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Difficulty: HardClocks and Calendars

An archivist intends to perfectly reuse a daily calendar planner originally printed for the year 20962096. In order for every single date in the planner to fall on the exact same day of the week for the entire year, what is the earliest subsequent year for which this planner can be used again?

  1. A
    21242124
  2. 21082108Answer
  3. C
    21022102
  4. D
    21362136

Answer

The earliest subsequent year with an identical calendar is 21082108.
A perfectly repeating calendar must satisfy two strict conditions: the starting weekday must be identical (meaning the accumulated odd days between them must be a multiple of 77), and both years must share the same leap status. The year 20962096 is a leap year. Because 21002100 is not a leap year (it is not divisible by 400400), the standard 2828-year repetition cycle is interrupted. We must count the odd days manually. Between 20962096 and 21072107, there are 1212 years in total: 22 leap years (2096,21042096, 2104) and 1010 normal years. The sum of odd days is (2×2)+(10×1)=14(2 \times 2) + (10 \times 1) = 14. Since 1414 is exactly divisible by 77, the year 21082108 starts on the same day as 20962096. Because 21082108 is also a leap year, every date including February 29 perfectly aligns.

Step-by-Step Solution

1
Determine the leap year status of the original year.
The year 20962096 is divisible by 44, so it is a leap year with 366366 days.
A leap year calendar can only perfectly repeat in another leap year. Otherwise, dates from March 1 onwards will shift out of alignment.
2
Identify the century boundary behavior for the upcoming century.
The upcoming century year is 21002100. Since 21002100 is not perfectly divisible by 400400, it is a normal year (not a leap year).
Crossing a non-leap century breaks the standard 2828-year repetition cycle for leap years, necessitating a manual calculation of odd days.
3
Calculate cumulative odd days year-by-year starting from 20962096.
Normal years contribute 11 odd day (365(mod7)=1365 \pmod 7 = 1); leap years contribute 22 odd days (366(mod7)=2366 \pmod 7 = 2). The sum must reach a multiple of 77 exactly at the start of a leap year.
A cumulative odd day sum that is a multiple of 77 ensures the starting day of the week exactly matches the original year.
4
Trace the odd days until both the weekday and leap year conditions are met.
From 20962096 to 21072107 (a span of 1212 years), there are 1010 normal years and 22 leap years (20962096 and 21042104). Total odd days = (10×1)+(2×2)=14(10 \times 1) + (2 \times 2) = 14. The sum 1414 is a multiple of 77.
Because the odd days perfectly balance out to 0(mod7)0 \pmod 7, the next year (21082108) starts on the identical weekday. Since 21082108 is also a leap year, its entire calendar matches 20962096.

Key Concept

Calendar Repetition Rules and Non-400 Century Boundaries
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Topics

Clocks and Calendars

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