Notice first that these days always fall on the same day of the week in any year (in 2012, on a Wednesday):
Apr. 4, June 6, Aug. 8, Oct. 10, Dec. 12 (or 4/4, 6/6, 8/8, 10/10, and 12/12)
May 9, July 11, Sept. 5, Nov. 7 (5/9, 7/11, 9/5, 11/7 … or: I work 9-to-5 at the 7-11)
also the last day of February, which you can also understand as March 0 (in both common years and leap years)
and in three out of four years Jan. 3, but in leap years (every four years) Jan. 4
So, how do you know which day of the week for a given year?
First, determine the century number. In the 2000s, it’s 2 (like Y2K); in the 1900s it’s 3 (Wednesday is the third day, and “we-in-dis-day” … ahem, most of us were born in the 1900s. If you weren’t, I guess your eff’ed.) There are other values for 1800 or 2100 and so on.
Next, determine how many full dozens of years have elapsed between 1900 or 2000 (or 1800 or 2100 and so on), inclusive. 2015? Then its 1. 2012? Also 1. 2004? 0. 2070? 5.
Then determine the remainder mod 12. So, 2015? 3. 2012? 0. 2004? 4. 2070? 10.
Then determine how many full quartets are in the remainder. 2015? Remainder ® is 3, no quartets (q), so 0. 2012? r=0, q=0. 2004? r=4, q=1. 2070? r=10, q=2.
Then sum these numbers up:
2015: 2+1+3+0=6
2012: 2+1+0+0=3
2004: 2+0+4+1=7
2070: 2+5+10+2=19
And take them mod 7
2015: 6 mod 7 = 6
2012: 3 mod 7 = 3
2004: 7 mod 7 = 0
2070: 19 mod 7 = 5
0 = Sunday (none-day)
1 = Monday (one-day)
2 = Tuesday (obvious)
3 = Wednesday (Threes-day kinda of sounds like Wednesday, right?)
4 = Thursday (Do.)
5 = Friday (Do.)
6 = Saturday (Six-urday, again, basically the same word!)
A trick, if you add in the offset between the day of interest and the appropriate day on the list I opened with before you take the sum mod 7, you will get the day of the week for your day of interest.
So, when is Apr. 12, 2012?
Offset for Apr. 4 = +8
Century number = 2
Dozens = 1
Remainder = 0
Quartets = 0
Sum = 11
11 mod 7 = 4
Thursday.