Calendar Practice Questions

1986 is an ordinary year. Add odd days year by year until the total is a multiple of 7 AND the next year is also an ordinary year: 1986: 1, 1987: 1, 1988: 2, 1989: 1, 1990: 1, 1991: 1, 1992: 2, 1993: 1, 1994: 1, 1995: 1, 1996: 2 → total 14 = 2×7, i.e. 0 odd days. So 1997 starts on the same day as 1986 and has the same number of days → same calendar.

Count odd days up to 15 Aug 2025: 2000 years → 0 odd days (every 400 years has 0 odd days); next 24 years (2001–2024): 6 leap + 18 ordinary = 12 + 18 = 30 days → 2 odd day(s); in 2025 (ordinary year): Jan 3, Feb 0, Mar 3, Apr 2, May 3, Jun 2, Jul 3, plus 15 days of Aug → 227 days → 3 odd day(s). Total odd days = 257 → 257 mod 7 = 5. Since 0 = Sunday, 1 = Monday, …, 6 = Saturday, the day is Friday.

Count odd days up to 26 Jan 1950: 1600 years → 0 odd days (every 400 years has 0 odd days); next 300 years → 1 odd day(s) (100 yrs = 5, 200 yrs = 3, 300 yrs = 1); next 49 years (1901–1949): 12 leap + 37 ordinary = 24 + 37 = 61 days → 5 odd day(s); in 1950 (ordinary year): 26 days of Jan → 26 days → 5 odd day(s). Total odd days = 88 → 88 mod 7 = 4. Since 0 = Sunday, 1 = Monday, …, 6 = Saturday, the day is Thursday.

Using the standard rule: 100 years = 5 odd days. Remaining 26 years contain 6 leap years and 20 ordinary years → 6×2 + 20 = 32 → 4 odd days. Total = 5 + 4 = 9 → 2 odd days. (If no non-leap century year falls inside the 126 years, there would be 31 leap years → 3 odd days.)

94 days = 13 weeks + 3 odd day(s). Weekdays repeat every 7 days, so only the 3 odd day(s) matter: Wednesday + 3 = Saturday.