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# How many ‘Friday the 13th’s are in this year? [Formula fun + challenge]

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Today is Friday the 13th. If you are a raging friggatriskaidekaphobiac, I suggest you to stop reading this post. For the rest of you, I have something fun.

Given a year in cell C3, let’s find out all the months with Friday the 13th. Something like this:

### Finding all Friday the 13ths in a year

Here is a formula to find the first Friday the 13th in a year.

{=IFERROR(SMALL(IF(WEEKDAY(DATE(\$C\$3,ROW(\$A\$1:\$A\$12),13))=6,ROW(\$A\$1:\$A\$12)),1),””)}

Let’s understand how it works, going from inside out:

DATE(\$C\$3,ROW(\$A\$1:\$A\$12),13) portion: This generates an array of 12 dates, one for each 13th of the month in the year C3.

WEEKDAY(DATE(…))=6 portion: This checks how many of those dates are Fridays. Returns an array of TRUE / FALSE values.

IF(WEEKDAY(DATE(…))=6,ROW(\$A\$1:\$A\$12)) portion: This returns an array of month numbers where we have Friday the 13th and FALSE values.

SMALL(IF(…),1) portion: This will give us the first month (ie 1st smallest value) with Friday the 13th in the year C3.

IFERROR() to suppress any errors.

To show all the Friday the 13ths in a year, simply replace 1 with an array of numbers (up to 3 should do).

Click here to download Friday the 13th finder workbook. Enter a year in cell C3 and see all the months with Friday the 13th instantly.

### A challenge for you…

If you are still reading, I have a challenge for you. Can you write a formula to find the next year with three Friday the 13ths? Assume the year is in C3. Post your formulas / VBA in the comment section.

So that’s all for now. Enjoy your Friday the 13th.

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### 14 Responses to “How many ‘Friday the 13th’s are in this year? [Formula fun + challenge]”

1. in C3=2016
in C4=3
in C5=1 (the first next year with three Friday the 13ths)

=SMALL(IF(MMULT(--(MOD(DATE(C3+ROW(1:1000),COLUMN(A:L),13),7)=6),ROW(1:12)^0)=C4,C3+ROW(1:1000)),C5)

formula check in the next 1000 years

2. Brian says:

This will generate a table of counts of Friday the 13th's by year. If I didn't screw it up the next year with three is 2026.

I created a simple parameter table with a start date and end date that I wanted to evaluate. That calculates the number of days and generates a list of those days. Then filter and group. The generation of the list in power query (i.e. without populating a date table in excel) is pretty cool, otherwise this isn't really doing anything than creating a big date and filtering/counting.

let
Source = List.Dates(StartDateAsDate, Days2, #duration(1,0,0,0)),
ConvertDateListToTable = Table.FromList(Source, Splitter.SplitByNothing(), null, null, ExtraValues.Error),
FilterFriday13 = Table.SelectRows(AddDayOfWeekColumn, each ([DayOfMonth] = 13) and ([Day of Week] = 5)),
Friday13thsByYear = Table.Group(FilterFriday13, {"Year"}, {{"Number of Friday the 13ths!", each Table.RowCount(_), type number}})
in
Friday13thsByYear

• Brian says:

With the parameters replaced by values should you want to play along at home. This runs for 20 years starting on 1/1/2016.

let
Source = List.Dates(#date(2016,1,1), 7300, #duration(1,0,0,0)),
ConvertDateListToTable = Table.FromList(Source, Splitter.SplitByNothing(), null, null, ExtraValues.Error),
FilterFriday13 = Table.SelectRows(AddDayOfWeekColumn, each ([DayOfMonth] = 13) and ([Day of Week] = 5)),
Friday13thsByYear = Table.Group(FilterFriday13, {"Year"}, {{"Number of Friday the 13ths!", each Table.RowCount(_), type number}})
in
Friday13thsByYear

3. Alex Groberman says:

=MATCH(3,MMULT(N(WEEKDAY(DATE(C3+ROW(1:100)-1,COLUMN(A:L),13))=6),1^ROW(1:12)),)+C3-1

• David N says:

It should be pointed out that Alex's solution, unlike some others, has the additional advantage of being non-array. My solution was nearly identical but with -- and SIGN instead of N and 1^.

=C3-1+MATCH(3,MMULT(--(WEEKDAY(DATE(C3-1+ROW(1:25),COLUMN(A:L),13))=6),SIGN(ROW(1:12))),0)

4. SunnyKow says:

Sub Friday13()

Dim StartDate As Date
Dim EndDate As Date
Dim x As Long
Dim r As Long

Range("C7:C12").ClearContents
StartDate = CDate("01/01/" & Range("C3"))
EndDate = CDate("31/12/" & Range("C3"))
r = 7
For x = StartDate To EndDate
If Day(x) = 13 And Weekday(x, vbMonday) = 5 Then
Cells(r, 3) = Month(x)
r = r + 1
End If
Next
End Sub

• SunnyKow says:

Calculate next year with 3 Friday 13th. Good for 100 years different from year entered in cell C3

Sub ThreeFriday13()

Dim StartDate As Date
Dim EndDate As Date
Dim x As Long
Dim WhatYear As Integer
Dim Counter As Integer

Range("E7").ClearContents
StartDate = CDate("01/01/" & Range("C3") + 1)
EndDate = CDate("31/12/" & Range("C3") + 100)
Counter = 0

For x = StartDate To EndDate
If WhatYear Year(x) Then
WhatYear = Year(x)
'Different year so reset counter
Counter = 0
End If
If Day(x) = 13 And Weekday(x, vbMonday) = 5 Then
Counter = Counter + 1
If Counter = 3 Then
WhatYear = Year(x)
Exit For
End If
End If
Next
Range("E7") = WhatYear

End Sub

• SunnyKow says:

*RE-POST as not equal did not show earliuer
Calculate next year with 3 Friday 13th. Good for 100 years different from year entered in cell C3

Sub ThreeFriday13()

Dim StartDate As Date
Dim EndDate As Date
Dim x As Long
Dim WhatYear As Integer
Dim Counter As Integer

Range("E7").ClearContents
StartDate = CDate("01/01/" & Range("C3") + 1)
EndDate = CDate("31/12/" & Range("C3") + 100)
Counter = 0

For x = StartDate To EndDate
If WhatYear NE Year(x) Then
WhatYear = Year(x)
'Different year so reset counter
Counter = 0
End If
If Day(x) = 13 And Weekday(x, vbMonday) = 5 Then
Counter = Counter + 1
If Counter = 3 Then
WhatYear = Year(x)
Exit For
End If
End If
Next
Range("E7") = WhatYear

End Sub

• goby says:

earlier*

5. Devesh says:

I've a doubt with using array formula here.
In sample workbook, I tried to replicate the formula again.
=IFERROR(SMALL(IF(WEEKDAY(DATE(\$C\$3,ROW(\$A\$1:\$A\$12),13))=6,ROW(\$A\$1:\$A\$12)),\$B7),"")
For this I selected C7 to C12, and typed the same formula and pressed ctrl+alt+Enter. But in all cells it is taking \$B7 (and not \$B7, \$B8, \$B9.... etc)
and since it is array formula I can't edit individual cell.
Thanks

6. Pablo says:

Hi Chandoo,
Cool stuff. You need to clarify that the answer of 5 represents the 1st month in the year that has a Friday the 13th, and not the number of Fridays the 13th in the year. Subtle, but important difference.
Thanks,
Pablo

7. Micah Dail says:

I like the MMULT() function far more, but here's how I would have tackled it. It uses an EDATE() base and MODE() over 100 years. I'm assuming that 100 years is enough time to catch the next year with 3 friday 13th's. Array entered, of course.

{=MODE(IFERROR(YEAR(IF((WEEKDAY(EDATE(DATE(C3, 1, 13), ROW(INDIRECT("1:1200"))))=6), EDATE(DATE(C3, 1, 13), ROW(INDIRECT("1:1200"))), "")), ""))}

8. Jason Morin says:

Finding all the Friday the 13ths in a Year:

=SUMPRODUCT((DAY(ROW(INDIRECT(DATE(C3,1,1)&":"&DATE(C3,12,31))))=13)*(TEXT(ROW(INDIRECT(DATE(C3,1,1)&":"&DATE(C3,12,31))),"ddd")="Fri"))

9. jmdias says:

{=sum(if(day.of.week(DATe(\$YEAR;{1;2;3;4;5;6;7;8;9;10;11;12};13);1)=6;1;0))}
just list the years

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