The first day of the year 2003 was Wednesday. What would have been the last…

Under the Gregorian calendar, a non-leap year (365 days) shifts the weekday of 1 January by exactly 1 day into the next year, because 365 = 52×7 + 1 — this extra day is called an ‘odd day’. A leap year (366 days) shifts it by 2 odd days, because 366 = 52×7 + 2. A further consequence of the same 52-complete-weeks structure: within any single non-leap year, the first day (1 January) and the last day (31 December) fall on the SAME weekday, since 31 December is exactly 364 = 52×7 days after 1 January.

1. 2003 is a non-leap year (not divisible by 4), so it contributes 1 odd day: 1 January 2004 = Wednesday + 1 day = Thursday.

2. 2004 is a leap year (divisible by 4), so it contributes 2 odd days: 1 January 2005 = Thursday + 2 days = Saturday.

3. 2005 is a non-leap year, so its first and last day coincide: 31 December 2005 = 1 January 2005 = Saturday.

Independently: the span from 1 January 2003 to 1 January 2006 covers three full years — 2003 (365) + 2004 (366) + 2005 (365) = 1096 days. 1096 mod 7 = 4, so 1 January 2006 = Wednesday + 4 days = Sunday. Since 31 December 2005 is exactly one day before 1 January 2006, 31 December 2005 = Saturday — confirming the same result.

So the last day of the year 2005 was Saturday.