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Question

Difficulty: HardClocks and Calendars

If January 10, 1896, was a Friday, what day of the week was January 10, 1908?

  1. A
    Saturday
  2. B
    Thursday
  3. FridayAnswer
  4. D
    Sunday

Answer

Friday
The correct day is Friday because evaluating the 12-year interval from January 10, 1896, to January 10, 1908, yields exactly 14 odd days. This breaks down into 12 baseline days (one for each year) plus 2 extra days for the leap years explicitly crossed (1896 and 1904). Importantly, the year 1900 is not a leap year, and the leap day of 1908 is not reached. Because 14 is perfectly divisible by 7 (remainder 0), the day of the week does not shift.

Step-by-Step Solution

1
Calculate the total number of years between the two dates.
The total span from January 10, 1896, to January 10, 1908, is exactly 12 years.
Each standard year contributes 1 odd day (since 365 mod 7 = 1) to the day of the week progression.
2
Identify the number of leap years crossed within this specific time frame.
The leap days crossed belong to the years 1896 and 1904. Total = 2 additional leap days.
1896 is a leap year, and we cross its February 29th because we start on January 10th. The year 1900 is a century year not divisible by 400, so it is a standard year (no leap day). 1904 is a leap year. Although 1908 is a leap year, the period ends on January 10th, so its February 29th is not crossed.
3
Calculate the total number of odd days and determine the final day of the week.
12 standard odd days + 2 leap odd days = 14 odd days. 14 divided by 7 leaves a remainder of 0. Friday + 0 days = Friday.
Adding the accumulated odd days to the original day provides the target day of the week. A remainder of 0 indicates a complete cycle.

Key Concept

Century leap year rules and date boundary logic in calendars
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Topics

Clocks and Calendars

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