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Prove that in each year, the 13th day of some month occurs on a Friday.

No clue... please help!

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    $\begingroup$ hint: piogenhole principle and little work. $\endgroup$
    – mesel
    Jul 2, 2014 at 19:40
  • $\begingroup$ @mesel I think the pigeonhole principle might be a good idea And 10 possibilities among 7 buckets days but previous year leap year might make it 8 buckets But not sure how to apply exactly from then on $\endgroup$ Jul 2, 2014 at 19:56
  • $\begingroup$ @mesel Please post your answer using the pigeonhole principle. $\endgroup$
    – bof
    Jul 3, 2014 at 9:37
  • $\begingroup$ @mesel Please elaborate by pigeonhole principle $\endgroup$ Jul 3, 2014 at 17:37
  • $\begingroup$ The pigeonhole principle doesn't help here. $\endgroup$
    – TonyK
    Jul 5, 2014 at 8:27

6 Answers 6

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Let us ignore leap years for a moment.

And let us assume that some date (e.g., 13th) falls on Monday in January.

What happens in February? January has $31=4\cdot 7+3$ days, so the days move by 3 to Thursday.

So if we denote the days in week by numbers $\{0,1,2,3,4,5,6\}$, we only have to consider the number of days in a month, compute modulo 7 and see what happens:

0 January (31 days)
3 February (28 days)
3 March (31 days)
6 April (30 days)
1 May (31 days)
4 June (30 days)
6 July (31 days)
2 August (31 days)
5 September (30 days)
0 October (31 days)
3 November (30 days)
5 December (31 days)
1 January next year 

(Computing day in January in the next year is irrelevant for this question; but comparing this with an actual calendar is a good sanity check.)

Notice, that all numbers 0,1,2,3,4,5,6 appear in the above table.


Now what happens if a year starts not on Monday (=0) but on Tuesday. You simply have to add $+1$ (computing modulo 7) in each row of the table. But since every number appeared at least one, we will get $6+1=0$ in some row. And the same is true for any other day.

So now you only have to create similar table for a leap year, check whether all numbers from 0 to 6 appear there, and you are done.


Somewhat related: Occurrence of Friday the 13th on Wikipedia.

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  • $\begingroup$ +1 Clever methode! I haven't though about that! I've learned something new. $\endgroup$
    – Bman72
    Jul 3, 2014 at 8:27
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    $\begingroup$ Good. Better yet, start with March and forget about leap years. (Conceivably, "every year has at least one Friday the 13th" could fail if we throw out January and February. Fortunately, it doesn't.) $\endgroup$
    – bof
    Jul 3, 2014 at 9:32
  • $\begingroup$ good point @bof $\endgroup$ Jul 3, 2014 at 9:36
  • $\begingroup$ Thnx :) I appreciate! Still I'd like to have a pigeonhole proof $\endgroup$ Jul 3, 2014 at 17:42
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In fact, every year will contain a Friday the 13-th between March and October (so leap years don't enter into it).

If March 13 is assigned $0 \pmod 7$, then the other moduli occur as indicated below: $$(\underbrace{\underbrace{\underbrace{\underbrace{\underbrace{\underbrace{\overbrace{31}^{\text{March}}}_{3 \pmod 7},\overbrace{30}^{\text{April}}}_{5 \pmod 7},\overbrace{31}^{\text{May}}}_{1 \pmod 7},\overbrace{30}^{\text{June}},\overbrace{31}^{\text{July}}}_{6 \pmod 7},\overbrace{31}^{\text{August}}}_{2 \pmod 7},\overbrace{30}^{\text{September}}}_{4 \pmod 7})$$

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    $\begingroup$ ...that is some nice $\LaTeX$. $\endgroup$
    – DanielV
    Jul 5, 2014 at 8:17
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Figure out where the 13th day of each month falls relative to January 1st (remember there are two possibilities, corresponding to leap year or non leap year). There are only seven possible values: 0, 1, 2, 3, 4, 5, 6 days after the day on which Jan. 1 falls. If each of those values occurs at least once, then at least one of those days must be a Friday. Can you take it from there?

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Just brute force:

There is a calendar for a common year (i.e. a non-leap year) beginning on Sunday.

There is a calendar for a common year beginning on Monday.

There is a calendar for a common year beginning on Tuesday.

. . . and so on. Seven calendars. Then seven more for leap years.

Go through all 14 of them and observe that each has at least one Friday the 13th. And some have two, and some have three, and none have more than three.

There may be no way to reach this conclusion except this kind of brute force, because the structure of the calendar (how many months, how many days in each month) is not defined by orderly rules.

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Look at a calendar and observe that the months of May, June, July, August, September, October, and November start on seven different days of the week. This will happen every year, because the number of days in each of those months is the same every year. Therefore one of those months will start on a Sunday, and so the 13th will fall on a Friday.

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I think this question has already good answers,

It is enough to show that there exist a month starting on Sunday.

Let $x$ be the starting day of the year then next month's first day is $x+31\equiv x+3$ and next month's first day, $x+3+28\equiv x+3$...

At the and we have, $$x,x+3,x+3,x+6,x+1,x+4,x+6,x+2,x+5..$$ now we reached seven different days, $x,x+1...,x+6$ by pigeonhole principle we must have all days which conclude the result.

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  • $\begingroup$ And a similar argument for leap years. Yes, of course, we all know how to prove it that way. But I would like to see you prove it using the pigeonhole principle, as you suggested in your comment yesterday. I don't believe that the pigeonhole principle is useful here, but seeing is believing. $\endgroup$
    – bof
    Jul 4, 2014 at 8:26
  • $\begingroup$ @bof: My intention with "little work" was that. I only use pigenhole principle to cocnclude. $\endgroup$
    – mesel
    Jul 4, 2014 at 13:18
  • $\begingroup$ The argument you posted did not use the pigeonhole principle in any way. $\endgroup$
    – bof
    Jul 4, 2014 at 22:36
  • $\begingroup$ understand it the way you want, its your choices. $\endgroup$
    – mesel
    Jul 4, 2014 at 22:41

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