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.

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    $\begingroup$ People don't agree about the first day of a century, e.g., is it January 01, 2000, or January 01, 2001? $\endgroup$ Sep 21 '15 at 11:05
  • $\begingroup$ Which century @ChristianBlatter? $\endgroup$ Sep 21 '15 at 11:06
  • $\begingroup$ Does this even matter? e.g. 1st Jan 2000-31st Dec 2099 is 25 leap years; 1st Jan 2100-31st Dec 2199 is 24 leap years; 1st Jan 2200-31st Dec 2299 is 24 leap years; 1st Jan 2300-31st Dec 2399 is 24 leap years. But 1st Jan 2001-31st Dec 2100 is 24 leap years; 1st Jan 2101-31st Dec 2200 is 24 leap years; 1st Jan 2201-31st Dec 2300 is 24 leap years; 1st Jan 2301-31st Dec 2400 is 25 leap years... $\endgroup$
    – sjb
    Sep 21 '15 at 11:10
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    $\begingroup$ @njzk2: sure, I know when tidy people think it started. Such people would rather eat their own face off than have a first century that's only 99 years long. I'm just pointing out that opportunities exist to wind such people up by choosing different definitions. $\endgroup$ Sep 21 '15 at 17:47
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    $\begingroup$ @njzk2: if anything, mathematicians should be happy to switch definitions and answer both cases, rather than referring to a dead Pope for the meanings of words ;-) Christian's correct, especially if you also consider it relevant when the millennium was celebrated around the world, that there's no consensus definition what the word means in general use. There should be a consensus, since that Wiki page is obviously correct, but that's beside the point. $\endgroup$ Sep 21 '15 at 17:51

$400$ 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.

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    $\begingroup$ So the correct answer to the question ("Which week day" where 'day' is singular) is surely Sunday, as that cannot be the first day of a century by either definition. $\endgroup$
    – abligh
    Sep 21 '15 at 20:50
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    $\begingroup$ @abligh: Indeed. Wikipedia's article on Sunday says in its introduction "No century in the Gregorian calendar starts on a Sunday, whether its first year is '00 or '01" $\endgroup$
    – Henry
    Sep 22 '15 at 6:46

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.

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    $\begingroup$ Why $4$ and which $4$? $\endgroup$ Sep 21 '15 at 10:54
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    $\begingroup$ +1 for craftily giving a solution where you avoid defining the 1st year of a century (Jan 1,2000 or Jan 1 2001?). And also for a correct solution! @AdityaAgarwal: The point here is that you should not study one century at a time. A consequence of this is that all you need to do is to figure out the first weekdays of the 21st, 22nd,23rd and 24th centuries. Because the calendar repeats every 400 years, the first day of the 20th century was on the same day of the week as the first day of the 24th century. Same with 21st and 25th and so on. $\endgroup$ Sep 21 '15 at 11:39
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    $\begingroup$ Your argument only proves "at most four", not "only four". $\endgroup$ Sep 21 '15 at 12:46
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    $\begingroup$ @njzk2: I beg to disagree. IIRC for example New York Times wished its readers a happy 20th century on Jan 1st, 1901. Yet some of my misguided fellow citizens thought that the 21st century began on Jan 1st, 2000 as opposed to Jan 1st, 2001. Unless you are willing to pick a side in that argument you should not name the 4 weekdays a century can begin with :-) $\endgroup$ Sep 21 '15 at 17:05
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    $\begingroup$ @ChristianBlatter: Good point. The centuries have numbers of days congruent to 5, 5, 5, 6 (mod 7). No combination of less than all of these gives a multiple of 7, and so the centuries will in fact start on four different days. $\endgroup$
    – paw88789
    Sep 21 '15 at 19:52

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!

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    $\begingroup$ This doesn't answer the question, which is about leap years in the Gregorian calendar, not some hypothetical calendar that nobody uses. Also, your analysis seems to ignore that years divisible by 100 are only leap years if they're also divisible by 400 (i.e., that we use the Gregorian calendar, not the Julian calendar). $\endgroup$ Sep 21 '15 at 17:05
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    $\begingroup$ Thanks for pointing out the difference between the calendars. I updated my answer to reflect that... agree to disagree on its relevance... $\endgroup$ Sep 21 '15 at 17:41
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    $\begingroup$ If the suggestion here is that every 4000 years should be another leap year, then the first day of the century that is a Sunday will be in the year 4300. $\endgroup$
    – Abel
    Sep 22 '15 at 12:32

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,

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


$W$ 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.


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.


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