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Difficulty: MediumClocks and Calendars

A commemorative time capsule was sealed by a city council on Tuesday, August 15, 1876. The municipal charter mandates that it must be opened exactly 50 years later, on August 15, 1926. Based on the standard Gregorian calendar, on what day of the week will the capsule be opened?

  1. SundayAnswer
  2. B
    Monday
  3. C
    Tuesday
  4. D
    Wednesday

Answer

The capsule will be opened on Sunday.
The correct answer accurately calculates the shift by accounting for 50 base years and exactly 11 leap days. It properly excludes 1900 as a leap year (due to the century rule) and correctly skips the 1876 leap day because the interval starts in August. Adding the resulting 61 odd days to Tuesday (61 mod 7 = 5) shifts the day forward by 5 days to Sunday.

Step-by-Step Solution

1
Determine the total number of base years in the interval.
1926 - 1876 = 50 years.
Each normal year adds exactly 1 odd day to the calendar shift.
2
Count the number of leap days crossed between August 15, 1876, and August 15, 1926.
There are 11 leap days (1880, 1884, 1888, 1892, 1896, 1904, 1908, 1912, 1916, 1920, 1924).
We must add 1 extra odd day for every February 29th crossed. The year 1876 is skipped because the start date is in August. The year 1900 is skipped because century years not divisible by 400 are not leap years.
3
Calculate the total number of odd days.
50 (from base years) + 11 (from leap days) = 61 total odd days.
Combining the normal yearly shift with the extra leap day shifts provides the total day shift.
4
Find the remainder when dividing the total odd days by 7 and apply it to the starting day.
61 divided by 7 leaves a remainder of 5. Tuesday + 5 days = Sunday.
The days of the week repeat in cycles of 7. Adding the remainder to the starting day gives the final day of the week.

Key Concept

Calculating odd days across century boundaries using the Gregorian calendar leap year rules.
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Topics

Clocks and Calendars

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