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First, you must find out the starting days for each month in the current year. 2005, for example, looks like this:
January - Saturday
February - Tuesday
March - Tuesday
April - Friday
May - Sunday
June - Wednesday
July - Friday
August - Monday
September - Thursday
October - Saturday
November - Tuesday
December - Thursday
Each weekday is then converted into a number from 0 to 6, where 0 equals a Sunday, and 6 equals a Saturday (See section 3.3 for help in remembering which number represents which day). In our 2005 example, this would result in the number:
622503514624
This number is then broken up into units of 3 digits, in order to make things easier to remember. Continuing with our 2005 example, the number would now look like this:
622 503 514 624
Each of these years is then converted into an image using your preferred system for memorizing 3-digit numbers ( MajorSystem or DominicSystem ), and then linking each image together, to form a chain that will quickly tell you on what weekday the 1st of each month occurs.
To figure out any particular date in the year that you have memorized, you first need to recall on which day the 1st falls. As an example, we'll use September 12, 2005. Since September is the 9th month, we use our mnemonic system to recall which is the 9th key number, which in this case is 4, which means that September 2005 begins on a Thursday.
Once the weekday of the 1st of the month is determined, we need to adjust for the date. September 12, 2005 is 11 days after the 1st, so we add 11 to 4 (our key number for the 1st of September), and we get 15. Casting out all multiples of 7 (the nearest of which is 14), we are left with a total of 1. 1, in our system represents Monday, so have determined that September 12, 2005, is on a Monday.
This is based on the fact that, in any given year, the dates 4/4, 6/6, 8/8, 10/10 and 12/12 will always fall on the same day of the week. This is also true of the easily remembered dates 5/9, 9/5, 7/11, 11/7, as well as the last day of February (regardless of whether it's a leap year or not).
Once you've figured out which day of the week this is, it's a simple matter to adjust to the particular date you need.
As an example, let's figure out on which weekday Halloween 2005 falls, given that the key weekday (by the above standard) for 2005 is Monday. Since we're looking for October, the 10th month, we know that October 10th falls on a Monday. The 31st is exactly 3 weeks after that, so we know that Halloween is a Monday!
Moving on to a tougher example, we'll try March 12th, 2005. There's no easy key date for March (the third month), but we do know that the last day of February is a Monday. That means March 1st is a Tuesday. The 12th is 1 week and 4 days later (11 days total), so all we have to do is jump ahead 4 days from Tuesday, giving us Saturday, as the day of the week for March 12, 2005.
The general rule is that, as you jump ahead one year, you move the key day ahead one day. Conversely, if you wish to move back a year, you move back one key day. 2005's key day is Monday, so 2006's key day would be Tuesday.
The exceptions to this, of course, are leap years. When moving from a non-leap year to a leap year, you have to move the key day ahead two days, instead of one. Similarly, when moving backward from a leap-year to a non-leap year, you have to move the key day two days back.
Knowing that the key day for 2005 is Monday, how can we figure out on what weekday will July 4th, 2007 fall? First, we move the key day ahead two days, for the intervening two years (with no leap years to worry about), giving us Wednesday as 2007's key day. July is the 7th month, so we know that 7/11/2007 will fall on a Wednesday. Since the 4th is one week before that, we know that 7/4/2007 will fall on a Wednesday, as well.
So that you can understand how to work with leap years, we'll use December 31st, 2002 as our next date. First, we step back one key day from 2005 to 2004 (Monday to Sunday). Moving from 2004 to 2003, we have to move back two key days from there (Sunday to Friday), and finally move back one more key day (Friday to Thursday) for the move from 2003 to 2002. Since we have our key day as Thursday for 2002, we know that 12/12/2002 is on a Thursday. The 31st is 2 weeks and 5 days after that, so we simply move 5 days ahead from a Thursday, giving Tuesday. Now we know that 12/31/2002 falls on a Tuesday.
January - 0
February - 3
March - 3
April - 6
May - 1
June - 4
July - 6
August - 2
September - 5
October - 0
November - 3
December - 5
You'll remember each of the numbers via a simple mnemonic, one for each month. We'll combine the first few letters of the month with a letter from the Major System that stands for the associated number. Here's the full list:
JANuS? (Two-faced Roman god after whom January is named)
FeMMe?
Ma'aM
APiSH?
MAiD?
JUNioR?
JUDGE
AUGUSTi``Ne
SELL
OCTOpuS? (Ignore P - which would be 9)
NOVuM? (Think of the birth control pill - Ortho Novum)
DECiMaL?
It should be a simple matter to associate each word with the corresponding month.
Our first example will be June 27, 1959.
Start by looking at the year. Take the last two digits only and divide them by 4. The technically accurate answer would be 14.75. You don't need to worry about any decimal places or remainders here, as 14 is good enough. To make dividing by 4 easier, just round down to the nearest even number, and divide by 2. If you get an odd number as a result, round down to the nearest even number, and divide by 2 again. For 1959, you round down to 58, divide by 2, which gives you 29. 29 is an odd number, so round down to 28, and divide by 2 to give you 14.
Take the result, and add it back to the year. So now, we have 59 + 14. Mentally simplify this by changing it into 59 + 10 + 4. 59 + 10 + 4 = 69 + 4 = 73. 73 will be your key number for the year.
The "mod" in the earlier formula refers to a number to be "cast out". Since the number 70 is an easily recognized multiple of 7, we can get rid of 70, and just remember 3. So if our key number wound up being, say, 81, we could subtract 77 (the nearest multiple of 7) we'd get 4 as the number to remember.
| Year | Key |
| 1900 | 0 |
| 1912 | 1 |
| 1924 | 2 |
| 1936 | 3 |
| 1948 | 4 |
| 1960 | 5 |
| 1972 | 6 |
| 1984 | 0 |
| 1996 | 1 |
Notice that that the last two digits of these years (all leap years, except for 1900), when divided by 12 (with casting out 7s), give the year key number. If you're given any of these years, and you know your multiples of 12, the key number is instantaneous.
If you're given any other year, there are two simple adjustments to make. As an example, we'll say you're given a date in 1982. First, you need to find the highest "12-year" below that date, which in our example, would be 1972. With 1972, we instantly know that the key year is 6. To this, we add number of years since that "12-year" (+10 in our example, giving us a running total of 16).
The last adjustment is to compensate for all the leap years that have happened since the "12-year". The adjustment is made as follows:
1) If the given year is 0, 1, 2 or 3 years past the "12-year", then add 0.
2) If the given year is 4, 5, 6 or 7 years past the "12-year", then add 1.
3) If the given year is 8, 9, 10 or 11 years past the "12-year", then add 2.
In our 1982 example, it is 10 years since 1972, so we would add a 2 to our running total, giving us 18. Casting out 7s, we get 4 as the key year for 1982.
| Year | Key |
| 1900 | 0 |
| 1911 | 6 |
| 1922 | 6 |
| 1933 | 6 |
| 1944 | 6 |
| 1955 | 5 |
| 1966 | 5 |
| 1977 | 5 |
| 1988 | 5 |
| 1999 | 4 |
Once you're given a year to work with, you simply need to find the highest "11 year" that is below it, and count forward from that year, adding one for each regular year and two for each leap year, to get the key number.
For example, if you're given the year 1969, you start counting from 1966 (with a key number of 5), 1967 gives you 6, 1968 (a leap year) gives you 8 and 1969 gives gives you 9. Casting out 7s, this gives a key number of 2 for 1969.
| 1900 | 0 |
| 1904 | 5 |
| 1908 | 3 |
| 1912 | 1 |
| 1916 | 6 |
| 1920 | 4 |
| 1924 | 2 |
It's a pretty simple pattern. The lower 3 leap years are descending odd numbers, and the higher 3 are descending even numbers. Once you've learned this, each of these key numbers will also give the key number for years that are multiples of 28 years away:
| 1900 | 0 | 1928 | 1956 | 1984 |
| 1904 | 5 | 1932 | 1960 | 1988 |
| 1908 | 3 | 1936 | 1964 | 1992 |
| 1912 | 1 | 1940 | 1968 | 1996 |
| 1916 | 6 | 1944 | 1972 | |
| 1920 | 4 | 1948 | 1976 | |
| 1924 | 2 | 1952 | 1980 |
This will give you the key number for any key year quickly. To get the key number for any other year, you simply find the nearest leap year before it, take that leap year's key number, and add the difference in years. For example, to get the key year for 1975, you simply recall the key for 1972 (which is the same as 1916, 6), and add 3 for the 3-year difference, giving 9. Casting out the 7's gives us 2 as the key number for 1975.
Another way to get the key number for a given leap year is to take twice the ten's digit and subtract half the one's digit, casting out 7's. Using 1972 as an example, twice the ten's digit(7) is 14 and half the one's digit(2) is 1. 14 - 1 = 13; 13 - 7 = 6.
| Year | Key | Year | Key | Year | Key | Year | Key | |||
| 1900 | 0 | 1901 | 1 | 1902 | 2 | 1903 | 3 | |||
| 1904 | 5 | 1905 | 6 | 1906 | 0 | 1907 | 1 | |||
| 1908 | 3 | 1909 | 4 | 1910 | 5 | 1911 | 6 | |||
| 1912 | 1 | 1913 | 2 | 1914 | 3 | 1915 | 4 | |||
| 1916 | 6 | 1917 | 0 | 1918 | 1 | 1919 | 2 | |||
| 1920 | 4 | 1921 | 5 | 1922 | 6 | 1923 | 0 | |||
| 1924 | 2 | 1925 | 3 | 1926 | 4 | 1927 | 5 | |||
| 1928 | 0 | 1929 | 1 | 1930 | 2 | 1931 | 3 | |||
| 1932 | 5 | 1933 | 6 | 1934 | 0 | 1935 | 1 | |||
| 1936 | 3 | 1937 | 4 | 1938 | 5 | 1939 | 6 | |||
| 1940 | 1 | 1941 | 2 | 1942 | 3 | 1943 | 4 | |||
| 1944 | 6 | 1945 | 0 | 1946 | 1 | 1947 | 2 | |||
| 1948 | 4 | 1949 | 5 | 1950 | 6 | 1951 | 0 | |||
| 1952 | 2 | 1953 | 3 | 1954 | 4 | 1955 | 5 | |||
| 1956 | 0 | 1957 | 1 | 1958 | 2 | 1959 | 3 | |||
| 1960 | 5 | 1961 | 6 | 1962 | 0 | 1963 | 1 | |||
| 1964 | 3 | 1965 | 4 | 1966 | 5 | 1967 | 6 | |||
| 1968 | 1 | 1969 | 2 | 1970 | 3 | 1971 | 4 | |||
| 1972 | 6 | 1973 | 0 | 1974 | 1 | 1975 | 2 | |||
| 1976 | 4 | 1977 | 5 | 1978 | 6 | 1979 | 0 | |||
| 1980 | 2 | 1981 | 3 | 1982 | 4 | 1983 | 5 | |||
| 1984 | 0 | 1985 | 1 | 1986 | 2 | 1987 | 3 | |||
| 1988 | 5 | 1989 | 6 | 1990 | 0 | 1991 | 1 | |||
| 1992 | 3 | 1993 | 4 | 1994 | 5 | 1995 | 6 | |||
| 1996 | 1 | 1997 | 2 | 1998 | 3 | 1999 | 4 |
Sunday -- 0 (Nothing to do on Sunday!)
Monday -- 1 (1st day of the traditional work week)
Tuesday -- 2 (TUESday = TWOSday)
Wednesday -- 3 (Halfway through the work week, halfway to 6)
Thursday -- 4 (Think "Foursday")
Friday -- 5 ("Friday" and "Five" both begin with F)
Saturday -- 6 ("Saturday" and "Six" both begin with S)
You'll note that more informal mnemonics are used here to associate the days with their corresponding key numbers.
As an example, let's say we want to know on what day of the week June 27th, 1959 fell. We'll add as follows:
+3 (Key number for 1959)
+4 (Key number for June)
+27 (Date in month)
This totals 34. If we cast out all the 7s, we get 6. 6 is the key number for Saturday, so June 27th, 1959, fell on a Saturday (which can be verified with a perpetual calendar).
+1 (Key number for 1968)
+0 (Key number for January)
+15 (Date in month)
This gives us 16, and we cast out 7s, leaving us with 2. To compensate for being in the January of a leap year, we subtract 1, giving us a final total of 1. This tells us that January 15, 1968 fell on a Monday.
It's important to know which years are leap years and which years are not. Most years that are evenly divisible by 4 are leap years. The exception to this rule is that years ending in "00" are only leap years if they are divisible by 400 (1600, 2000 and 2400 are leap years, while 1800, 1900, and 2100 are not).
1700s: add 4
1800s: add 2
1900s: add 0
2000s: add 6
2100s: add 4
2200s: add 2
2300s: add 0
...and so on, repeating regularly.
If you wanted to check on what day of the week on which July 4, 1776 fell, you would calculate it as follows:
+4 (key number for 1976)
+4 (compensate for the 1700s)
+6 (key number for July)
+4 (date)
This totals 18, which is 4, once the 7s are cast out. This gives us Thursday.
Among the trickiest dates are leap years in other centuries. On what day of the week did February 14th, 1852 fall?
+2 (key number for 1952)
+2 (compensate for 1800s)
+3 (key number for February)
+14 (date)
This gives us 21, but it's a leap year, so we need to subtract 1, giving us 20. Casting out 7s gives us 6, which indicates it was a Saturday.
On your left hand assign the month to each nuckle as follows
| thumb | index | middle | ring | pink |
| april, july | february, march, november | january, october | ||
| juni | may | |||
| september, december | augustus |
To get a date on your fingers, go with your thumb to the proper month. You are now on the 1st, 7th, 14th, 21st, 28th day of that week. Going top down first from ring, middle, and index finger you get the proper day of the week. For this year the top of the ring finger is sunday. This will change after a year.
So to calculate the 20th of March, your thumb goes to the top of the middle finger (which means that in this year the first day of March is on a wednesday). After subtracting 14, you get 6th day in march as the same day. You are currently at the fifth so continue with 5 days, resulting in the middle of the ring finger. This is the day after sunday, so monday.
The following feats no longer focus on the day of the week itself. Instead, the formulas in the following feats usually start with the key number for the day of the week as a known variable, and simply subtract other known information to compute a key for the month, year or date.
To avoid working with negative numbers, you'll need to increase the key number for the day of the week. Since multiples of 7 don't affect the formula, and may need to subtract as much as 31 for the date and 6 for the month or year key (but never the year and the month), it's best to start by adding 42 to the day of the week key.
You can perform this feat as an apparent afterthought after giving the weekday, or as a stand-alone feat.
This feat seems much more difficult than it really is. To figure out when other days fall on the same day of the week, all you need to do is find other months (if any) that have the same key number.
First, remember that there will never be 4 or more occurrences of the same date on the same day of the week in any given year.
Here are a list of years, in order of their key dates:
0) January (non-leap year) and October
1) May (If May 12th in a given year is on, say, a Tuesday, you can say definitively that there are no other months with Tuesday the 12th in that year)
2) February (leap year) and August
3) February (non-leap year), March and November.
4) June (Similar to May)
5) September and December
6) January (leap year), April and July
These relationships may seem complex, but they can be mastered. You can connect them using the MajorSystem:
0) JANuS? being eaten by an OCTOpuS?
1) MAiD?, all alone in a big empty room
2) FiBBiN?' August-iNe
3) FeMMe? MA'aM taking ortho NOVuM?
4) JUNioR?, all alone in a big empty room
5) You're trying to SELL to a giant DECiMaL? point
6) CHaNGe? ("Chanuary") from an APiSH? creature into a Judge
Even without mnemonics, these relationships can also be learned easily through sheer repetition.
As a stand-alone feat, there are calculations required, but they're not any more difficult than the regular calendar feat, as states earlier.
Let's say you're asked how many Friday the 13ths occur in the year 1997. To answer this, we start with the key for the day of the week. In our example, Friday is a 5. Start by adding 42 to it, giving 47.
From this number, we're going to subtract the year key and the date key. Since we're asked about 1997, which has a key of 2, we subtract 2 giving 45. Next, we subtract 13 for the day, giving us 32. Finally, we cast out the 7s, and we're left with 4.
The remaining number gives us the a month key. In this case, the only 4 month is June (remember "JUNioR?, alone in a big empty room"?), so we can reply that the only Friday the 13th in 1997 happens in June.
Before performing this feat, you want to be familiar with which months have the same key numbers, as in "Similar Dates in the Same Year".
We start by taking the requested day of the week, and adding 42 to its key. As an example, we'll say you're asked which months in 1963 had 5 Tuesdays. Since tuesday has a key number of 2, we add 42 for a starting total of 44.
Next, we subtract the key for the year. Since 1963's key is 1, we subtract 1 from 44, giving us 43. Now we subtract 1 for the date number. At this step, 1 is always subtracted because the 1st of the month will always have 5 weeks for any month, except a non-leap year February. Applying this final step, we get 42, which becomes 0 after casting out 7's.
The remaining number gives us the month key for months in that year, with at least 29 days (every month but a non-leap year February) in which 5 of the given weekday can be found. This number will also be further adjusted for months with 30 and 31 days.
In our example, we came up with a total of 0, which tells us that, in 1963, the months with a key of 0 and at least 29 days will have 5 Tuesdays in them. These months are January and October.
Next, we subtract 1 to compensate for months consisting of at least 30 days. All the months with this key (except February) will also have 5 of the given day. 0 minus 1 (plus 7, of course) is 6, so the "6" months, April and July, will also have 5 Tuesdays in 1963.
Finally, 1 is subtracted again to compensate for 31 days months. Be careful to not include February, September, April, June and November at this step, even if they have the right keys, because they all have less than 31 days. In our example, 6 minus 1 is 5, so the "5" months with 31 days will also have 5 Tuesdays in 1963. December and September are both "5" months, but September only has 30 days.
So, the answers to which months in 1963 had 5 Tuesdays are January, October, April, July and December.
To do this, you need to have learned the version of this feat where you have memorized the key numbers for the years (section 1.2.2).
Hand a willing participant a perpetual calendar, and ask them to give their birthday, without the year, and the day of the week on which it fell. While they're looking up the day of the week for their birthday, estimate how old they are. For this to work correctly, your guess needs to be within +/- 3 to 4 years of their actual age.
If you need practice estimating people's ages, check out:
The Age Project
AgeGuess.com
Google Search: Guessing a Person's Age
For an example, let's say the person says "June 15th, Tuesday", and you estimate that the person looks to be around 30, and you're performing this in 2004 (yes, the current year is important).
Earlier, you learned that the date plus the month key number plus the year key number equals the day of week key number (casting out 7s as you go). In this feat, we're going to solve for the year key number. To start, take the weekday key number and add 42 to it. From this, subtract the date and the key number for the month. After casting out 7s, this will give you the year key number for which you will be searching.
In our example, the key for Tuesday is 2. Adding 42 to this gives 44. 44 minus 15 (the date) is 29. The key number for 4 is June, so we subtract 4 from 29, giving us 25. Casting out 7s, this gives us 4. We now know we're searching for a year with a key number of 4.
Using your estimate of the person's age, you need to recall which years around that time have the appropriate key number.
Since the person in our example looks to be 30, and this is being performed in 2004, we're looking for a year around 1974 with a key number of 4. A quick mental run from 4 years before to 4 years after 1974 show that both 1971 and 1976 will fit the bill. From here, you have to determine which age looks more appropriate for the person. We know the person is probably either 33 or 28, so how does the person carry themselves?
If the person's birthday is in January or February of a leap year, the key number you get from the equation will be incorrect. But their are ways to figure this out and adjust.
If someone who looks to be in their mid- to late-30s (giving us an estimate of, say, 26) in 2004 says, "February 14th, Saturday", you would calculate the key year as above. Saturday is 6, to which we add 42, giving us 48. 48 minus 14 (the date) is 34. 34 minus 3 (for February) is 31. Casting out 7s, this leaves us with a key year of 3.
Looking for years with a key number of 3 around 1978, you'll get either 1981 or 1970. Neither of these would seem to make sense in this case. 1981 would make him 23, and this guy seemed to be older than 23. 1970 would make him 34, and this guy seems younger than 34.
When you're given a January or February date that results in two answers that are way off (assuming your age-estimation skills are good), then you may be dealing with a leap year. To adjust, add 1 to your year key number (again, casting out 7s if necessary), and look through just nearby leap years.
Near 1978 would be 1980 and 1976. 1980 has a key number of 2, so that doesn't work. 1976, however, does have a key number of 4, which would make this person 28 - right about where you estimated his age!
Fortunately, you only need to worry about the possibility of leap years if you're given a January or February.
Some tips:
What if we had a 364-day year? This is still pretty good, because there would be exactly 52 weeks in the years. Unfortunately, 364 isn't evenly divisible by 12, so some months would have more days, and others would have fewer (or, we could simplify everything and have thirteen 28-day months!). January 1st would always fall on the same day each year, but we'd need to remember key numbers for each month, in order to adjust for the fact that not all months begin on that same day.
Having a 365-day year makes things even trickier. 365 isn't evenly divisible by 7 or 12! If one year were to begin on a Sunday, the next year would begin on a Monday, and so on. This means that, in addition to the month adjustments, we also have to have adjustments for each year. Notice that, so far, leap years still haven't come into question.
Now, even a 365.25-day year wouldn't be too bad. We'd simply have a leap year every 4 years to compensate for the missing day that accumulates in that time. This is effectively the original Julian calendar.
What we actually have, however, is a 365.24237404-day year (the time from one vernal equinox to the next).
Our current Gregorian calendar approximates this by removing 3 leap years every 400 years by eliminating "00" years that aren't divisible by 400, giving an average year of 365.2425 days.
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