Naked Science Memory Course - Copyright Michael Curtis 2007
Calculate the day of the week
Here is a method for stating the day of the week for a large number of dates:
Consider a table of records such as this:
| First Of Jan | Year Number | Day | Leap Year | |
| 01-Jan-1990 | 2 | Monday |
Notice that it has a Leap Year column. Since the LeapYear box is empty, I know that 1990 was not a leap year. That matters because the existence or absence of 29th February has an effect on the calculations which follow below. If someone asks me what the date was on the 18th February 1990, I would need to think about that 1990 row. In the 1990 row (above), the first day of the year is associated with the digit 2. I would also want to break the "18th February" into two small numbers for me to do clever mathematics to: whenever you are told a month, you turn it into a special digit and whenever you see a date like 3rd or 4th or 23rd, you turn that into a special digit also.
Here are the special digits which I use for months. Note that an extra benefit of having a visual representation of any month is that you can memorise months as simple images in a visual story. Below, you see a digit which will be used in math whenever someone reads out a date for which you need to calculate the day of the week:
January 6 Ox
February 2 Bengal tiger
March 2 Bunny / buck
April 5 Monster
May 0 Python
June 3 Horse
July 5 Mouflon [a type of sheep]
August 1 Diana monkey
September 4 Iconic rooster
October 6 Retriever dog
November 2 Boar [wild pig]
December 4 Tame rat
[Those images are based on Chinese astrology images which have a month cycle aspect.]
Since we are interested in February, we need to note the '2' digit which we see next to the word 'February'.
Next, let us turn the '18th' of the date into a single digit. You 'ALWAYS' subtract 7s from the number until you are left with a remainder: 18-7 = 11. 11 - 7 = 4. So digit 4 is our other special digit.
Next, add the month digit to the day digit: 2 + 4 = 6. If the answer had been more than 7 then I would have subtracted 7 again. (I like to work with small numbers because I can deal with them more easily mentally). Let us call this result 'the day+month total'.
Now we need a table so that I can find a digit to match the year (the year in the question is 1990) .
Here is a table of year values for you to find a year's digit:
| First Of Jan | Year Number | Day | Leap Year | |
| 01-01-2000 | 7 | Saturday | Leap Year [even though it ends in 00: it is divisible by 400] | |
| 01-01-2001 | 2 | Monday | ||
| 01-01-2002 | 3 | Tuesday | ||
| 01-01-2003 | 4 | Wednesday | ||
| 01-01-2004 | 5 | Thursday | Leap Year | |
| 01-01-2005 | 7 | Saturday | ||
| 01-01-2006 | 1 | Sunday | ||
| 01-01-2007 | 2 | Monday | ||
| 01-01-2008 | 3 | Tuesday | Leap Year | |
| 01-01-2009 | 5 | Thursday | ||
| 01-01-2010 | 6 | Friday | ||
| 01-01-2011 | 7 | Saturday | ||
| 01-01-2012 | 1 | Sunday | Leap Year | |
| 01-01-2013 | 3 | Tuesday | ||
| 01-01-2014 | 4 | Wednesday | ||
| 01-01-2015 | 5 | Thursday | ||
| 01-01-2016 | 6 | Friday | Leap Year | |
| 01-01-2017 | 1 | Sunday | ||
| 01-01-2018 | 2 | Monday | ||
| 01-01-2019 | 3 | Tuesday | ||
| 01-01-2020 | 4 | Wednesday | Leap Year | |
| 01-01-2021 | 6 | Friday | ||
| 01-01-2022 | 7 | Saturday | ||
| 01-01-2023 | 1 | Sunday | ||
| 01-01-2024 | 2 | Monday | Leap Year | |
| 01-01-2025 | 4 | Wednesday | ||
| 01-01-2026 | 5 | Thursday | ||
| 01-01-2027 | 6 | Friday | ||
| 01-01-2028 | 7 | Saturday | Leap Year | |
| 01-01-2029 | 2 | Monday | ||
| 01-01-2030 | 3 | Tuesday | ||
| 01-01-2031 | 4 | Wednesday | ||
| 01-01-2032 | 5 | Thursday | Leap Year | |
| 01-01-2033 | 7 | Saturday | ||
| 01-01-2034 | 1 | Sunday | ||
| 01-01-2035 | 2 | Monday | ||
| 01-01-2036 | 3 | Tuesday | Leap Year | |
| 01-01-2037 | 5 | Thursday | ||
| 01-01-2038 | 6 | Friday | ||
| 01-01-2039 | 7 | Saturday | ||
| 01-01-2040 | 1 | Sunday | Leap Year | |
| 01-01-2041 | 3 | Tuesday | ||
| 01-01-2042 | 4 | Wednesday | ||
| 01-01-2043 | 5 | Thursday | ||
| 01-01-2044 | 6 | Friday | Leap Year | |
| 01-01-2045 | 1 | Sunday | ||
| 01-01-2046 | 2 | Monday | ||
| 01-01-2047 | 3 | Tuesday | ||
| 01-01-2048 | 4 | Wednesday | Leap Year | |
| 01-01-2049 | 6 | Friday | ||
| 01-01-2050 | 7 | Saturday | ||
| 01-01-2051 | 1 | Sunday | ||
| 01-01-2052 | 2 | Monday | Leap Year | |
| 01-01-2053 | 4 | Wednesday | ||
| 01-01-2054 | 5 | Thursday | ||
| 01-01-2055 | 6 | Friday | ||
| 01-01-2056 | 7 | Saturday | Leap Year | |
| 01-01-2057 | 2 | Monday | ||
| 01-01-2058 | 3 | Tuesday | ||
| 01-01-2059 | 4 | Wednesday | ||
| 01-01-2060 | 5 | Thursday | Leap Year | |
| 01-01-2061 | 7 | Saturday | ||
| 01-01-2062 | 1 | Sunday | ||
| 01-01-2063 | 2 | Monday | ||
| 01-01-2064 | 3 | Tuesday | Leap Year | |
| 01-01-2065 | 5 | Thursday | ||
| 01-01-2066 | 6 | Friday | ||
| 01-01-2067 | 7 | Saturday | ||
| 01-01-2068 | 1 | Sunday | Leap Year | |
| 01-01-2069 | 3 | Tuesday | ||
| 01-01-2070 | 4 | Wednesday | ||
| 01-01-2071 | 5 | Thursday | ||
| 01-01-2072 | 6 | Friday | Leap Year | |
| 01-01-2073 | 1 | Sunday | ||
| 01-01-2074 | 2 | Monday | ||
| 01-01-2075 | 3 | Tuesday | ||
| 01-01-2076 | 4 | Wednesday | Leap Year | |
| 01-01-2077 | 6 | Friday | ||
| 01-01-2078 | 7 | Saturday | ||
| 01-01-2079 | 1 | Sunday | ||
| 01-01-2080 | 2 | Monday | Leap Year | |
| 01-01-2081 | 4 | Wednesday | ||
| 01-01-2082 | 5 | Thursday | ||
| 01-01-2083 | 6 | Friday | ||
| 01-01-2084 | 7 | Saturday | Leap Year | |
| 01-01-2085 | 2 | Monday | ||
| 01-01-2086 | 3 | Tuesday | ||
| 01-01-2087 | 4 | Wednesday | ||
| 01-01-2088 | 5 | Thursday | Leap Year | |
| 01-01-2089 | 7 | Saturday | ||
| 01-01-2090 | 1 | Sunday | ||
| 01-01-2091 | 2 | Monday | ||
| 01-01-2092 | 3 | Tuesday | Leap Year | |
| 01-01-2093 | 5 | Thursday | ||
| 01-01-2094 | 6 | Friday | ||
| 01-01-2095 | 7 | Saturday | ||
| 01-01-2096 | 1 | Sunday | Leap Year | |
| 01-01-2097 | 3 | Tuesday | ||
| 01-01-2098 | 4 | Wednesday | ||
| 01-01-2099 | 5 | Thursday |
1990 is not in that table. However, its value can be worked out by finding the year 90 in the table above and then:
Firstly: Take the digit of the 2090 year row: 1 ;
Secondly: For a 1900s digit value, you need to add 1. Well, 1 + 1 = 2. So, 1990 is a 2 digit.
Here is another example: For 1994, you look up year 94 of the 2000s table: value 6. You then add 1 to get a 1990s value. 6+1 = 7. So 1994 is a 7 value.
However, sometimes the value of 7 is gone over and then you need to subtract 7. Here is an example: year 1995 needs to be worked out from year 2095 which is 7; then, I add 1 to 7 to get 8. But 8 is over my 'one to seven' number range; therefore, I subtract 7 to get 1 (8-7=1) so that I have a value inside my 'one to seven' number range.
Anyway, we discovered that 1990 uses a special 2 digit. We add that 2 to the month+day total from earlier: we add 2 to the 6 and get 8.
Finally, I consult a table which tells me what day of the week the digit 8 signifies: a Sunday:
| Sum of Digits | Day of Week |
| 2 | Monday |
| 3 | Tuesday |
| 4 | Wednesday |
| 5 | Thursday |
| 6 | Friday |
| 7 | Saturday |
| 8 | Sunday |
| 9 | Monday |
| 10 | Tuesday |
| 11 | Wednesday |
| 12 | Thursday |
| 13 | Friday |
| 14 | Saturday |
| 15 | Sunday |
The year table method provides digits from 1 to 7 and the day/month table method provides digits from 1 to 7; so, at most, their total is 7 + 7 = 14 . For example, the 24th December 2005 involved me finding the digit 7 for the year 2005 and the digit 7 for the '24th December'. I see that 14 equates to a Saturday.
So why does the 'sum of digits' table go up to 15 ? Because, on a leap year, any date from the 1st March onwards needs a 1 digit added to it in order for the answer to come out correctly. And, if you are asked for the 29th February 2000 then you need to pretend that you have been asked for the 1st March on a non-leap year: the answer will be correct in that way.
29th Feb 2000 is like the 1st March for that year but WITHOUT adding the leap year 1 digit:
The answer should be Tuesday for 29th February 2000. In order to arrive at that answer, let us look up the digit that belongs to "1st March" [the day after is looked at when we handle this exceptional date] : 1st is simply digit 1. And the month of March has a 2 special digit. The day+month total is: 1 + 2 = 3. It is NOT more than 7 . So it is fine and needs no extra subtraction step done to it.
The 3 is added to the 1/1/2000 digit which is a 7 digit:
3 + 7 = 10 .
In the sum of digits table, a 10 is a Tuesday.
In fact, different centuries have their different offsets just like the 1900s need a '+1' effect.
1700s: add 5. Note that different countries adopted the Gregorian calendar in different years. So, for that reason, there is a cut-off point for what year the system starts in. eg. British Empire in 1752.
1800s: add 3
1900s: add 1
2100s: add 5
Memorising the table of year special digits
There are 100 years in the table you saw above. I wanted to make a nice way to remember them.
There is a list of 100 'math people' in this course. They are drawn people with 7 possible colours of the math people's faces [see math articles]. Not by coincidence, each colour matches a century's digit. Person 00, for example, has a face colour that is for the year 2000 .
Also, in the calendar math history system of numbers 000-999, numbers 001-100 have people with surnames where the 'acrostic digit' of the start of the surname matches the century table. Person 100 is Will Self. The 00 special digit therefore is like the 'S' of 'Self. In the acrostic digits lesson and 'numbers in images' lesson, we saw that S = 7. So, a calendar date involving the year 2000 uses special digit 7.
Further Reading
Searching the internet, you will find similar ways of finding the day of a year. By memorising all 366 'day+month' results, you could become a little faster. But that is a lot of work. There are other mathematical strategies for finding the day of the year. Look in the 'Art of Memory' Yahoo group for Late June 2008 to read about Gauss.
