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Homework Help: Prove there is at least 1 friday the 13th between march and october

  1. Nov 11, 2007 #1
    Prove that in every year, there is at least 1 friday the 13th in the period from march to october.

    Well.. the only thing I can find about this is that it might use Zellers forumla:


    Apart from that I've no idea :) I can't seem to relate it to the problem. I've always failed at proofs. :p

  2. jcsd
  3. Nov 11, 2007 #2


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    Try looking at it this way. What day of the week does the First of the month fall on if that month has a Friday the Thirteenth? This is what you'll want to watch for. Now, put the First of March on Sunday. In that case, the 6th is on Friday, and a week later, so is the 13th. There's your Friday the Thirteenth for that case.

    Suppose March 1st is on Monday. March has 31 days, which is 4 weeks + 3 days. That means March 29th is on Monday, so the 31st is on Wednesday. The First of April would be on Thursday, which is shifted 3 days over from Monday (when March 1st was). So we really want to just look at how much each succeeding month exceeds 28 days and thus how much it will shift the First of the next month. April will shift dates by two days (30 = 28 + 2), so the First of May is on Saturday. May shifts dates by three days, so June 1st is on Tuesday; June shifts by two, putting July 1st on Thursday; and July shifts by three, making August 1st a Sunday. So there is a Friday the Thirteenth in August in this case.

    You should then try this with March 1st on each of the remaining five days of the week to find which month has its First on Sunday and therefore has a Friday the Thirteenth. It will work for all seven possibilities, thus proving the claim.

    This is a bit "brute force", but it will be clearer than something subtler if you're not used to doing proofs... (If you know about modular arithmetic, I can give you a more concise approach.)
    Last edited: Nov 11, 2007
  4. Nov 12, 2007 #3
    Awesome, I do understand that, it would be great as an intro to my report :P

    Considering the report is to do with modular arithmetic could you give me some of the concise approact? :P

    Would it be about something to do with the 'amount of days in a month' mod 7?

  5. Nov 12, 2007 #4


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    Just so. You'd start by assigning the days of the week to a "clock" with seven positions, with Sunday as 0, Monday as 1, ... , and Saturday as 6. Fridays fall at 5, but since we want to look at Friday the Thirteenth, we could also treat the problem as the same as looking for Sunday the First.

    You would then assign a value to each month, according its number of days (mod 7). So March is 3, April is 2, ... , and October is 3. You are then just looking for the sum of consecutive months that add to 0 (mod 7), starting from each "position on the clock". The month after the one that brings the sum to 0 is one with a Friday the Thirteenth.

    In the first example, we started at 0, so March is a month with a Friday the Thirteenth. (There may be others, but the proposition only asks that we show there is at least one such month. We are doing what is called an "existence proof".)

    In the second example where we started at 1, our sum ran

    1 + 3 + 2 + 3 + 2 + 3 = 14 = 0 (mod 7) ,

    as we went through March, April, May, June, and July. So the First of August is Sunday and thus August has a Friday the Thirteenth.

    From there, you would look at the remaining starting positions and see how many months must be summed to get you to zero. It will work in each case by the time you reach October. This proves the proposition.

    This is also handy for figuring out what day of the week a date will fall on, if you know which one it was on in some year. For instance, Christmas and New Year's Day fall on the same day of the week, being exactly a week apart. A year has 365 days, so 365 = 1 (mod 7); a leap year acts like 2 (mod 7). Christmas was on a Monday (1) in 2006, so it will fall on 1 + 1 = 2 or Tuesday this year, and 1 January 2008 will also be a Tuesday. Since 2008 is a leap year, it will shift the dates after 29 February by two, so Christmas 2008 will fall on 2 + 2 = 4 or Thursday, and so will New Year's 2009.

    Something like this is also used to deal with calculating the "moveable feasts" like Easter, which brings you back to Zeller...
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