The last day of a century cannot be ?

Concept

The day of the week that a period of years ends on is found using the odd-days method: an ordinary year (365 days) leaves 1 odd day and a leap year (366 days) leaves 2 odd days, since 365 = 52 weeks + 1 day and 366 = 52 weeks + 2 days. Adding the odd days over any span of years and reducing the total modulo 7 gives a remainder from 0 to 6, which maps onto Sunday through Saturday (0 = Sunday, 1 = Monday, 2 = Tuesday, 3 = Wednesday, 4 = Thursday, 5 = Friday, 6 = Saturday) and fixes the day the span ends on.

Application

1. 100 years contain 76 ordinary years and 24 leap years, giving 76(1) + 24(2) = 124 odd days; 124 mod 7 = 5, so the 1st century (year 100) ends on the day with remainder 5 — Friday.

2. 200 years double this count: (5 x 2) = 10 odd days; 10 mod 7 = 3, so the 2nd century (year 200) ends on the day with remainder 3 — Wednesday.

3. 300 years triple it: (5 x 3) = 15 odd days; 15 mod 7 = 1, so the 3rd century (year 300) ends on the day with remainder 1 — Monday.

4. 400 years is special: because a year divisible by 400 is always a leap year, 400 years contain 97 leap years (not 96), giving 303(1) + 97(2) = 497 odd days; 497 mod 7 = 0, so the 4th century (year 400) ends on the day with remainder 0 — Sunday.

Cross-check

A 0-odd-day result over 400 years means the calendar resets exactly, so the same 4-step pattern of remainders 5, 3, 1, 0 repeats for every following block of four centuries. Across all centuries, the remainder therefore only ever takes the values 5, 3, 1, or 0 — mapping only onto Friday, Wednesday, Monday, or Sunday. The remaining remainders (2, 4, 6 — Tuesday, Thursday, Saturday) never occur as a century's last day.

Result

Among the options, Wednesday, Friday, and Monday each occur in the repeating cycle above, but Thursday's remainder (4) never appears — so the last day of a century cannot be Thursday.