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My Puzzle # 3 -- rmac22

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My Puzzle # 3

 

Considering the calendar system we now use, neglecting the corrections that have to be done every century, or even less often, but including leap years, how many unique calendars are there not considering the year? 

 

That is, remove the year from a calendar, and in the course of time there will be another year with exactly the same arrangement of days.  You could put the different year on it and it would be exactly right.

 

You can solve how many unique calendars there are with pure reasoning, by working through it year by year, or some combination of both.    

 

Related to that, what is the number of years it takes to use them all if you start counting the first year after a leap year?  That too you can solve with pure reasoning, etc.

 

This may be entirely too easy for a bunch as intelligent as you all, but we shall see. 

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Quoting funtimes0007:

So my answer for your part is correct, I said 14.

 

I didn't attempt part 2.

14 is correct, no one last time attempted to sort out the 28 years but some stated it.  By hand year by year it is a pain -- 28 steps and easy to make a mistake.  Easier with math.  I checked all this against a number of calendars I have.  Just to make sure.  No I don't have 28, but enough to check the progressions.  

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Solution Unique Calendars

 

The two things that complexly define a calendar are the day of the week January 1st happens and if it is or is not a leap year.  So, there are seven unique leap year calendars and seven unique non leap year calendars -- fourteen (14) in all. 

 

How many years will it take to use them all?  Well to cover seven leap years will, take 28 years as a minimum starting with the first year after a leap year.   A minimum because the shifting of days year to year may leave some out.   

 

So number the day of the week as usual Sunday being 1, Monday being 2 and so on.  Say the first leap year in the 28 years has January 1st on Sunday  or day number one.  Now the next leap year January will be on day six, 2 days shift from the first new year and one each from the three years between them.  Thus the sequence is as follows

 

1,  6,  11 or 4,  9 or 2,  7,  12 or 5,   10 or 3  and we are done.  Rewriting that we have

 

1,  6,  4,  2,  7,  5,  3 all seven numbers are used so we are done.  28 years is adequate to cover and use all the unique leap years. 

 

We can do the same thing for the 3 years in between proving the unique non leap year calendars will be covered and used 3 times 7 covering the 21 non leap years. 

 

 

Done

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Quoting funtimes0007:

:) I posted this riddle some long time ago on MM site. You have a good memory if you still remember the answers.

I have added Tarzan as an added twist. Lets see who can answer the whole riddle. I have answered the first one.

 

@ funtimes

 

re the calendar puzzle -- let me go ahead and post my solution.  Clumsy fingers wiped out almost all of it this morning.  So, I will have to redo.  No promises, but I should have it done in a few days, maybe tomorrow. 

 

 

re the Volkswagen, Mercedes and elephants puzzle -- since what I think I remember has nothing to do with the blogs I will concede I don’t know the answers.  I will happily anticipate finding out what those are.  I like my clues nonetheless.     

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Quoting funtimes0007:

You need to solve my 4 and 8 elephant problem on AnnStacy blog :)) No one is taking a shot at it.

@ funtimes 

 

I would give two clues since I know the answers.  Or at least those I remember from when I was last exposed to them.  Who knows where they went from there.  

 

1.  The elephant riddles completely ignore physical realities as well as logic.

 

 

2.  The thing that made them funny was the fact that they were ridiculous. 

 

Perhaps everyone else knows the answers too and like me is refraining from posting them.

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Quoting funtimes0007:

Let me see if I can find the time to solve it. AnnStacy took all my free time.

Thanks

Good, I was about to write up the solution and post it.  
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@ funtimes  

 

If you know how do so. 

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Last go around Dak solved this in ome day.  Perhaps it is too easy.