Natural Blogarithms

I heard someone mention the fact that this Friday is the 13th and it reminded me of a blog entry I was intending to post. It was a little puzzle provided and answered by Eon.

What is the most number of Friday the 13th’s that a single year may have?

Clearly, its less than 13, which happened to be the first answer offered by a student this morning in Calculus IV. But how do you find the answer without a brute force search through the next several decades. Consider a sequence of values modulo 7.

Jan - 0 ; Jul - 6
Feb - 3 ; Aug - 2
Mar - 3 ; Sep - 5
Apr - 6 ; Oct - 0
May - 1 ; Nov - 3
Jun - 4 ; Dec - 5

This tells you the number a days displacement you have from one month to the next. For example, if the 13th falls on a Tuesday in January, that will be displaced by 3 in February, namely it will fall on Friday in February and in March it is still displaced by 3 so it is still on Friday, then in April it is displaced by 6 so it is on Monday, etc. From this list you see that the most occuring displacement is 3 and it occurs three times. This means that if the 13th falls on a Tuesday in January, then we will have a Friday the 13th in February, March and November. The most number of Friday the 13th’s in a year is 3. Coincidently, that occurs next in 2009.

All of the above is for non-leap years, so how many are possible in a leap year? Well the sequence for a leap year is 0, 3, 4, 0, 2, 5, 0, 3, 6, 1, 4, 6. So, it still has a maximum number of three. (displacement 0). So if Friday the 13th occurs in January of a leap year, it will also occur in April and July. This occurs next in 2012 but not again until 2040.

How about that?

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