For the part that depends on the last two digits of the year I use a method that I've not seen anywhere else. It is algorithmically more work than Conway's meth

For the part that depends on the last two digits of the year I use a method that I've not seen anywhere else. It is algorithmically more work than Conway's method (Y + Y//4, where Y is the last two digits of the year and // is integer division rounding positive results down), but I think is easier for quick mental computation for most people. Let Y = 10 T + U, i.e., T is the first digit of the two digit year and U is the second digit. If T is even, the year part is: 2 T + U if U = 0, 1, 2, or 3 2 T + U + 1 if U = 4, 5, 6, or 7 2 T + U + 2 if U = 8 or 9 If T is odd: 3 + 2 T + U if U = 0 or 1 3 + 2 T + U + 1 if U = 2, 3, 4, or 5 3 + 2 T + U + 2 if U = 6, 7, 8 or 9 Algorithmically I compute it something like this: Compute 2 T + U If T is even Add 1 if T >= 4 and add another 1 if T >= 8 else Add 3 Add 1 if T >= 2 and add another 1 if T >= 6 Those add 1s come from the way leap years are distributed within decades. The 2 T + U takes into account any leap years when U = 0. The add 1s are to take care of the leap years that happen when U != 0. In even decades those occur at 4 and 8, and in odd decades they occur at 2 and 6. Example: for 2024, T=2 and U=4. Mentally my train of thought would be: 2 doubles to 4, plus 4 = 8 = 1 mod 7, even decade so add 1 for 4 <= T < 8 giving 2. For that example there is not really easier than Conway's Y + Y//4, which would be 24 + 6 = 30 = 2 mod 7. Example: for 2099, T = 9 and U = 9. Mentally that goes something like this: 9 = 2 mod 7 doubles to 4, add 9 = 2 mod 7 giving 6 T was odd so add 3 giving 9 = 2 mod 7 T was odd so the so there were leap years at 2 and 6, which each add 1 That gives us 4 I find that easier that Conway's 99 + 99/4. I can do that, and at decent speed, but I can make mistakes. With my method you never have to deal with a number above 32, and that's only if you do no reductions mod 7 along the way. If you reduce mod 7 whenever you can you never have to deal with anything above 12. Note: when reducing mod 7 along the way it is crucial to remember that whether you take the T is even case or the T is odd case depends on the actual value of T. E.g., when I am doing 2099 I reduce both T and U to 2 right away but first note that 9 is odd so will need to take the odd case.