Question 25

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)

Correct Answer: 2024

Solution

Every non-leap year have 365 days. i.e (7x52+1) days.

Leap year have 366 days i.e ( 7x52+2) days

A cycle of four years will have 1+1+1+2 days i.e 5 days shift

For the days to repeat, the shift should be a multiple of 7.

A cycle of four years if repeated 7 times will have a 35 days shift. Every day will repeat after 4x7 (28 years).

After 1996 the cycle will be in 1996+28 = 2024

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