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16. Finding the Day of the Week, given its date

The seventh printing of the book  Mathematics for Practical Use by Kaj L. Neilsen, Barnes and Nobles,( c) 1962, on page includes the following formula 

to calculate a whole number  N. 

In this formula, 

• d between 1 and 31 is the day of the month, 
• m between 3 and 14 is the number of the month in the year, with January and February being included as the  13th and 14th months of the previous year, and
• Y is the number of the year in the Western Calendar.

and the square brackets in this formula take the greatest integer or whole number part of a number. For example, 

         [141.85] = 141   - for unsigned or positive numbers drop the decimal or fractional part if present.

         [34¼] = 34   - ditto
         [3.1416] = 3 -  ditto

         [1453]  = 1453  -  there is nothing to drop.

Neilsen's Book on page 207 says that the remainder on the division by N by 7 gives the day of week, and then shows the day of the week for the date January 1st, 1961 is a Sunday.  There-in lies the reason for calculating N. 

We will consider a few further examples and explain the use and parts of the formula in greater detail.

Now the remainder on division of the whole number N by  7 will be a number in the set 

{0, 1, 2, 3, 4, 5, 6}

with the day N being a Sunday if the remainder is 1.

Example:  Use the formula to find the day of the week is July 9, 2010? 

Solution:    We will use

d = 9 is the day of the month, 
 m = 7 since July is the seventh month of the year, and
 Y = 2010 is the number of the year.

to calculate 

Now with the aid of my calculator:

Therefore 

=  9 + 14 + 4 + 2010 +502 - 20 +5 + 2

= 2526 

Now my calculator gives   N  ÷ 7  =  360.857 with 360 being the greatest integer part. Now   The remainder on division of this N by 7 is

N -  360 × 7 = 2526 - 2520 = 6

So from the above table,  July 9th, 2010 (today's date as I write) should be a Friday, and it is. 

Exercises: 

1. Use the above formula for N to find the day of week of   January 15, 2012. Hint: d = 15, m =13 and Y =2012-1 because January and February are taken to be part of the previous year. .
2. Use the above formula for N to find the day of week of   February 29, 2016. Hint: d = 29, m =14 and Y =2016-1 because January and February are taken to be part of the previous year. .
3. Use the above formula for N to find the day of week of   March 10, 2025. Hint: d= 10,  m =3 and Y =2025.
4. Calculate the day of the week on which you were born.
5. Calculate the day of the week from today's date.

Going Further  

 The same results as above would follow from calculating

in place of 

Later, if you obtain and algebraic understanding of calculations with remainders, you can explain why. For that note the remainder on division of 30 by 7 is 2. The remainder on division of 365 by 7 is 1. 

In the present Occidental (Gregorian) calendar, in use since 1582 (??) for several centuries,  each year has 365 days, except for leap years which have one extra day. 

The term 365Y in M in principle counts the number of days since the formal year Y = 0 in the current occidental calendar. 

counts the number of leap years since the formal Y = 0 in the current calendar.  This expression is also equal to the number extra days that each leap year add since the formal year Y = 0.  So the expression 

counts the number of days from the formal year  0 in the current calendar to the current year Y.

Going Further II

By playing with part of the N formula, that is, through numerical experimentation, I filled in the above table.

Column II lists the days in each month.  Column III lists the number of days to the end of the current month  since the start of the previous February. By counting from the 1st of March, the extra day in any leap year, will become day 366 and be the last day numbered. 

Column (IV) list the month number. Here January and February are taken to be the 13th and 14th of the previous year to make the calculation simpler.  Numerical experimentation led to the following.  Namely, if q = m-3 then the column (VIII) expression 

gives the values in Column (III), shifted vertically by one.  The foregoing yield the following.

Day d of month  m in year Y (where 3 < m < 14) is the 

day after the last day of February in Year Y when q = m-3.  

Examples

Let's us do a few examples to check the foregoing.

• For March 1st,  d =1, m = 3 and q =3.  That gives

  d+ 30q

  +

  [

  3(q+1) 5

  ]

  = 1+ 0 =1 as expected

  ---

• For December 31st,  d =31, m = 12 and q = 9 and so

  d+ 30q

  +

  [

  3(q+1) 5

  ]

  = 31 + 276 = 307 in agreement with column (III)

  ---

• Twenty day later, gives January 20th. For it, d =20, m = 13 and q = 10 and so

  d+ 30q

  +

  [

  3(q+1) 5

  ]

  = 20 + 306 = 327 which 20 more  than the Dec 1st value 307

  ---

• For February 28th,  d =28, m = 14 and q = 11 and so

  d+ 30q

  +

  [

  3(q+1) 5

  ]

  = 28 + 337 = 365

  Thus Feb 28th is always 365 days after the end of the previous February.

  ---

• For February 29th, whenever it occurs,  d =29, m = 14 and q = 11 and so

  d+ 30q

  +

  [

  3(q+1) 5

  ]

  = 29 + 337 = 366

  Thus Feb 28th is 366 days after the end of the previous February.

Counting the Days since the last day of February, 2000. 

The formula 

gives the number of days since the end of February in the formal year 0 the Gregorian Calendar 
when 

• d = the day of the currant month,
• q = m -3 where m = the number of the current month (3 < m < 14) with January and February be counted as the 13th and 14th month of the previous year, and
• Y = the current year for month m, with the convention that the current year for January and February is the last calendar year.

Note in the foregoing, if we take  Y to be the {current year - 2000} then the above formula gives the number of days since Friday, February 29, 2000.   That will provide smaller numbers.

Example:  For Friday, July 9th, 2010, 

• d = 9
• m = 7 and q = m - 3 = 4
• Y = 2010 -2000 = 10.

That gives the following number of days since the start of March, 2000:

=  9 + 4 (30) +  [12/5] + 365 (10) + [10/4] - [10/100] + [10/400]

= 9 + 120  +[2.2]+ 3650 +[2.5] - 0 + 0

= 9 + 120  + 2 + 3650 + 2   (since square bracket indicate 
                      greatest integer part)
 = 3783

Extra: In the calculation of the number of days since March 29th, 2000, the remainder after the division of M+3 by 7 gives the day of the week if we use the following table

Here  M+ 6 =3783 +3 = 3786    My calculator gives 3786 divided by 7 is 540.857 (approx). The greatest integer part of the latter is 529.  So the

  remainder = 3786 - 540 × 7 = 3786 - 2780 = 6.

That makes  July 9th, 2010 a Friday (today at the time of writing). And that is correct..

Question:  How many calendar days are there in a project which begins on July 9th, 2010 and if it ends on June 15th, 2011? 

Solution:

For July 9th, 2010, the above calculation gives M = 3783.

For June 15th, 2011, we have 

• d = 15
• m = 6 and q = m - 3 = 3
• Y = 2011 -2000 = 11

That gives the following number of days since the start of March, 2000:

 =  15 + 30 × 3  +  [12/5] + 365 (11) + [11/4] - [11/100] + [11/400]

   = 15 + 90  +[2.2]+ 4015 +[2 ¾] - 0 + 0

     = 15 + 90  + 2 + 4015 + 2   (since square bracket indicate 
                      greatest integer part)
    =  4124

Now  the number of calendar days in the project, including the start and end dates,  is  

     4124 - 3783 + 1  =  342

End Notes

The last formulas for M followed my playing with the formula for N in the Neilsen's book.  If I have made a mistake, I hope that some reader will catch it and tell me.  That being said, before or besides starting this adventure, I should searched on the world wide web (or in a library) for origins of formulas for counting days in the year.  That might have shortened the adventure above and provided a more scholarly approach.

The webpage http://www.convertunits.com/dates/ provides some background (briefly) and provides an online option to calculate the number of days between two dates - excluding the end date.  It also gives the number of hours, minutes and seconds. For the example above, that webpage gives 341 without the end date and 342 with.  

To learn more, see search on number the days in the year or years

Search on   Bing,  Search on Google or another internet search engine. 

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