Question:medium
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)
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)
Updated On: Jul 16, 2026
Show Solution
Correct Answer: 2024
Solution and Explanation
Step 1: A common year shifts a date's weekday by 1 day, a leap year by 2 days, once its own Feb 29 has passed.
Step 2: From 1996 to 2024 is 28 years; the leap years among 1996-2023 are 1996, 2000, 2004, 2008, 2012, 2016, 2020 — 7 leap years, 21 common years.
Step 3: Total odd-day shift \( = 21(1) + 7(2) = 35 \), and \( 35 \div 7 = 5 \) exactly, so the net shift is 0 days — every date lands back on its original weekday. \[ \boxed{2024} \]
Step 2: From 1996 to 2024 is 28 years; the leap years among 1996-2023 are 1996, 2000, 2004, 2008, 2012, 2016, 2020 — 7 leap years, 21 common years.
Step 3: Total odd-day shift \( = 21(1) + 7(2) = 35 \), and \( 35 \div 7 = 5 \) exactly, so the net shift is 0 days — every date lands back on its original weekday. \[ \boxed{2024} \]
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