What is the minimum number days between one Friday the $13$th and the next Friday the $13$th?(Assume that the year is a leap year)
What is the minimum number days between one Friday the $13$th and the next Friday the $13$th?
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I typed "shortest time between friday 13" into Google and got several useful hits.
I followed a link, and came across this page. Check out PART B.
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Key concept used: Calendar concepts specifically number of days between dates on the basis of odd days as remainder after dividing a number of days by $7$. If number of odd days is $0$ or a multiple of $7$, after the intervening number of days, we get the same day (Monday or Tuesday etc).
Answer: Between $13$th to next $13$th the minimum number of days is $28$ and being multiple of $7$, that would also be a Friday if the difference between the two dates were $28$. This would have been the case only if the starting month were February and the year not a leap year. But the year being a leap year, from February $13$th to next month $13$th difference is $29$ days, one day extra and so February starting point does not give us any advantage towards the minimum number of days. We need to adopt a more general approach.
We know each $30$ day month gives $2$ days extra after dividing by $7$ and $31$ day month gives $3$ days extra while the $29$ day February gives $1$ day extra. These extra number of days are called odd days and is the most important concept in calendar sums as this is directly related to weekdays. For example, if odd days between two dates is $3$ days and the starting day was a Sunday, the ending date would be Sunday $+ 3$ days, that is, a Wednesday. Thus we need to collect the odd days of each month and go on dividing the sum of odd days by $7$ again to always know the resultant number of odd days.
In this problem as February failed as the starting month, in a general approach, we need to collect number of days of each intervening month from the starting month to detect when the sum of odd days becomes $7$ with minimum number of intervening months. In other words, the problem is transformed to breaking up $7$ into a sum of minimum number of integers that are the odd days of the intervening months.
With known values of odd days of months, this can only be, $7 = 2 + 3 + 2$, that is, starting month a $30$ day month and next two months a $31$ day month and a $30$ day month. Thus to have the earliest next $13$th the Friday from the starting at $13$th the Friday, the number of intervening days would be,
$$30 + 31 + 30 = 91~.$$
Modulo $7$, the number of days in the months of a leap year are $$ 3, 1, 3, 2, 3, 2, 3, 3, 2, 3, 2, 3. $$ You want to find the shortest subsequences (made of consecutive numbers) for which the sum is divisible by $7$.
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