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

Question

I think the question says everything. What I want is, a very short approach.
What I did: Lets call the day which is not the part of a whole week, a free day. So in a normal year, there is $1$ free day.
And in a leap year, there are $2$ free days.
So in $100$ years (From the $0^{\text{th}}$ year to the $99^{\text{th}}$ year), there will be
(i) $25$ leap years, if the century is divisible by $400$, else
(ii) $24$ leap years.
Lets deal with the first case, first.
First Case: In a century, there will be $25*2=50$ free days $+$ $75*1=75$ free days $\Large\begin{cases}\end{cases}=125$.
So we know that $125 \equiv 6(\text{mod} \ 7)$. So there will be $5$ free days in a century.
Now in $2$ centuries, there will be ($250 \equiv 5(\text{mod }7)$), $5$ free days.
In $3$ centuries, there will be $4$.
In $4$ centuries, $3$.
In $5$, $2$.
In $6$, $1$
And in $7$, $0$.
But I don't seem to understand what to do forward, please point me in the right direction.
Edit: Finally this question has got enough attention. Someone was asking for options here they come:
(i) Wednesday, Friday and Sunday
(ii) Wednesday, Friday and Saturday
(iii) Wednesday, Thursday and Sunday.
@ChristianBlatter Good Solution. But I didn't understand your last part. And please all, I urge you to give a fully "self" calculated result, because this a contest question, so it would obviously not be allowed there.

Answer 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.

Answer 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.

Answer 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!

Answer 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.

Answer 5

scubasteve623 is correct in that there will most likely be adjustments in the future because the calendar is not perfect, but the answer you're looking for is Tuesday, Thursday and Sunday.

Assuming those adjustments don't happen anytime soon...
1-1-2100 will be Friday
1-1-2200 will be Wednesday
1-1-2300 will be Monday
1-1-2400 will be Saturday
...and the cycle will repeat from 2500 onwards.