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Difficulty: Very hardClocks and Calendars

An analog 1212-hour clock is set to the exact correct standard time at 12:00 NOON on Sunday, February 26, 1896. This particular clock consistently gains exactly 44 minutes every 2424 hours. Assuming standard time follows the Gregorian calendar without any daylight saving adjustments, what time will this faulty clock display, and what will be the true day of the week, when exactly 88 standard years have passed (i.e., on February 26, 1904 at 12:00 NOON standard time)?

  1. 2:44 PM, TuesdayAnswer
  2. B
    2:48 PM, Wednesday
  3. C
    2:52 PM, Thursday
  4. D
    2:40 PM, Monday

Answer

The faulty clock will display 2:44 PM, and the true day of the week will be Tuesday.
Between February 26, 1896, and February 26, 1904, there is exactly one leap day crossed (February 29, 1896). The year 1900 is not a leap year due to the century rule, and the period ends before February 29, 1904. This results in precisely 29212921 elapsed days. Adding 22 odd days (2921(mod7)2921 \pmod 7) to Sunday determines the true day is Tuesday. The clock gains 1168411684 minutes (2921×42921 \times 4), which translates to 1616 full 1212-hour cycles plus an extra 164164 minutes (22 hours and 4444 minutes). Advancing 12:00 NOON by this remainder yields exactly 2:44 PM.

Step-by-Step Solution

1
Calculate the total number of true days elapsed between February 26, 1896, and February 26, 1904.
Identify that the period covers exactly 8 years, but requires careful evaluation of leap days.
Calendar and clock drift problems require the exact number of 24-hour periods that have passed.
2
Determine how many leap days (February 29ths) fall inside this exact date range.
Exactly 1 leap day is included (February 29, 1896).
1896 is a leap year and its leap day occurs after Feb 26. 1900 is a century year not divisible by 400, so it is NOT a leap year. 1904 is a leap year, but the period ends on Feb 26, before its leap day occurs.
3
Compute the total elapsed days.
8 years×365 days+1 leap day=2921 days8 \text{ years} \times 365 \text{ days} + 1 \text{ leap day} = 2921 \text{ days}.
This establishes the precise duration for both the day-of-week shift and the total minutes gained by the clock.
4
Calculate the true day of the week.
Tuesday.
2921(mod7)=22921 \pmod 7 = 2 odd days. Adding 2 days to the starting day (Sunday) yields Tuesday.
5
Calculate the total time gained by the faulty clock.
2921 days×4 minutes/day=11684 minutes2921 \text{ days} \times 4 \text{ minutes/day} = 11684 \text{ minutes}.
The clock continuously gains 4 minutes every 24 hours over the exact elapsed duration.
6
Determine the time displayed on the 12-hour analog dial.
2:44 PM.
A 12-hour clock resets every 720 minutes. 11684(mod720)=16411684 \pmod{720} = 164 minutes. 164164 minutes is exactly 2 hours and 44 minutes. Adding this shift to 12:00 NOON gives 2:44 PM.

Key Concept

Integration of Calendar Leap Exceptions with Continuous Clock Drift
Estimated Time:4m 0s
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Topics

Clocks and Calendars

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