Frequently Asked Questions

1: Could you possibly give me a formula to calculate the weekday, which I can put into my HP calculator?

In order to calculate the weekday using a standard calculator try the following method:

To calculate the weekday for 21st May 2006 (or any date on or between 1/3/1900 and 28/2/2100)

d = 21
m = 5
y = 2006

∴ f(m) = f(5) = 1 (from table number 3)

The formula is:

$\frac{1.25\times y+f\left( m \right)+d-1}{7}$

$=\frac{1.25\times 2006+1+21-1}{7}$

$=\frac{\text{2528}.\text{5}}{7}$

=361.21428571428571428571428571429

Now subtract 361 from the above number (as it is the integer part coming before the decimal point) and then multiply by 7 and round it off to two decimal places.

You will get 1.50

Ignore the decimal part and only retain the integer part to get 1 which corresponds to Sunday ( 0 = Sat, 1 = Sun, 2 = Mon, ..)


2: What are these Heap years, Seap years, etc. What is the significance of its classification?

The names of these years have been specially coined by me. Like we know that the Olympic games are held in Leap years. That is in the years divisible by 4. But what about the other years. For instance, the FIFA (soccer) World Cup are held only in Seap years. Means the years which have remainder 2 when divided by 4. Similarly, the Cricket World Cup is held in Teap years with the exception of 1992 and 1996 World Cup.

Secondly, this classification is especially useful and imperative when we wish to find the years that satisfy a particular condition.

 
3: Is it true that everyone celebrate their 28th birthday on the same weekday on which they were born?

About 79 % of the people celebrate their 28th birthday on the weekday they were born. Refer chapter number 7 to know the 5-6-11-6 cycle. Which means that your birthday falls on the same weekday of your birth in the cycle of 5 years, 6 years, 11 years and 6 years and again 5 years.... depending upon the type of year of your birth. Therefore, one can infer that one's birthday repeats on the same weekday of birth after 28 years. This takes place if the years follow the pattern of 1 leap year after every three non-leap years. (One in every four years)

But this cycle is obstructed by the Heap year. (Heap years are those years which is divisible by 100 but not divisible by 400 E.g. 1800, 1900, 2100, 2200, etc.) A Heap year though looks like a Leap year because of its divisibility by 4, does not have 29 days in February. Therefore if a person is born in the year 1887 will not have his/her 28th birthday on the same weekday of his/her birth. Because he/she goes across through the Heap year where one day (29th February 1900) was missing thus disturbing the entire 5-6-11-6 cycle.

Thus any one born on or in between 29 Feb 1872 to 28 Feb 1900 will not have his/her 28th birthday on the same weekday.

However, one born on or in between 1 Mar 1900 to 28 Feb 2172 will have his/her 28th birthday on the same weekday. This includes every one of us. Thus, for practical purposes (for this present time) we all (about 100 %) have our 28th birthdays on the same weekday of our birth. This is because 2000 was not a Heap year but a normal Leap year.


4: How is the Babwani's Congruence useful in our everyday life?

We use calendars in our everyday life and thus Babwani's Congruence can be used to a great extent. One can orally calculate the weekday of any given date especially the one in the present year.

For example, if we are currently in 2007, we calculate in advance what is x\[\left\lfloor 1.25\times 2007 \right\rfloor =\left\lfloor \text{2508}\text{.75} \right\rfloor \,=\left\lfloor 2.75 \right\rfloor =2\text{(mod7)}\]

Now correlate the above mentioned formula as

$day=\left\lfloor 1.25\times y \right\rfloor +f\left( m \right)+d-1$

$=2+f\left( m \right)+d-1$

$=1+f\left( m \right)+d$

Now use this formula, remembering the table number 3 showing f(m), calculate in seconds the weekday.

E.g. 10th April 2007; day = 1 + 10 + 6 (mod 7) = 17 (mod 7) = 3 = Tuesday. (remember f(4) = 6)

26th September 2007; day = 1 + 26 + 5 (mod 7) = 32 (mod 7) = 4 = Wednesday. (remember f(9) = 5)

Secondly, consider a situation where the year 2007 is about to get over and you don't possess a calendar for the year 2008. So temporarily, till you get a new calendar of 2008; which calendar month of 2007 can you use instead of January 2008?

The answer is May 2007 can be used temporarily used. How? Let us see.

Let us generalize the current year (not necessarily 2007) as y and the next year as (y + 1).

The weekday of 1st January of the year (y + 1) = the weekday of 1st (unknown month) of the year y

$\left\lfloor 1.25\times \left( y+1 \right) \right\rfloor +1+f\left( 1 \right)-1=\left\lfloor 1.25\times y \right\rfloor +1+f\left( m \right)-1$

$\therefore \left\lfloor 1.25y \right\rfloor +\left\lfloor 1.25 \right\rfloor +1+0-1=\left\lfloor 1.25y \right\rfloor +1+f\left( m \right)-1$

$\therefore f\left( m \right)=\left\lfloor 1.25 \right\rfloor $

$\therefore f\left( m \right)=1$

but f(5) = 1

∴ the required month is May.

This is true irrespective of the nature (leap or non-leap) of the current or the upcoming year. Try it out.

Similarly, you can find many other personalized applications.


5: Which weekday cannot come at the end of the century?

There can be two cases to answer your questions.

CASE 1: Many (including me) consider 31st December xx00 to be the last day of the century (e.g. 31st December 2000) and the next day to be the first day of the next century.

CASE 2: While others consider 31st December xx99 to be the last day of the century (e.g. 31st December 1999) and the next day to be the first day of the next century.

You may directly refer the case you are interested in. However both are solved in the same manner, of course the answers will differ.

CASE 1:

Let us first find the weekday that can be the last day of the century (31st December xx00).

According to the congruence formula, substituting the values: d = 31, m = 12, c = c, y = 0

$w=\left\lfloor \frac{5y}{4} \right\rfloor +f\left( m \right)+d-2\times \bmod \left( c,4 \right)$

$=\left\lfloor \frac{5\times 0}{4} \right\rfloor +f\left( 12 \right)+31-2\times \bmod \left( c,4 \right)$

$=0+5+31-2\times \bmod \left( c,4 \right)$

$=36-2\times \bmod \left( c,4 \right)$

$=1-2\times \bmod \left( c,4 \right)\left( \bmod 7 \right)$

Now, be whatever the value of c, mod (c,4) can have only four values { 0, 1, 2, 3 )

w = 1 – 2 × 0 = 1 = Sunday (mod 7)

or w = 1 – 2 × 1 = – 1 = 6 = Friday (mod 7)

or w = 1 – 2 × 2 = – 3 = 4 = Wednesday (mod 7)

or w = 1 – 2 × 3 = – 5 = 2 = Monday (mod 7)

So except for the above weekdays, the other weekdays cannot be the last day of the century. So the answer is Tuesday, Thursday and Saturday cannot be the last day of any Gregorian century.

CASE 2:

Let us first find the weekday that can be the last day of the century (31st December xx99).

According to the congruence formula, substituting the values: d = 31, m = 12, c = c, y = 99

$w=\left\lfloor \frac{5y}{4} \right\rfloor +f\left( m \right)+d-2\times \bmod \left( c,4 \right)$

$=\left\lfloor \frac{5\times 99}{4} \right\rfloor +f\left( 12 \right)+31-2\times \bmod \left( c,4 \right)$

$=\left\lfloor 123.75 \right\rfloor +5+31-2\times \bmod \left( c,4 \right)$

$=123+5+31-2\times \bmod \left( c,4 \right)$

$=159-2\times \bmod \left( c,4 \right)$

$=5-2\times \bmod \left( c,4 \right)\left( \bmod 7 \right)$

Now, be whatever the value of c, mod (c,4) can have only four values { 0, 1, 2, 3 )

w = 5 – 2 × 0 = 5 = Thursday (mod 7)

or w = 5 – 2 × 1 = 3 = Tuesday (mod 7)

or w = 5 – 2 × 2 = 1 = Sunday (mod 7)

or w = 5 – 2 × 3 = – 1 = 6 = Friday (mod 7)

So except for the above weekdays, the other weekdays cannot be the last day of the century. So the answer is Monday, Wednesday, Saturday cannot be the last day of any Gregorian century.

In case your question refers to the Julian century, then there is no weekday that doesn’t come at the end of the century.