Identical calendars

Question

We can see, for example, that the years 2009 and 2015 have identical calendars. Similarly, 2000 and 2028.

I read once that given any year X at most 28 years later there will be another year Y, with the calendar identical to that of X.

Here I am referring to our usual calendar, the Gregorian.

I have not yet been able to prove such an assertion. I ask for help.

That's not surprising, because it's not true in general, though the next exception will be only 2072 (because 2100 will not be a leap year). — , Jan 08 '18 at 11:14
See also https://math.stackexchange.com/questions/33277/how-often-will-the-same-date-be-on-the-same-day-of-the-week — lhf, Jan 08 '18 at 12:02

score 1 · Answer 1 · answered Jan 08 '18 at 11:14

1

That's because:

There's a leap year in every $4$ years (the first exception to this rule will take place in the year $2\,100$).
Weeks have $7$ days.
The least common multiple of $7$ and $4$ is $28$.

answered Jan 08 '18 at 11:14

José Carlos Santos

440,053

A doubt:
Can we guarantee that the succession of years (X +1, X +2, ..., X + 28) presents some year with the calendar identical to that of X, even if the succession has some year that indicates end of century and is not leap (1700, 1800, 1900, 2100, etc.)?
– Paulo Argolo Jan 08 '18 at 11:34
@PauloArgolo I don't think so. – José Carlos Santos Jan 08 '18 at 11:37
I have not got a counterexample yet. – Paulo Argolo Jan 08 '18 at 11:40
Correction: Professor Vector, above, is right. The first year with a calendar identical to that of 2072 will be 2112 (coming 40 years later). – Paulo Argolo Jan 08 '18 at 11:52

score 1 · Answer 2 · answered Jan 08 '18 at 12:08

As José points out, if there were no irregularities in leap years, the calendar would repeat every 28 years, so there certainly will be another year with the same calendar within the next 28 (the 28th, if not before). In fact, it is only the leap years which take that long: if you started from a non-leap year, assuming no irregularities in leap years, you would get another year with the same calendar within $11$ years (it is $6$ or $11$ years later, depending on the year).

So the only years we need to worry about are when there is a skipped leap year coming up. If the year we start from, $X$, is not a multiple of $4$, then we are ok unless the skipped leap year occurs within the next $11$ years. But if it does, the $12$ years starting from $X$ contain $10$ years of $365$ days and $2$ of $366$. Since $10\times 365+2\times 366$ is a multiple of $7$, $X+12$ will have the same calendar as $X$.

If $X$ is a skipped leap year, then $X+6$ will have the same calendar.

If $X$ is a leap year, and one of the next two leap years is skipped, then again $X+12$ will start $10\times 365+2\times 366$ days later, and it will also be a leap year, so will have the same calendar as $X$. However, you can check that if a leap year in $X+12,...,X+28$ is skipped, the next year with the same calendar as $X$ will be more than $28$ years later.

So the only years which don't work are years ending 72, 76, 80, 84 or 88, where the next century year is not a leap year. This is $15$ failures in every $400$-year period.

@ Especially Lime: We can then conclude that given any year X, at most 40 then there will be a year Y with the calendar identical to that of X. Is not it true? — Paulo Argolo, Jan 08 '18 at 12:22
Correction: We can then conclude that given any year X, at most 40 years later there will be a year Y with the calendar identical to that of X. Is not it true? — Paulo Argolo, Jan 08 '18 at 12:24
@PauloArgolo Yes, that's true. If there is a skipped leap year coming up, then the next $40$ years will consist of $31$ normal years and $9$ leap years, which gives a multiple of $7$ days, so $X+40$ has the same calendar as $X$. — Especially Lime, Jan 08 '18 at 13:09

Identical calendars

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