Determine the day of the week on which you were born

[Math100]

Homework Statement

Determine the day of the week on which you were born.

Relevant Equations

None.

The date with month $m$, day $d$, year $Y=100c+y$ where $c\geq 16$ and $0\leq y<100$, has weekday number
$w\equiv d+[(2.6)m-0.2]-2c+y+[\frac{c}{4}]+[\frac{y}{4}]\pmod {7}$ provided that March is taken as the first month of the year and January and February are assumed to be the eleventh and twelfth months of the previous year.
To determine the day of the week on which I was born, I will use my birthday as an example.
$w\equiv 31+[(2.6)(8)-0.2]-2c+y+[\frac{c}{4}]+[\frac{y}{4}]\pmod {7}$

 

[fresh_42]

What is the first day of the week?

 

[Math100]

What is the first day of the week?

I believe it's Sunday.

 

[Math100]

I entered $c=19, y=95$ into the equation above but I am not getting the correct answer.

 

[fresh_42]

I got two correct answers for dates that I checked if Sunday counts as number zero (or seven).

Let's see what happens with today (9/25/2022):

\begin{align*}
w&\equiv d+[(2.6)m-0.2]-2c+y+[\frac{c}{4}]+[\frac{y}{4}]\pmod {7} \\
&=25+[2.6 \cdot 7 -0.2]-2\cdot 20 + 22+[20/4] +[22/4]\pmod{7}\\
&=25+18-40+22+5+5=35\equiv 0\pmod{7}
\end{align*}

I haven't seen that formula before. But three out of the three tests I made produced the correct result.

We had to prove by induction that the 13th of a month is more often a Friday than any other weekday.

 

[Math100]

I got two correct answers for dates that I checked if Sunday counts as number zero (or seven).

Let's see what happens with today (9/25/2022):

\begin{align*}
w&\equiv d+[(2.6)m-0.2]-2c+y+[\frac{c}{4}]+[\frac{y}{4}]\pmod {7} \\
&=25+[2.6 \cdot 7 -0.2]-2\cdot 20 + 22+[20/4] +[22/4]\pmod{7}\\
&=25+18-40+22+5+5=35\equiv 0\pmod{7}
\end{align*}

I haven't seen that formula before. But three out of the three tests I made produced the correct result.

We had to prove by induction that the 13th of a month is more often a Friday than any other weekday.

I got $137.1$, but isn't that equal to $137.1\equiv 4\pmod {7}$? Given the fact that my birthday is on October 31st, 1995.

 

[fresh_42]

Let's see:
\begin{align*}
w&\equiv d+[(2.6)m-0.2]-2c+y+[\frac{c}{4}]+[\frac{y}{4}]\pmod {7} \\
&=31+[2.6 \cdot 8 -0.2]-2\cdot 19 + 95+[19/4] +[95/4]\pmod{7}\\
&=31+[20.6]-38+95+4+3=115\equiv 3\pmod{7}
\end{align*}

You were born on a Wednesday.

 

[Math100]

Let's see:
\begin{align*}
w&\equiv d+[(2.6)m-0.2]-2c+y+[\frac{c}{4}]+[\frac{y}{4}]\pmod {7} \\
&=31+[2.6 \cdot 8 -0.2]-2\cdot 19 + 95+[19/4] +[95/4]\pmod{7}\\
&=31+[20.6]-38+95+4+3=115\equiv 3\pmod{7}
\end{align*}

You were born on a Wednesday.

But why did my Korean hospital says Tuesday then? That's weird.

 

[fresh_42]

But why did my Korean hospital says Tuesday then? That's weird.

No, that's true. Wednesday is wrong. I made a mistake. I calculated $[95/4]$ as the remainder, not the integer part.

Correction:
\begin{align*}
w&\equiv d+[(2.6)m-0.2]-2c+y+[\frac{c}{4}]+[\frac{y}{4}]\pmod {7} \\
&=31+[2.6 \cdot 8 -0.2]-2\cdot 19 + 95+[19/4] +[95/4]\pmod{7}\\
&=31+[20.6]-38+95+4+23=135\equiv 2\pmod{7}
\end{align*}
... and $2$ is Tuesday.

 

[pbuk]

Let's see what happens with today (9/25/2022):

I can understand why an American would use that ridiculous way of writing a date, but surely you know better ?

 

[fresh_42]

I can understand why an American would use that ridiculous way of writing a date, but surely you know better ?

I would have written it 200220925175845, but communication means understanding the set-up of your dialogue partner.

 

[pbuk]

Is this intended to be a restatement of Zeller's congruence? Note that the roundings should be floors e.g. $\left [ \frac y 4 \right ]$ should be $\lfloor \frac y 4 \rfloor$.

 

[fresh_42]

Is this intended to be a restatement of Zeller's congruence? Note that the roundings should be floors e.g. $\left [ \frac y 4 \right ]$ should be $\lfloor \frac y 4 \rfloor$.

My textbook defines $[x]=\lfloor x\rfloor$ so it doesn't make a difference.

 

[Buzz Bloom]

Pick your year.
https://www.timeanddate.com/calendar/?year=2022&country=1

 

[WWGD]

I'd think we need to know this is the Gregorian Calendar and not Chinese, or otherwise, which may not be 365.25 days long. If its Gregorian, you can use that 365=52.7+1, 366=52.7+2. Then find the number of leap years and "regular" ones in-between.

 

[fresh_42]

I'd think we need to know this is the Gregorian Calendar and not Chinese, or otherwise, which may not be 365.25 days long. If its Gregorian, you can use that 365=52.7+1, 366=52.7+2. Then find the number of leap years and "regular" ones in-between.

It's not exactly 365.25. It gets complicated every 100 (no leap) and 400 (leap) years IIRC.

 

[WWGD]

It's not exactly 365.25. It gets complicated every 100 (no leap) and 400 (leap) years IIRC.

I guess we use simplified assumptions. Hey, if physicists can approximate a basketball player with a cylinder, why not?

 

[fresh_42]

I guess we use simplified assumptions. Hey, if physicists can approximate a basketball player with a cylinder, why not?

Where is that d*** cylinder?

Uh! Found it!

 

[pbuk]

It's not exactly 365.25. It gets complicated every 100 (no leap) and 400 (leap) years IIRC.

I guess we use simplified assumptions.

No, the $\left \lfloor \frac c 4 - 2c \right \rfloor$ term (where as in the OP $c$ is the century number i.e. year mod 100) takes care of centuries. As I said above this is Zeller's congruence.