A couple weeks ago I wrote a post about the year share component of calculating the day of the week. To calculate the day of the week, you need to add the day of the week, a constant for the month, and the year share. Calculating year share is not that hard, but it’s the hardest step to carry out in your head.

The year share of a date *y* is

⌊5*y*/4⌋ % 7

where *y* is the last two digits of a year. There’s no need to multiply by 5, since you could compute

(*y* + ⌊*y*/4⌋) % 7

but this still takes some effort if *y* is large.

The earlier post gave nine different ways to compute the year share, and the easiest in my opinion is method #6:

- Round
*y*down to the most recent leap year and remember how much you had to round down. Call that amount*r*. - Let
*t*be the tens value and*u*the units value. - Return (2
*t*–*u*/2 +*r*) % 7.

In symbols,

- Let
*y*‘ = 4 ⌊*y*/4⌋ and*r*=*y*–*y*‘. - Let
*t*= ⌊*y’*/10⌋ and*u*=*y*‘ % 10. - Return (2
*t*–*u*/2 +*r*) % 7.

This method has a longer description but is easier to carry out mentally as the following examples illustrate.

## Examples

Let *y* = 97. Here are the calculations required by the direct method.

- ⌊
*y*/4⌋ = 24 - 97 + 24 = 121
- 121 % 7 = 2

Here are the calculations required by the second method.

*y*‘ = 96 and*r*= 1.*t*= 9 and*u*= 6.- (2*9 – 3 + 1) % 7 = 2

The former requires adding two 2-digit numbers. The latter requires doubling a single digit number and subtracting another single digit number.

The former requires reducing a 3-digit number mod 7 and the latter requires reducing a 2-digit number mod 7. Furthermore, the 2-digit number is never more than 18.

## Proof

To prove that the two methods are equivalent we have to prove

⌊5*y*/4⌋ ≡ 2*t* – *u*/2 + *r* mod 7

which is kinda messy. The way *t*, *u*, and *r* are defined prevent this from being a simple algebraic calculation.

We can verify that

⌊5*y*/4⌋ ≡ 2*t* – *u*/2 + *r* mod 7

for *y* = 0, 1, 2, … 19 by brute force then prove the rest by induction. We’ll show that if the congruence holds for *y* then it holds for *y* + 20.

Suppose you increase *y* by 20. Then ⌊5*y*/4⌋ increases by 25. The value of *t* increases by 2, and the values of *u* and *r* remain unchanged, so the right side increases by 4. Since 25 % 7 = 4, the congruence still holds.

Another proof:

– Firstly, it’s enough to prove this for r=0, because the remainder r simply gets added in both cases. (We can write this down more formally, but it’s easier to just “see” than to explain.)

– So we can assume that y is a multiple of 4, namely y = 10t + u, and we want to prove that 5y/4 ≡ 2t − u/2 (mod 7), or equivalently (multiplying by 4) that 5y ≡ 8t − 2u, or equivalently (substituting y) that 5(10t + u) ≡ 8t – 2u. Well, mod 7, we can see that 50≡8 (≡1), and 5≡-2, so this is true.

% equal remainder?

Yes