5 or 11.


And that's the standard four-offset method.

Robert



--- In MentalCalculation@yahoogroups.com, "Jan van Koningsveld"
<jan.van.koningsveld@e...> wrote:
>
> Hi Robert,
>
> what exactly is the standard four-offset method ?
>
> Jan
>
>
>   After experimenting a little speaking the days insytead of writing
>   them, I discovered that a score of 40+ can be achieved using the
>   standard four-offset method, after all. RobF
>
>
>
> [Non-text portions of this message have been removed]
>












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---------------------------------
Bring words and photos together (easily) with
  PhotoMail  - it's free and works with Yahoo! Mail.

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

hi,i wanna know the names of the participants frm ind


---------------------------------
  Jiyo cricket on Yahoo! India cricket

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

Thank YOU Issam - I don't think you have been wasting your time in
discovering a calendar method. By "exploring all those combinations"
you must have acquired a genuine feel for the calendar, something
that is lacking in someone who has simply learnt a method by rote
from a book or from a message posted to this group without any
understanding of why the method works.

31-12-2999??? Is there a shortage of dentists in Spain as well?

Best wishes,
Robert


--- In MentalCalculation@yahoogroups.com, issam khneisser
<issamn_1@y...> wrote:
>
> hi rob,
>
>   WOW, thank you from every body in this group for the
information. but unfortunately, i had discover that alone. i should
had waited till i read all these details from you.
>   and not waisted my time to explore all those combinations.
>
>   using your method to compute Alberto Question:
>   Why 31-12-2999 is a tuesday?
>
>   it is true to be a tuesday just read Robert explanation below
>
>   best regards
>   Issam KHNEISSER
>
>
> rob221b <rob221b@y...> wrote:
>   Hi Jan,
>
> By "standard four-offset method" I mean the method of assigning
> offsets to the day of the month, the month, the century and the
year,
> and adding them up to get the offset for the day of the week. You
know
> all about this of course, but in case anyone is not familiar,
here's a
> summary of the principles underlying the method used by most
calendar
> calculators:
>
> Base date: Sunday 0 January 1900
> 1900 is convenient for the base century/year because most people
> living were born in the 20th century, and so the "On which day of
the
> week was I born?" questions come mainly from them.
> There is no 0th day of the month, but that doesn't matter (really
the
> base date is the 31st Dec 1899, but better to have zero point
> comprising zeros).
>
> The 0th is the base day of the month, and has offset zero.
> If the 0th of a particular month is a Tuesday, then the 24th day
of
> the same month must be a Friday, because 24 mod 7 = 3, and so the
24th
> falls on a day of the week that is 3 days later than the 0th. The
24th
> is therefore assigned an offset 3 relative to the base day of the
> month (the 0th).
>
> January, the base month has offset zero.
> January has 31 days, so February is assigned offset 3 (31 mod 7 =
3),
> i.e. date in Feb is 3 days later than the same date in January.
> Example: Today is Saturday 21 Jan, so 21 Feb this year will be a
> Tuesday.
> February has 28 days (for now ignoring leap years), so a date in
March
> will fall on the same day of the week as it does in February (28
mod 7
> = 0). Therefore March has offset zero relative to February and an
> offset 3 relative to January. The 21 Mar this year will be a
Tuesday,
> same as February.
> March has 31 days and 31 mod 7 = 3, so dates in April fall on day
of
> the week that is 3 days later than the same date in March. So
April
> has offset 3 relative to March, and therefore an offset 6 relative
to
> January. So 21 April this year will be a Friday.
> Continuing this, offsets relative to January can be defined for
every
> month:
>
> Jan 0
> Feb 3
> Mar 3
> Apr 6
> May 1
> Jun 4
> Jul 6
> Aug 2
> Sep 5
> Oct 0
> Nov 3
> Dec 5
>
> It is not necessary to commit this table to memory if you know the
> number of days in a month, but it speeds up the calc if it IS
> memorised. Possibly at first just remember (besides Jan 0) two or
> three offsets and generate the rest (I remember May 1,
because "May
> day" and Oct 0 because Oct begins with 0). Example, if know offset
for
> October is zero and know September has 30 days, then offset for
> September must be 5, since 30 mod 7 + 5 = 0.
>
> The base year of any century is 00, which is assigned offset 0.
> 365 mod 7 =1, so if a certain date one year falls on a Friday,
then
> the same date will fall on a Saturday the following year. If it
were
> not for leap years, then the year offsets would simply increase by
1
> each year: 00 would have offset 0, 01 would have offset 1, 02
would
> have offset 2. 03 offset 3, 04 offset 4, and so on up to 99, which
has
> offset 1 (=99 mod 7). Leap years complicate matters slightly and
the
> simple offset is increased by 1 every February the 29th. Rather
than
> have two offsets for leap years (one applicable up to Feb 29th and
one
> after Feb 29th) it is simpler to have one offset for the whole
leap
> year, and since Jan-Feb is a much shorter period than Mar-Dec,
then it
> is better to use the offset applicable to the latter period and
treat
> Jan/Feb in Leap years as a special case (deducting one from the
final
> offset – more on this later).
>
> The offsets for the years are therefore:
>
> 00 0
> 01 1
> 02 2
> 03 3
> 04 5
> 05 6
> 06 0
> 07 1
> 08 3
> 09 4
> .
> .
> .
> 98 3
> 99 4
>
> I should have emphasised earlier that Offset is identical to
(Offset
> mod 7), because the days have a 7 day cycle -a week, and that all
> arithmetic can be performed modulo 7.
>
> It is of course very easy to calculate the year offsets as
explained
> above instead of memorizing them -for year Y, evaluate (Y mod 7 +
Y
> div 4) mod 7, or (Y + Y div 4) mod 7 or an equivalent expression,
but
> the fastest calculations are those that involve the least
calculating.
> Also can exploit the fact that Y+28 has the same year offset as Y
(28
> is the smallest multiple of 7 and 4).
>
> The base century is the 20th century (or the 1900s), which has
offset
> 0.
> A date in the 21st century falls on day of the week that is one
day
> before the same date in the 20th century. To see why, look at the
> above table of year offsets. The year 99 has an offset 4. If
continue
> the table, the next year – year 100 would have offset 6, so the
year
> 2000 has offset 6 relative to the base year, 1900. This means the
21st
> century has offset 6 relative to the 20th century. In the same
way,
> offsets can be assigned to other centuries.
> If a year cc00 is such that cc is not divisible by 4, then it is
not a
> leap year. So, 1700,1800,1900, 2100, 2200, 2300 are not leap
years,
> but 1600, 2000 and 2400 are leap years. The offsets for the
centuries
> are therefore:
>
> 20 0
> 21 6
> 22 4
> 23 2
> 24 0
> 25 6
> and the pattern continues.
>
> The pattern extends in the other direction also, so the 19th
century
> has offset 2, the 18th century has offset 4, etc. The calendar
clearly
> repeats itself every 400 years.
>
>
> With all 4 offsets now defined, an example is in order:
>
> 16 April 1758
> Day of the month 16 has offset 2
> Month April has offset 6
> Century 18 (i.e. the 1700s) has offset 4
> Year 58 has offset 2
>
> Total offset 0, i.e. the day of the week is zero days after the
day of
> the week of our base date of 0 January 1900 – a Sunday.
>
>
> An example to illustrate special case for Jan/Feb in Leap Year..
>
> 11 February 13852
> 11 has offset 4
> Month Feb has offset 3
> Century 139 (the 13800s) has offset 2 (use fact that calendar
repeats
> every 400 years, so just subtract multiple of 4 from 38 to get
near
> familiar territory – here 38 – 20 = 18, so offset is same as that
for
> the 1800s, i.e. 2)
> Year 52 has offset 2
>
> Total offset 4, so Thursday. BUT 52 is a leap year and month is
> February, so, as explained before, should be day earlier, i.e.
> Wednesday. Wednesday is correct.
>
>
> That's the theory. Here's the practice..
>
> If the date is proposed verbally (not written), then the
calculation
> can begin before the full date has been given.
> Example:
> 22 July 1960
> When hear "22" immediately have in mind 1 (offset for 22)
> When hear "July", calculate in no time at all 0 (either do 1+6 or
1-1,
> i.e. offset for July is either 6 or –1, they are equal modulo 7).
> When hear "19", do nothing because that's the base century, so
total
> offset will just be whatever the year offset is (month is not
Jan/Feb,
> so no leap year correction to make whatever the year turns out to
be).
> When hear "60", can say immediate "Friday" (year 60 has offset 5).
>
> If the date is proposed in written form, then it might be better
not
> to cast out sevens at all (the final offset will never exceed 49
and
> with practice the association `offset-day', e.g. 49-Sunday, will
> become automatic). I don't know this for sure, since I
automatically
> cast out sevens – cannot change a habit of 20 years!
>
>
> There are a number of advantages in having the year offsets
memorised
> rather than calculate them. They can be used as a look-up table
when
> answering questions such as, in which years between 1920 and 1975
did
> the 18th of August fall on a Friday?
>
> Rain man? It is easy, especially if you know the magic sequence 6,
5,
> 6, 11, 6, 5, 6, 11, ...
>
> 18 August has offset 6 (4+2).
> If the Year offset is Y, then 6 + Y = 5 (Friday).
> So Y is 6.
>
> Now we search for the first 6 in our memorised table of year
offsets
> starting with the year 20 (i.e. the year 1920).
> Year 20 has offset 4
> Year 21 has offset 5
> Year 22 has offset 6
> So the first year after 1920 for which 18 Aug was Friday, is 1922.
> Now, years, including leap years, with identical calendars follow
the
> sequence 6, 11, 6, 5, 6, 11, 6, 5, …11. Example: 1900, 1906, 1917,
> 1923, 1928, 1934,..., so the next year after 1920 for which the
18th
> Aug fell on a Friday is either 1927, 1928 or 1933. We test first
the
> offset for the year 27, which is 5. We can then reel off all the
years
> (the next must be 6 years later, then 11 years, and so on
according to
> 5 6 11 6 5 6 11).
> Hence, the years are:
> 1922, 1927, 1933, 1944, 1950, 1955, 1961, 1972.
> Notice the pattern 5, 6, 11, 6, 5, 6, 11.
>
> If the first offset had been a 6 instead of 5, then the second
year in
> the sequence must be identified from the look-up table, since 6
can be
> followed by either 5 or 11.
>
>
> And that's the standard four-offset method.
>
> Robert
>
>
>
> --- In MentalCalculation@yahoogroups.com, "Jan van Koningsveld"
> <jan.van.koningsveld@e...> wrote:
> >
> > Hi Robert,
> >
> > what exactly is the standard four-offset method ?
> >
> > Jan
> >
> >
> >   After experimenting a little speaking the days insytead of
writing
> >   them, I discovered that a score of 40+ can be achieved using
the
> >   standard four-offset method, after all. RobF
> >
> >
> >
> > [Non-text portions of this message have been removed]
> >
>
>
>
>
>
>
>
>
>
>
>
>
>   SPONSORED LINKS
>         Mathematics and computer science   Mathematics
Mental     Computer programs
>
> ---------------------------------
>   YAHOO! GROUPS LINKS
>
>
>     Visit your group "MentalCalculation" on the web.
>
>     To unsubscribe from this group, send an email to:
>  MentalCalculation-unsubscribe@yahoogroups.com
>
>     Your use of Yahoo! Groups is subject to the Yahoo! Terms of
Service.
>
>
> ---------------------------------
>
>
>
>
>
>
> ---------------------------------
> Bring words and photos together (easily) with
>  PhotoMail  - it's free and works with Yahoo! Mail.
>
> [Non-text portions of this message have been removed]
>

I also have found out method below described by Robert;
   Emmanuël

rob221b <rob221b@...> a écrit :
   Thank YOU Issam - I don't think you have been wasting your time in
discovering a calendar method. By "exploring all those combinations"
you must have acquired a genuine feel for the calendar, something
that is lacking in someone who has simply learnt a method by rote
from a book or from a message posted to this group without any
understanding of why the method works.

31-12-2999??? Is there a shortage of dentists in Spain as well?

Best wishes,
Robert


--- In MentalCalculation@yahoogroups.com, issam khneisser
wrote:
>
> hi rob,
>
> WOW, thank you from every body in this group for the
information. but unfortunately, i had discover that alone. i should
had waited till i read all these details from you.
> and not waisted my time to explore all those combinations.
>
> using your method to compute Alberto Question:
> Why 31-12-2999 is a tuesday?
>
> it is true to be a tuesday just read Robert explanation below
>
> best regards
> Issam KHNEISSER
>
>
> rob221b wrote:
> Hi Jan,
>
> By "standard four-offset method" I mean the method of assigning
> offsets to the day of the month, the month, the century and the
year,
> and adding them up to get the offset for the day of the week. You
know
> all about this of course, but in case anyone is not familiar,
here's a
> summary of the principles underlying the method used by most
calendar
> calculators:
>
> Base date: Sunday 0 January 1900
> 1900 is convenient for the base century/year because most people
> living were born in the 20th century, and so the "On which day of
the
> week was I born?" questions come mainly from them.
> There is no 0th day of the month, but that doesn't matter (really
the
> base date is the 31st Dec 1899, but better to have zero point
> comprising zeros).
>
> The 0th is the base day of the month, and has offset zero.
> If the 0th of a particular month is a Tuesday, then the 24th day
of
> the same month must be a Friday, because 24 mod 7 = 3, and so the
24th
> falls on a day of the week that is 3 days later than the 0th. The
24th
> is therefore assigned an offset 3 relative to the base day of the
> month (the 0th).
>
> January, the base month has offset zero.
> January has 31 days, so February is assigned offset 3 (31 mod 7 =
3),
> i.e. date in Feb is 3 days later than the same date in January.
> Example: Today is Saturday 21 Jan, so 21 Feb this year will be a
> Tuesday.
> February has 28 days (for now ignoring leap years), so a date in
March
> will fall on the same day of the week as it does in February (28
mod 7
> = 0). Therefore March has offset zero relative to February and an
> offset 3 relative to January. The 21 Mar this year will be a
Tuesday,
> same as February.
> March has 31 days and 31 mod 7 = 3, so dates in April fall on day
of
> the week that is 3 days later than the same date in March. So
April
> has offset 3 relative to March, and therefore an offset 6 relative
to
> January. So 21 April this year will be a Friday.
> Continuing this, offsets relative to January can be defined for
every
> month:
>
> Jan 0
> Feb 3
> Mar 3
> Apr 6
> May 1
> Jun 4
> Jul 6
> Aug 2
> Sep 5
> Oct 0
> Nov 3
> Dec 5
>
> It is not necessary to commit this table to memory if you know the
> number of days in a month, but it speeds up the calc if it IS
> memorised. Possibly at first just remember (besides Jan 0) two or
> three offsets and generate the rest (I remember May 1,
because "May
> day" and Oct 0 because Oct begins with 0). Example, if know offset
for
> October is zero and know September has 30 days, then offset for
> September must be 5, since 30 mod 7 + 5 = 0.
>
> The base year of any century is 00, which is assigned offset 0.
> 365 mod 7 =1, so if a certain date one year falls on a Friday,
then
> the same date will fall on a Saturday the following year. If it
were
> not for leap years, then the year offsets would simply increase by
1
> each year: 00 would have offset 0, 01 would have offset 1, 02
would
> have offset 2. 03 offset 3, 04 offset 4, and so on up to 99, which
has
> offset 1 (=99 mod 7). Leap years complicate matters slightly and
the
> simple offset is increased by 1 every February the 29th. Rather
than
> have two offsets for leap years (one applicable up to Feb 29th and
one
> after Feb 29th) it is simpler to have one offset for the whole
leap
> year, and since Jan-Feb is a much shorter period than Mar-Dec,
then it
> is better to use the offset applicable to the latter period and
treat
> Jan/Feb in Leap years as a special case (deducting one from the
final
> offset – more on this later).
>
> The offsets for the years are therefore:
>
> 00 0
> 01 1
> 02 2
> 03 3
> 04 5
> 05 6
> 06 0
> 07 1
> 08 3
> 09 4
> .
> .
> .
> 98 3
> 99 4
>
> I should have emphasised earlier that Offset is identical to
(Offset
> mod 7), because the days have a 7 day cycle -a week, and that all
> arithmetic can be performed modulo 7.
>
> It is of course very easy to calculate the year offsets as
explained
> above instead of memorizing them -for year Y, evaluate (Y mod 7 +
Y
> div 4) mod 7, or (Y + Y div 4) mod 7 or an equivalent expression,
but
> the fastest calculations are those that involve the least
calculating.
> Also can exploit the fact that Y+28 has the same year offset as Y
(28
> is the smallest multiple of 7 and 4).
>
> The base century is the 20th century (or the 1900s), which has
offset
> 0.
> A date in the 21st century falls on day of the week that is one
day
> before the same date in the 20th century. To see why, look at the
> above table of year offsets. The year 99 has an offset 4. If
continue
> the table, the next year – year 100 would have offset 6, so the
year
> 2000 has offset 6 relative to the base year, 1900. This means the
21st
> century has offset 6 relative to the 20th century. In the same
way,
> offsets can be assigned to other centuries.
> If a year cc00 is such that cc is not divisible by 4, then it is
not a
> leap year. So, 1700,1800,1900, 2100, 2200, 2300 are not leap
years,
> but 1600, 2000 and 2400 are leap years. The offsets for the
centuries
> are therefore:
>
> 20 0
> 21 6
> 22 4
> 23 2
> 24 0
> 25 6
> and the pattern continues.
>
> The pattern extends in the other direction also, so the 19th
century
> has offset 2, the 18th century has offset 4, etc. The calendar
clearly
> repeats itself every 400 years.
>
>
> With all 4 offsets now defined, an example is in order:
>
> 16 April 1758
> Day of the month 16 has offset 2
> Month April has offset 6
> Century 18 (i.e. the 1700s) has offset 4
> Year 58 has offset 2
>
> Total offset 0, i.e. the day of the week is zero days after the
day of
> the week of our base date of 0 January 1900 – a Sunday.
>
>
> An example to illustrate special case for Jan/Feb in Leap Year..
>
> 11 February 13852
> 11 has offset 4
> Month Feb has offset 3
> Century 139 (the 13800s) has offset 2 (use fact that calendar
repeats
> every 400 years, so just subtract multiple of 4 from 38 to get
near
> familiar territory – here 38 – 20 = 18, so offset is same as that
for
> the 1800s, i.e. 2)
> Year 52 has offset 2
>
> Total offset 4, so Thursday. BUT 52 is a leap year and month is
> February, so, as explained before, should be day earlier, i.e.
> Wednesday. Wednesday is correct.
>
>
> That's the theory. Here's the practice..
>
> If the date is proposed verbally (not written), then the
calculation
> can begin before the full date has been given.
> Example:
> 22 July 1960
> When hear "22" immediately have in mind 1 (offset for 22)
> When hear "July", calculate in no time at all 0 (either do 1+6 or
1-1,
> i.e. offset for July is either 6 or –1, they are equal modulo 7).
> When hear "19", do nothing because that's the base century, so
total
> offset will just be whatever the year offset is (month is not
Jan/Feb,
> so no leap year correction to make whatever the year turns out to
be).
> When hear "60", can say immediate "Friday" (year 60 has offset 5).
>
> If the date is proposed in written form, then it might be better
not
> to cast out sevens at all (the final offset will never exceed 49
and
> with practice the association `offset-day', e.g. 49-Sunday, will
> become automatic). I don't know this for sure, since I
automatically
> cast out sevens – cannot change a habit of 20 years!
>
>
> There are a number of advantages in having the year offsets
memorised
> rather than calculate them. They can be used as a look-up table
when
> answering questions such as, in which years between 1920 and 1975
did
> the 18th of August fall on a Friday?
>
> Rain man? It is easy, especially if you know the magic sequence 6,
5,
> 6, 11, 6, 5, 6, 11, ...
>
> 18 August has offset 6 (4+2).
> If the Year offset is Y, then 6 + Y = 5 (Friday).
> So Y is 6.
>
> Now we search for the first 6 in our memorised table of year
offsets
> starting with the year 20 (i.e. the year 1920).
> Year 20 has offset 4
> Year 21 has offset 5
> Year 22 has offset 6
> So the first year after 1920 for which 18 Aug was Friday, is 1922.
> Now, years, including leap years, with identical calendars follow
the
> sequence 6, 11, 6, 5, 6, 11, 6, 5, …11. Example: 1900, 1906, 1917,
> 1923, 1928, 1934,..., so the next year after 1920 for which the
18th
> Aug fell on a Friday is either 1927, 1928 or 1933. We test first
the
> offset for the year 27, which is 5. We can then reel off all the
years
> (the next must be 6 years later, then 11 years, and so on
according to
> 5 6 11 6 5 6 11).
> Hence, the years are:
> 1922, 1927, 1933, 1944, 1950, 1955, 1961, 1972.
> Notice the pattern 5, 6, 11, 6, 5, 6, 11.
>
> If the first offset had been a 6 instead of 5, then the second
year in
> the sequence must be identified from the look-up table, since 6
can be
> followed by either 5 or 11.
>
>
> And that's the standard four-offset method.
>
> Robert
>
>
>
> --- In MentalCalculation@yahoogroups.com, "Jan van Koningsveld"
> wrote:
> >
> > Hi Robert,
> >
> > what exactly is the standard four-offset method ?
> >
> > Jan
> >
> >
> > After experimenting a little speaking the days insytead of
writing
> > them, I discovered that a score of 40+ can be achieved using
the
> > standard four-offset method, after all. RobF
> >
> >
> >
> > [Non-text portions of this message have been removed]
> >
>
>
>
>
>
>
>
>
>
>
>
>
> SPONSORED LINKS
> Mathematics and computer science Mathematics
Mental Computer programs
>
> ---------------------------------
> YAHOO! GROUPS LINKS
>
>
> Visit your group "MentalCalculation" on the web.
>
> To unsubscribe from this group, send an email to:
> MentalCalculation-unsubscribe@yahoogroups.com
>
> Your use of Yahoo! Groups is subject to the Yahoo! Terms of
Service.
>
>
> ---------------------------------
>
>
>
>
>
>
> ---------------------------------
> Bring words and photos together (easily) with
> PhotoMail - it's free and works with Yahoo! Mail.
>
> [Non-text portions of this message have been removed]
>








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

Dear Robert,
I would not call this four-offset method "standard" because the numbering 1
= mo, 2 = tue etc. breaks with a tradition which goes well back to the
middle ages.
I prefer to keep in contact with tradition and stick to 1 = su, 2 = mo
(like Portuguese mo = secunda feira,... , fri = sexta feira).
This concerns only the numbers associated with the months.
Ulrich

On 1/29/06, rob221b <rob221b@...> wrote:
>
> Thank YOU Issam - I don't think you have been wasting your time in
> discovering a calendar method. By "exploring all those combinations"
> you must have acquired a genuine feel for the calendar, something
> that is lacking in someone who has simply learnt a method by rote
> from a book or from a message posted to this group without any
> understanding of why the method works.
>
> 31-12-2999??? Is there a shortage of dentists in Spain as well?
>
> Best wishes,
> Robert
>
>
> --- In MentalCalculation@yahoogroups.com, issam khneisser
> <issamn_1@y...> wrote:
> >
> > hi rob,
> >
> >   WOW, thank you from every body in this group for the
> information. but unfortunately, i had discover that alone. i should
> had waited till i read all these details from you.
> >   and not waisted my time to explore all those combinations.
> >
> >   using your method to compute Alberto Question:
> >   Why 31-12-2999 is a tuesday?
> >
> >   it is true to be a tuesday just read Robert explanation below
> >
> >   best regards
> >   Issam KHNEISSER
> >
> >
> > rob221b <rob221b@y...> wrote:
> >   Hi Jan,
> >
> > By "standard four-offset method" I mean the method of assigning
> > offsets to the day of the month, the month, the century and the
> year,
> > and adding them up to get the offset for the day of the week. You
> know
> > all about this of course, but in case anyone is not familiar,
> here's a
> > summary of the principles underlying the method used by most
> calendar
> > calculators:
> >
> > Base date: Sunday 0 January 1900
> > 1900 is convenient for the base century/year because most people
> > living were born in the 20th century, and so the "On which day of
> the
> > week was I born?" questions come mainly from them.
> > There is no 0th day of the month, but that doesn't matter (really
> the
> > base date is the 31st Dec 1899, but better to have zero point
> > comprising zeros).
> >
> > The 0th is the base day of the month, and has offset zero.
> > If the 0th of a particular month is a Tuesday, then the 24th day
> of
> > the same month must be a Friday, because 24 mod 7 = 3, and so the
> 24th
> > falls on a day of the week that is 3 days later than the 0th. The
> 24th
> > is therefore assigned an offset 3 relative to the base day of the
> > month (the 0th).
> >
> > January, the base month has offset zero.
> > January has 31 days, so February is assigned offset 3 (31 mod 7 =
> 3),
> > i.e. date in Feb is 3 days later than the same date in January.
> > Example: Today is Saturday 21 Jan, so 21 Feb this year will be a
> > Tuesday.
> > February has 28 days (for now ignoring leap years), so a date in
> March
> > will fall on the same day of the week as it does in February (28
> mod 7
> > = 0). Therefore March has offset zero relative to February and an
> > offset 3 relative to January. The 21 Mar this year will be a
> Tuesday,
> > same as February.
> > March has 31 days and 31 mod 7 = 3, so dates in April fall on day
> of
> > the week that is 3 days later than the same date in March. So
> April
> > has offset 3 relative to March, and therefore an offset 6 relative
> to
> > January. So 21 April this year will be a Friday.
> > Continuing this, offsets relative to January can be defined for
> every
> > month:
> >
> > Jan 0
> > Feb 3
> > Mar 3
> > Apr 6
> > May 1
> > Jun 4
> > Jul 6
> > Aug 2
> > Sep 5
> > Oct 0
> > Nov 3
> > Dec 5
> >
> > It is not necessary to commit this table to memory if you know the
> > number of days in a month, but it speeds up the calc if it IS
> > memorised. Possibly at first just remember (besides Jan 0) two or
> > three offsets and generate the rest (I remember May 1,
> because "May
> > day" and Oct 0 because Oct begins with 0). Example, if know offset
> for
> > October is zero and know September has 30 days, then offset for
> > September must be 5, since 30 mod 7 + 5 = 0.
> >
> > The base year of any century is 00, which is assigned offset 0.
> > 365 mod 7 =1, so if a certain date one year falls on a Friday,
> then
> > the same date will fall on a Saturday the following year. If it
> were
> > not for leap years, then the year offsets would simply increase by
> 1
> > each year: 00 would have offset 0, 01 would have offset 1, 02
> would
> > have offset 2. 03 offset 3, 04 offset 4, and so on up to 99, which
> has
> > offset 1 (=99 mod 7). Leap years complicate matters slightly and
> the
> > simple offset is increased by 1 every February the 29th. Rather
> than
> > have two offsets for leap years (one applicable up to Feb 29th and
> one
> > after Feb 29th) it is simpler to have one offset for the whole
> leap
> > year, and since Jan-Feb is a much shorter period than Mar-Dec,
> then it
> > is better to use the offset applicable to the latter period and
> treat
> > Jan/Feb in Leap years as a special case (deducting one from the
> final
> > offset – more on this later).
> >
> > The offsets for the years are therefore:
> >
> > 00 0
> > 01 1
> > 02 2
> > 03 3
> > 04 5
> > 05 6
> > 06 0
> > 07 1
> > 08 3
> > 09 4
> > .
> > .
> > .
> > 98 3
> > 99 4
> >
> > I should have emphasised earlier that Offset is identical to
> (Offset
> > mod 7), because the days have a 7 day cycle -a week, and that all
> > arithmetic can be performed modulo 7.
> >
> > It is of course very easy to calculate the year offsets as
> explained
> > above instead of memorizing them -for year Y, evaluate (Y mod 7 +
> Y
> > div 4) mod 7, or (Y + Y div 4) mod 7 or an equivalent expression,
> but
> > the fastest calculations are those that involve the least
> calculating.
> > Also can exploit the fact that Y+28 has the same year offset as Y
> (28
> > is the smallest multiple of 7 and 4).
> >
> > The base century is the 20th century (or the 1900s), which has
> offset
> > 0.
> > A date in the 21st century falls on day of the week that is one
> day
> > before the same date in the 20th century. To see why, look at the
> > above table of year offsets. The year 99 has an offset 4. If
> continue
> > the table, the next year – year 100 would have offset 6, so the
> year
> > 2000 has offset 6 relative to the base year, 1900. This means the
> 21st
> > century has offset 6 relative to the 20th century. In the same
> way,
> > offsets can be assigned to other centuries.
> > If a year cc00 is such that cc is not divisible by 4, then it is
> not a
> > leap year. So, 1700,1800,1900, 2100, 2200, 2300 are not leap
> years,
> > but 1600, 2000 and 2400 are leap years. The offsets for the
> centuries
> > are therefore:
> >
> > 20 0
> > 21 6
> > 22 4
> > 23 2
> > 24 0
> > 25 6
> > and the pattern continues.
> >
> > The pattern extends in the other direction also, so the 19th
> century
> > has offset 2, the 18th century has offset 4, etc. The calendar
> clearly
> > repeats itself every 400 years.
> >
> >
> > With all 4 offsets now defined, an example is in order:
> >
> > 16 April 1758
> > Day of the month 16 has offset 2
> > Month April has offset 6
> > Century 18 (i.e. the 1700s) has offset 4
> > Year 58 has offset 2
> >
> > Total offset 0, i.e. the day of the week is zero days after the
> day of
> > the week of our base date of 0 January 1900 – a Sunday.
> >
> >
> > An example to illustrate special case for Jan/Feb in Leap Year..
> >
> > 11 February 13852
> > 11 has offset 4
> > Month Feb has offset 3
> > Century 139 (the 13800s) has offset 2 (use fact that calendar
> repeats
> > every 400 years, so just subtract multiple of 4 from 38 to get
> near
> > familiar territory – here 38 – 20 = 18, so offset is same as that
> for
> > the 1800s, i.e. 2)
> > Year 52 has offset 2
> >
> > Total offset 4, so Thursday. BUT 52 is a leap year and month is
> > February, so, as explained before, should be day earlier, i.e.
> > Wednesday. Wednesday is correct.
> >
> >
> > That's the theory. Here's the practice..
> >
> > If the date is proposed verbally (not written), then the
> calculation
> > can begin before the full date has been given.
> > Example:
> > 22 July 1960
> > When hear "22" immediately have in mind 1 (offset for 22)
> > When hear "July", calculate in no time at all 0 (either do 1+6 or
> 1-1,
> > i.e. offset for July is either 6 or –1, they are equal modulo 7).
> > When hear "19", do nothing because that's the base century, so
> total
> > offset will just be whatever the year offset is (month is not
> Jan/Feb,
> > so no leap year correction to make whatever the year turns out to
> be).
> > When hear "60", can say immediate "Friday" (year 60 has offset 5).
> >
> > If the date is proposed in written form, then it might be better
> not
> > to cast out sevens at all (the final offset will never exceed 49
> and
> > with practice the association `offset-day', e.g. 49-Sunday, will
> > become automatic). I don't know this for sure, since I
> automatically
> > cast out sevens – cannot change a habit of 20 years!
> >
> >
> > There are a number of advantages in having the year offsets
> memorised
> > rather than calculate them. They can be used as a look-up table
> when
> > answering questions such as, in which years between 1920 and 1975
> did
> > the 18th of August fall on a Friday?
> >
> > Rain man? It is easy, especially if you know the magic sequence 6,
> 5,
> > 6, 11, 6, 5, 6, 11, ...
> >
> > 18 August has offset 6 (4+2).
> > If the Year offset is Y, then 6 + Y = 5 (Friday).
> > So Y is 6.
> >
> > Now we search for the first 6 in our memorised table of year
> offsets
> > starting with the year 20 (i.e. the year 1920).
> > Year 20 has offset 4
> > Year 21 has offset 5
> > Year 22 has offset 6
> > So the first year after 1920 for which 18 Aug was Friday, is 1922.
> > Now, years, including leap years, with identical calendars follow
> the
> > sequence 6, 11, 6, 5, 6, 11, 6, 5, …11. Example: 1900, 1906, 1917,
> > 1923, 1928, 1934,..., so the next year after 1920 for which the
> 18th
> > Aug fell on a Friday is either 1927, 1928 or 1933. We test first
> the
> > offset for the year 27, which is 5. We can then reel off all the
> years
> > (the next must be 6 years later, then 11 years, and so on
> according to
> > 5 6 11 6 5 6 11).
> > Hence, the years are:
> > 1922, 1927, 1933, 1944, 1950, 1955, 1961, 1972.
> > Notice the pattern 5, 6, 11, 6, 5, 6, 11.
> >
> > If the first offset had been a 6 instead of 5, then the second
> year in
> > the sequence must be identified from the look-up table, since 6
> can be
> > followed by either 5 or 11.
> >
> >
> > And that's the standard four-offset method.
> >
> > Robert
> >
> >
> >
> > --- In MentalCalculation@yahoogroups.com, "Jan van Koningsveld"
> > <jan.van.koningsveld@e...> wrote:
> > >
> > > Hi Robert,
> > >
> > > what exactly is the standard four-offset method ?
> > >
> > > Jan
> > >
> > >
> > >   After experimenting a little speaking the days insytead of
> writing
> > >   them, I discovered that a score of 40+ can be achieved using
> the
> > >   standard four-offset method, after all. RobF
> > >
> > >
> > >
> > > [Non-text portions of this message have been removed]
> > >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >   SPONSORED LINKS
> >         Mathematics and computer science   Mathematics
> Mental     Computer programs
> >
> > ---------------------------------
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> >
> >
> >     Visit your group "MentalCalculation" on the web.
> >
> >     To unsubscribe from this group, send an email to:
> >  MentalCalculation-unsubscribe@yahoogroups.com
> >
> >     Your use of Yahoo! Groups is subject to the Yahoo! Terms of
> Service.
> >
> >
> > ---------------------------------
> >
> >
> >
> >
> >
> >
> > ---------------------------------
> > Bring words and photos together (easily) with
> >  PhotoMail  - it's free and works with Yahoo! Mail.
> >
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> >
>
>
>
>
>
>
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[Non-text portions of this message have been removed]

Dear Friends,
             According to standard four offset method
4/november/1998
offset of date is 4
offset of month is 3
offset of year is 3
offset of century is 0 (1900 as base century)

Total offset =10 so,wednesday. (This is fine)

Now of 4/November/1978
Everything this same....

So what day will be on this day....

Regards
Deepak
   


On Tue, 31 Jan 2006 Ulrich Voigt wrote :
>Dear Robert,
>I would not call this four-offset method "standard" because the numbering 1
>= mo, 2 = tue etc. breaks with a tradition which goes well back to the
>middle ages.
>I prefer to keep in contact with tradition and stick to 1 = su, 2 = mo
>(like Portuguese mo = secunda feira,... , fri = sexta feira).
>This concerns only the numbers associated with the months.
>Ulrich
>
>On 1/29/06, rob221b <rob221b@...> wrote:
> >
> > Thank YOU Issam - I don't think you have been wasting your time in
> > discovering a calendar method. By "exploring all those combinations"
> > you must have acquired a genuine feel for the calendar, something
> > that is lacking in someone who has simply learnt a method by rote
> > from a book or from a message posted to this group without any
> > understanding of why the method works.
> >
> > 31-12-2999??? Is there a shortage of dentists in Spain as well?
> >
> > Best wishes,
> > Robert
> >
> >
> > --- In MentalCalculation@yahoogroups.com, issam khneisser
> > <issamn_1@y...> wrote:
> > >
> > > hi rob,
> > >
> > >   WOW, thank you from every body in this group for the
> > information. but unfortunately, i had discover that alone. i should
> > had waited till i read all these details from you.
> > >   and not waisted my time to explore all those combinations.
> > >
> > >   using your method to compute Alberto Question:
> > >   Why 31-12-2999 is a tuesday?
> > >
> > >   it is true to be a tuesday just read Robert explanation below
> > >
> > >   best regards
> > >   Issam KHNEISSER
> > >
> > >
> > > rob221b <rob221b@y...> wrote:
> > >   Hi Jan,
> > >
> > > By "standard four-offset method" I mean the method of assigning
> > > offsets to the day of the month, the month, the century and the
> > year,
> > > and adding them up to get the offset for the day of the week. You
> > know
> > > all about this of course, but in case anyone is not familiar,
> > here's a
> > > summary of the principles underlying the method used by most
> > calendar
> > > calculators:
> > >
> > > Base date: Sunday 0 January 1900
> > > 1900 is convenient for the base century/year because most people
> > > living were born in the 20th century, and so the "On which day of
> > the
> > > week was I born?" questions come mainly from them.
> > > There is no 0th day of the month, but that doesn't matter (really
> > the
> > > base date is the 31st Dec 1899, but better to have zero point
> > > comprising zeros).
> > >
> > > The 0th is the base day of the month, and has offset zero.
> > > If the 0th of a particular month is a Tuesday, then the 24th day
> > of
> > > the same month must be a Friday, because 24 mod 7 = 3, and so the
> > 24th
> > > falls on a day of the week that is 3 days later than the 0th. The
> > 24th
> > > is therefore assigned an offset 3 relative to the base day of the
> > > month (the 0th).
> > >
> > > January, the base month has offset zero.
> > > January has 31 days, so February is assigned offset 3 (31 mod 7 =
> > 3),
> > > i.e. date in Feb is 3 days later than the same date in January.
> > > Example: Today is Saturday 21 Jan, so 21 Feb this year will be a
> > > Tuesday.
> > > February has 28 days (for now ignoring leap years), so a date in
> > March
> > > will fall on the same day of the week as it does in February (28
> > mod 7
> > > = 0). Therefore March has offset zero relative to February and an
> > > offset 3 relative to January. The 21 Mar this year will be a
> > Tuesday,
> > > same as February.
> > > March has 31 days and 31 mod 7 = 3, so dates in April fall on day
> > of
> > > the week that is 3 days later than the same date in March. So
> > April
> > > has offset 3 relative to March, and therefore an offset 6 relative
> > to
> > > January. So 21 April this year will be a Friday.
> > > Continuing this, offsets relative to January can be defined for
> > every
> > > month:
> > >
> > > Jan 0
> > > Feb 3
> > > Mar 3
> > > Apr 6
> > > May 1
> > > Jun 4
> > > Jul 6
> > > Aug 2
> > > Sep 5
> > > Oct 0
> > > Nov 3
> > > Dec 5
> > >
> > > It is not necessary to commit this table to memory if you know the
> > > number of days in a month, but it speeds up the calc if it IS
> > > memorised. Possibly at first just remember (besides Jan 0) two or
> > > three offsets and generate the rest (I remember May 1,
> > because "May
> > > day" and Oct 0 because Oct begins with 0). Example, if know offset
> > for
> > > October is zero and know September has 30 days, then offset for
> > > September must be 5, since 30 mod 7 + 5 = 0.
> > >
> > > The base year of any century is 00, which is assigned offset 0.
> > > 365 mod 7 =1, so if a certain date one year falls on a Friday,
> > then
> > > the same date will fall on a Saturday the following year. If it
> > were
> > > not for leap years, then the year offsets would simply increase by
> > 1
> > > each year: 00 would have offset 0, 01 would have offset 1, 02
> > would
> > > have offset 2. 03 offset 3, 04 offset 4, and so on up to 99, which
> > has
> > > offset 1 (=99 mod 7). Leap years complicate matters slightly and
> > the
> > > simple offset is increased by 1 every February the 29th. Rather
> > than
> > > have two offsets for leap years (one applicable up to Feb 29th and
> > one
> > > after Feb 29th) it is simpler to have one offset for the whole
> > leap
> > > year, and since Jan-Feb is a much shorter period than Mar-Dec,
> > then it
> > > is better to use the offset applicable to the latter period and
> > treat
> > > Jan/Feb in Leap years as a special case (deducting one from the
> > final
> > > offset – more on this later).
> > >
> > > The offsets for the years are therefore:
> > >
> > > 00 0
> > > 01 1
> > > 02 2
> > > 03 3
> > > 04 5
> > > 05 6
> > > 06 0
> > > 07 1
> > > 08 3
> > > 09 4
> > > .
> > > .
> > > .
> > > 98 3
> > > 99 4
> > >
> > > I should have emphasised earlier that Offset is identical to
> > (Offset
> > > mod 7), because the days have a 7 day cycle -a week, and that all
> > > arithmetic can be performed modulo 7.
> > >
> > > It is of course very easy to calculate the year offsets as
> > explained
> > > above instead of memorizing them -for year Y, evaluate (Y mod 7 +
> > Y
> > > div 4) mod 7, or (Y + Y div 4) mod 7 or an equivalent expression,
> > but
> > > the fastest calculations are those that involve the least
> > calculating.
> > > Also can exploit the fact that Y+28 has the same year offset as Y
> > (28
> > > is the smallest multiple of 7 and 4).
> > >
> > > The base century is the 20th century (or the 1900s), which has
> > offset
> > > 0.
> > > A date in the 21st century falls on day of the week that is one
> > day
> > > before the same date in the 20th century. To see why, look at the
> > > above table of year offsets. The year 99 has an offset 4. If
> > continue
> > > the table, the next year – year 100 would have offset 6, so the
> > year
> > > 2000 has offset 6 relative to the base year, 1900. This means the
> > 21st
> > > century has offset 6 relative to the 20th century. In the same
> > way,
> > > offsets can be assigned to other centuries.
> > > If a year cc00 is such that cc is not divisible by 4, then it is
> > not a
> > > leap year. So, 1700,1800,1900, 2100, 2200, 2300 are not leap
> > years,
> > > but 1600, 2000 and 2400 are leap years. The offsets for the
> > centuries
> > > are therefore:
> > >
> > > 20 0
> > > 21 6
> > > 22 4
> > > 23 2
> > > 24 0
> > > 25 6
> > > and the pattern continues.
> > >
> > > The pattern extends in the other direction also, so the 19th
> > century
> > > has offset 2, the 18th century has offset 4, etc. The calendar
> > clearly
> > > repeats itself every 400 years.
> > >
> > >
> > > With all 4 offsets now defined, an example is in order:
> > >
> > > 16 April 1758
> > > Day of the month 16 has offset 2
> > > Month April has offset 6
> > > Century 18 (i.e. the 1700s) has offset 4
> > > Year 58 has offset 2
> > >
> > > Total offset 0, i.e. the day of the week is zero days after the
> > day of
> > > the week of our base date of 0 January 1900 – a Sunday.
> > >
> > >
> > > An example to illustrate special case for Jan/Feb in Leap Year..
> > >
> > > 11 February 13852
> > > 11 has offset 4
> > > Month Feb has offset 3
> > > Century 139 (the 13800s) has offset 2 (use fact that calendar
> > repeats
> > > every 400 years, so just subtract multiple of 4 from 38 to get
> > near
> > > familiar territory – here 38 – 20 = 18, so offset is same as that
> > for
> > > the 1800s, i.e. 2)
> > > Year 52 has offset 2
> > >
> > > Total offset 4, so Thursday. BUT 52 is a leap year and month is
> > > February, so, as explained before, should be day earlier, i.e.
> > > Wednesday. Wednesday is correct.
> > >
> > >
> > > That's the theory. Here's the practice..
> > >
> > > If the date is proposed verbally (not written), then the
> > calculation
> > > can begin before the full date has been given.
> > > Example:
> > > 22 July 1960
> > > When hear "22" immediately have in mind 1 (offset for 22)
> > > When hear "July", calculate in no time at all 0 (either do 1+6 or
> > 1-1,
> > > i.e. offset for July is either 6 or –1, they are equal modulo 7).
> > > When hear "19", do nothing because that's the base century, so
> > total
> > > offset will just be whatever the year offset is (month is not
> > Jan/Feb,
> > > so no leap year correction to make whatever the year turns out to
> > be).
> > > When hear "60", can say immediate "Friday" (year 60 has offset 5).
> > >
> > > If the date is proposed in written form, then it might be better
> > not
> > > to cast out sevens at all (the final offset will never exceed 49
> > and
> > > with practice the association `offset-day', e.g. 49-Sunday, will
> > > become automatic). I don't know this for sure, since I
> > automatically
> > > cast out sevens – cannot change a habit of 20 years!
> > >
> > >
> > > There are a number of advantages in having the year offsets
> > memorised
> > > rather than calculate them. They can be used as a look-up table
> > when
> > > answering questions such as, in which years between 1920 and 1975
> > did
> > > the 18th of August fall on a Friday?
> > >
> > > Rain man? It is easy, especially if you know the magic sequence 6,
> > 5,
> > > 6, 11, 6, 5, 6, 11, ...
> > >
> > > 18 August has offset 6 (4+2).
> > > If the Year offset is Y, then 6 + Y = 5 (Friday).
> > > So Y is 6.
> > >
> > > Now we search for the first 6 in our memorised table of year
> > offsets
> > > starting with the year 20 (i.e. the year 1920).
> > > Year 20 has offset 4
> > > Year 21 has offset 5
> > > Year 22 has offset 6
> > > So the first year after 1920 for which 18 Aug was Friday, is 1922.
> > > Now, years, including leap years, with identical calendars follow
> > the
> > > sequence 6, 11, 6, 5, 6, 11, 6, 5, …11. Example: 1900, 1906, 1917,
> > > 1923, 1928, 1934,..., so the next year after 1920 for which the
> > 18th
> > > Aug fell on a Friday is either 1927, 1928 or 1933. We test first
> > the
> > > offset for the year 27, which is 5. We can then reel off all the
> > years
> > > (the next must be 6 years later, then 11 years, and so on
> > according to
> > > 5 6 11 6 5 6 11).
> > > Hence, the years are:
> > > 1922, 1927, 1933, 1944, 1950, 1955, 1961, 1972.
> > > Notice the pattern 5, 6, 11, 6, 5, 6, 11.
> > >
> > > If the first offset had been a 6 instead of 5, then the second
> > year in
> > > the sequence must be identified from the look-up table, since 6
> > can be
> > > followed by either 5 or 11.
> > >
> > >
> > > And that's the standard four-offset method.
> > >
> > > Robert
> > >
> > >
> > >
> > > --- In MentalCalculation@yahoogroups.com, "Jan van Koningsveld"
> > > <jan.van.koningsveld@e...> wrote:
> > > >
> > > > Hi Robert,
> > > >
> > > > what exactly is the standard four-offset method ?
> > > >
> > > > Jan
> > > >
> > > >
> > > >   After experimenting a little speaking the days insytead of
> > writing
> > > >   them, I discovered that a score of 40+ can be achieved using
> > the
> > > >   standard four-offset method, after all. RobF
> > > >
> > > >
> > > >
> > > > [Non-text portions of this message have been removed]
> > > >
> > >
> > >
> > >
> > >
> > >
> > >
> > >
> > >
> > >
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