Doomsday Rule - Finding A Year's Doomsday

Finding A Year's Doomsday

We first take the anchor day for the century. For the purposes of the Doomsday rule, a century starts with '00 and ends with '99. The following table shows the anchor day of centuries 1800–1899, 1900–1999, 2000–2099 and 2100–2199.

Century Anchor day Mnemonic Index (day of week)
1800–1899 Friday - 5 (Fiveday)
1900–1999 Wednesday We-in-dis-day
(most living people were born in that century)		 3 (Treblesday)
2000–2099 Tuesday Y-Tue-K or Twos-day
(Y2K was at the head of this century)		 2 (Twosday)
2100–2199 Sunday Twenty-one-day is Sunday
(2100 is the start of the next century)		 0 (Noneday)

Next, we find the year's Doomsday. To accomplish that according to Conway:

  1. Divide the year's last two digits (call this y) by 12 and let a be the floor of the quotient.
  2. Let b be the remainder of the same quotient.
  3. Divide that remainder by 4 and let c be the floor of the quotient.
  4. Let d be the sum of the three numbers (d = a + b + c). (It is again possible here to divide by seven and take the remainder. This number is equivalent, as it must be, to the sum of the last two digits of the year taken collectively plus the floor of those collective digits divided by four.)
  5. Count forward the specified number of days (d or the remainder of d/7) from the anchor day to get the year's Doomsday.

For the twentieth-century year 1966, for example:

\begin{matrix}\left({\left\lfloor{\frac{66}{12}}\right\rfloor+66 \bmod 12+\left\lfloor{\frac{66 \bmod 12}{4}}\right\rfloor}\right) \bmod 7+\rm{Wednesday} & = & \left(5+6+1\right) \bmod 7+\rm{Wednesday} \\ \ & = & \rm{Monday}\end{matrix}

As described in bullet 4, above, this is equivalent to:

\begin{matrix}\left({66 + \left\lfloor{\frac{66}{4}}\right\rfloor}\right) \bmod 7+\rm{Wednesday} & = & \left(66+16\right) \bmod 7+\rm{Wednesday} \\ \ & = & \rm{Monday}\end{matrix}

So Doomsday in 1966 fell on Monday.

Similarly, Doomsday in 2005 is on a Monday:

Read more about this topic:  Doomsday Rule

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