If 21 January 1998 is a Wednesday, then what will be the day of the week on 31…

Concept — odd days The days of the week repeat in a cycle of 7, so when a stretch of days is added to a known weekday, only the remainder of that stretch on division by 7 matters. That remainder is the number of odd days: a common year of 365 days carries 1 odd day, a leap year of 366 days carries 2, and adding n odd days moves the weekday forward by n places.

Applying it to these two dates Split the stretch from 21 January 1998 to 31 December 2000 into two whole year-to-year legs and one tail inside 2000.

1. 21 January 1998 to 21 January 1999 spans a common year, 365 days. Since 365 = 52 × 7 + 1, this leg carries 1 odd day.

2. 21 January 1999 to 21 January 2000 spans February 1999, again a common year, 365 days. So this leg also carries 1 odd day.

3. 2000 is divisible by 400, so it is a leap year of 366 days. 21 January is its 21st day, so the tail from 21 January 2000 to 31 December 2000 is 366 − 21 = 345 days. Since 345 = 49 × 7 + 2, this tail carries 2 odd days.

4. Total odd days = 1 + 1 + 2 = 4.

5. Move the anchor weekday forward by 4 places from Wednesday: Thursday, Friday, Saturday, Sunday.

Cross-check Re-anchor the count at 1 January 1998 instead. 21 January is the 21st day of 1998, so 1 January 1998 sits 20 days earlier, and 20 = 2 × 7 + 6, which places it 6 steps behind the given Wednesday, on a Thursday. From 1 January 1998 to 31 December 2000 is 365 + 365 + 365 = 1095 days, because 31 December is the 366th day of 2000 and so 365 days have elapsed within that year. Since 1095 = 156 × 7 + 3, the target date lands 3 places after Thursday, giving the same weekday as the first route.

Where the count usually slips Two whole years from 21 January 1998 reach 21 January 2000 after 730 days, and 730 = 104 × 7 + 2, which is only 2 odd days. The remaining 345 days of 2000 add 2 more and carry the total to 4, so stopping at the two-year mark leaves the count short.

So 31 December 2000 falls on a Sunday.