Which week day(s) cannot be the first day of a century?

Solution 1:

$400$ years together have $400\cdot365+97=146\,097$ days, which is divisible by $7$. Therefore everything repeats after $400$ years. According to Wolfram Alpha January $01$ of the years $2000$, $2100$, $2200$, and $2300$ are a Saturday, Friday, Wednesday, and Monday, respectively. It follows that no century begins with a Tuesday, Thursday, or Sunday.

Letting the centuries begin on January $01$ of year $100k+1$, the missing days would be Wednesday, Friday, and Sunday.

Solution 2:

In four hundred years there $97$ leap years and $303$ non leap years. This contributes a total of $97\cdot 2+303=497$ 'free days'. This is a multiple of $7$, which means that the weekday cycle repeats every $400$ years. Thus only four weekdays (out of seven) will ever be first days of a century.

Solution 3:

The leap year formula is not perfect. More precisely a year is 365 days, 5 hours, 48 minutes and 46 seconds long. On the average year of the Gregorian calendar, we fall behind 27 seconds. So every 3236 years or so, we will have to skip a leap year to remain on target. Due to this, every day will be possible to start a century.

History has shown on multiple occasions that we will tweak our calendar to maintain the familiar seasons. It's also easier to change one day than to shift the equinoxes, solstices, crop planting dates, school opening/closings, etc. So even though this is a hypothetical situation, I 100% guarantee the leap years will be adjusted before July is allowed to become chilly on the Northern Hemisphere!

Solution 4:

Perhaps you'd like to know a bit more about the mathematics used. This seems pretty good.

The book, CRC Standard Mathmematical Tables and Formulae gives the day of the week $W$ as a number: $0$ is Sunday and $6$ is Saturday,

$$(1) \quad W \equiv d+\lfloor2.6 \cdot m-0.2 \rfloor +Y+\lfloor Y/4 \rfloor +\lfloor C/4 \rfloor-2 \cdot C \mod 7$$

Where,

$W$ is the day of the week

$d$ is the day of the month ($1$ to $31$)

$m$ is the month where January and February are treated as months of the preceding year:

March $\Rightarrow 1$, April $\Rightarrow 2$, $\cdots$, December $\Rightarrow 10$, January $\Rightarrow 11$, February $\Rightarrow 12$.

$C$ is the century minus one ($1997$ has $C=19$ while $2025$ has $C=20$).

$Y$ is the year ($1997$ has $Y=97$ except $Y=96$ for January and February).

This formula looks quite intimidating, but if you practice, memorize a bit, and modify the method a bit, you can calculate dates in real-time! $^1$

To Answer the Question:

Let's apply $(1)$ to answer your question,

Which week day cannot be the first day of a century?

We set the variables for January 1 using $m=11$, $d=1$, $Y=99$, and $C$ is kept as a variable.

Substituting into $(1)$ and simplifying, we get,

$$W \equiv 153+\lfloor C/4 \rfloor -2 \cdot C \mod 7$$

The only values that $W$ can be are $5,3,1,6$: Friday, Wednesday, Monday or Saturday.

Nothing fancy here needs to be done to prove this, just note that the floor function goes up one for every four increase in $C$. Then note that the term $2 \cdot C$ goes up eight for every four increase in $C$. Notice that this results in a decrease of $7$, but '$7 \mod 7$' is zero. Therefore, there are only four unique values of $W$.

$^1$ I can attest to its usefulness. I use the method daily when I'm planning things in my Calendar. It's also a novel way to ask about someone's birthday.