Question

There were 12 friends A, B, C, D, E, F, G, H, I, J, K, & L. In the year 1996, A celebrated his birthday on January 11 and it was Thursday. B celebrated his birthday on February 20, which was a Tuesday and C, D, E, F, G, H and I, celebrated on April 05 (Friday), May 05 (Sunday), June 05 (Wednesday), July 05 (Friday), August 05 (Monday), September 05 (Thursday) and October 05 (Saturday) respectively. J, K and L celebrated their birthday on November 15 (Friday), March 15, (Friday) and December 15 (Sunday) respectively. Before the year 2025, when will all of them celebrate their birthdays again on the same day as they did in 1996? (Note:- DO NOT include spaces in your answer)

Answer 1

Correct Answer: 2024

To determine the next year when all friends' birthdays will fall on the same day of the week as they did in 1996, we need to consider the pattern of the Gregorian calendar and leap years.
Here's a step-by-step process to solve the problem:
1. Identifying Key Information:
- A's birthday on January 11 was a Thursday in 1996.
  - B's birthday on February 20 was a Tuesday in 1996.
  - The other friends' birthdays and their respective weekdays in 1996 are provided.
2. Analyzing the Days of the Week in 1996:
  - 1996 is a leap year. In a leap year, February has 29 days.
  - Days of the week repeat every 7 days.
3. Finding the Cycle of Repetition:
  - The calendar for a non-leap year repeats itself every 6 or 11 years. For a leap year, the repetition cycle is different.
  - The complete pattern of day and date repeats every 28 years for the Gregorian calendar, accounting for the leap year cycle.
4. Calculating the Next Same Day:
  - Add 28 years to 1996 to determine when the same day-date combination will occur again.
\( 1996 + 28 = 2024 \)
Thus, all friends will celebrate their birthdays on the same day of the week as in 1996 in the year 2024

Answer 2

Instead of quoting the general "calendar repeats every 28 years" rule, verify the repeat year directly by counting odd days — the standard technique for finding when a weekday-date pairing recurs.

1. Rule used: Moving from one year to the next, a common (non-leap) year shifts every date's weekday by 1 day, and a leap year shifts it by 2 days, but only once its own February 29 has actually passed in the calendar.
2. Dates falling after February 29 (C, D, E, F, G, H, I in April through October, and J, K, L in November, March, December): from 1996 to a candidate year, every year from 1996 up to the year before it contributes its full odd-day count (1 for common years, 2 for leap years), because the leap day of each of those years has already been crossed by the time the same date recurs.
3. Dates falling before February 29 (A on January 11, B on February 20): the leap day of the destination year itself has not yet occurred by January or February, so only the years 1996 up to the year before it contribute, exactly as above — meaning both groups of dates realign in the same year.
4. Counting from 1996 to 2024 (the 28 years 1996 up to 2023 each contribute): the leap years in this span are 1996, 2000, 2004, 2008, 2012, 2016 and 2020 — 7 leap years — and the remaining 21 years are common years. Total odd days \( = 21 \times 1 + 7 \times 2 = 21 + 14 = 35 \). \( 35 \div 7 = 5 \) exactly, so the total shift is a whole number of weeks — a shift of 0 odd days.

Since the accumulated weekday shift from 1996 to 2024 comes out to exactly 0 (a multiple of 7), every one of the twelve friends' birthdays falls back on the same day of the week as it did in 1996, and no earlier year before 2025 gives a zero remainder for all of them together.

Therefore, the correct answer is 2024.