Lucky friday the thirteenth

Number Theory Level 2

Being born on the thirteenth of the month, Calvin considers the 13th to be a lucky day. What is the most number of Friday the 13th's that can occur in a calendar year?


The answer is 3.

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Arron Kau by Arron Kau

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6 solutions

We use the rule for number of days for the months (e.g. January has 31 days, February has 28 days, and so forth.)

We seek for the accumulating number of days in which the date is 13 for each month. January has 13; February has 44; March has 72; April has 103; May has 133; June has 164; July has 194; August has 225; September has 256; October has 286; November has 317; December has 347.

Computing their modulo 7's, we have: (In order from January to December): 6, 2, 2, 5, 0, 3, 5, 1, 4, 6, 2, 4. Since the number 2 occurs the most number of times which is three times, we can assume that Friday the 13th's can occur during February, March, and November. Hence, it has at most three Friday the 13th's that can occur in one calendar year.

Suppose the year has 29 days in February or we have the leap year. We simply add 1 to each remainders from March to December since there are 29 days in February. (In order from January to December): 6, 2, 3, 6, 1, 4, 6, 2, 5, 0, 3, 5.

It has the same case as the calendar with normal days wherein there are at most three Friday the 13th's. Hence, the answer.

6 Helpful 0 Interesting 0 Brilliant 0 Confused
Bhargav Das
Dec 14, 2013

We need to use here Modular Arithmetic .

We need to look for 2 2 cases, for a leap year and for a non-leap year .

CASE 1: First, note for any non-leap year that,

Jan.- 10 t h 10th day,

Feb.- 7 t h 7th day,

March- 7 t h 7th day,

April- 4 t h 4th day,

May- 9 t h 9th day,

June- 6 t h 6th day,

July- 11 t h 11th day,

Aug.- 8 t h 8th day,

Sep.- 5 t h 5th day,

Oct.- 10 t h 10th day,

Nov.- 7 t h 7th day and for

Dec.- 12 t h 12th day

we have the same day .

CASE 2: And for any leap year we have,

Jan.- 11 t h 11th day,

Feb.- 8 t h 8th day,

March- 7 t h 7th day,

April- 4 t h 4th day,

May- 9 t h 9th day,

June- 6 t h 6th day,

July- 11 t h 11th day,

Aug.- 8 t h 8th day,

Sep.- 5 t h 5th day,

Oct.- 10 t h 10th day,

Nov.- 7 t h 7th day and for

Dec.- 12 t h 12th day

we have the same day .

Let that desired same day be Friday in our case.

Now, working out modulo 13 we get,

First CASE 1 gives,

Jan.- 10 t h 10th day, 10 ≡ − 3 m o d 13 10\equiv -3 \bmod{13}

Feb.- 7 t h , 14 t h . . . 7th,14th... day, 14 ≡ 1 m o d 13 14\equiv 1 \bmod{13}

March- 7 t h , 14 t h . . . 7th,14th... day, 14 ≡ 1 m o d 13 14\equiv 1 \bmod{13}

April- 4 t h , 11 t h , . . . 4th,11th,... day, 11 ≡ − 2 m o d 13 11\equiv -2 \bmod{13}

May- 9 t h 9th day, 9 ≡ − 4 m o d 13 9\equiv -4 \bmod{13}

June- 6 t h , 13 t h . . . 6th,13th... day, 13 ≡ 0 m o d 13 13\equiv 0 \bmod{13}

July- 11 t h 11th day, 11 ≡ − 2 m o d 13 11 \equiv -2 \bmod{13}

Aug.- 8 t h , 15 t h , . . 8th,15th,.. day, 15 ≡ 2 m o d 13 15\equiv 2 \bmod{13}

Sep.- 5 t h , 12 t h , . . . 5th,12th,... day, 12 ≡ − 1 m o d 13 12\equiv -1 \bmod{13}

Oct.- 10 t h 10th day, 10 ≡ − 3 m o d 13 10\equiv -3 \bmod{13}

Nov.- 7 t h , 14 t h , . . . 7th,14th,... day 14 ≡ 1 m o d 13 14\equiv 1 \bmod{13} and for

Dec.- 12 t h 12th day 12 ≡ − 1 m o d 13 12\equiv -1 \bmod{13} .

We observe that 1 m o d 13 1 \bmod{13} occurs the most no. of times ( 3 times).Hence, maximum no. of Friday the 13th's is 3 3 in this case.

Now, CASE 2: gives us,

Jan.- 11 t h 11th day, 11 ≡ − 2 m o d 13 11\equiv -2 \bmod{13}

Feb.- 8 t h , 15 t h . . . 8th,15th... day, 15 ≡ 2 m o d 13 15\equiv 2 \bmod{13}

March- 7 t h , 14 t h . . . 7th,14th... day, 14 ≡ 1 m o d 13 14\equiv 1 \bmod{13}

April- 4 t h , 11 t h , . . . 4th,11th,... day, 11 ≡ − 2 m o d 13 11\equiv -2 \bmod{13}

May- 9 t h 9th day, 9 ≡ − 4 m o d 13 9\equiv -4 \bmod{13}

June- 6 t h , 13 t h . . . 6th,13th... day, 13 ≡ 0 m o d 13 13\equiv 0 \bmod{13}

July- 11 t h 11th day, 11 ≡ − 2 m o d 13 11 \equiv -2 \bmod{13}

Aug.- 8 t h , 15 t h , . . 8th,15th,.. day, 15 ≡ 2 m o d 13 15\equiv 2 \bmod{13}

Sep.- 5 t h , 12 t h , . . . 5th,12th,... day, 12 ≡ − 1 m o d 13 12\equiv -1 \bmod{13}

Oct.- 10 t h 10th day, 10 ≡ − 3 m o d 13 10\equiv -3 \bmod{13}

Nov.- 7 t h , 14 t h , . . . 7th,14th,... day 14 ≡ 1 m o d 13 14\equiv 1 \bmod{13} and for

Dec.- 12 t h 12th day 12 ≡ − 1 m o d 13 12\equiv -1 \bmod{13} .

We now observe that − 2 m o d 13 -2 \bmod{13} occurs the most no. of times ( 3 times ). Hence, maximum no. of Friday the 13th's is 3 3 in this case too.

Hence, we can conclude that the maximum no. of Friday the 13th's that can occur in a calendar year is 3 \boxed{3} .

4 Helpful 0 Interesting 0 Brilliant 0 Confused

daheq is this?

Yash Bhagwat - 7 years, 4 months ago

Friday the thirteenth can be occurred thrice in Thursday year (example 2009, 2015) and Sunday-Monday year (example 2012) Common year: February, March, November; example year: 1998, 2009, 2015, 2026, 2037, 2043. Leap year: January, April, July; example year: 1984, 2012, 2040, 2068, 2096, 2108.

1 Helpful 0 Interesting 0 Brilliant 0 Confused

I am also born on 13 but on a Tuesday...

Philips Zephirum Lam - 7 years, 6 months ago

its a wild guess !!!!

Xum Man - 7 years, 5 months ago
Ajala Singh
Dec 13, 2013

I guessed 3!

1 Helpful 0 Interesting 0 Brilliant 0 Confused

LOL. same here

Agnishom Chattopadhyay - 7 years, 6 months ago

how did u come up with the guess ???

Byla Bella - 7 years, 6 months ago
Ramiel To-ong
Jan 14, 2016

that happened on Friday, 13th of January 2012

0 Helpful 0 Interesting 0 Brilliant 0 Confused
Mujeeb Ur Rahman
Dec 14, 2013

Used Excel Spreadsheet Two sheets: One for normal and another for leap year. Here is the solution https://docs.google.com/spreadsheet/ccc?key=0AhsEBE9xtIJ1dGxudGFRc2ROTnpoZkdsOGRkZm1OY1E&usp=drive_web#gid=0

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