Question:
John
was born on Feb 29th of 2012 which happened to be a Wednesday. If he
lives to be 101 years old, how many birthdays would he celebrate on a
Wednesday?
A. 3
B. 4
C. 5
D. 1
Correct Answer : (B)
Explanatory Answer:
Let
us do this iteratively. Feb 29th 2012 = Wednesday => Feb 28th
2012 = Tuesday
Feb
28th 2013 = Thursday (because 2012 is a leap year, there will be 2
odd days)
Feb
28th 2014 = Friday, Feb 28th
2015 = Saturday, Feb 28th 2016 = Sunday, Feb 29th 2016 =
Monday
Or,
Feb 29th to Feb 29th after 4 years, we have 5 odd days
So,
every subsequent birthday, would come after 5 odd days.
2016
birthday – 5 odd days
2020
birthday – 10 odd days = 3 odd days
2024
birthday – 8 odd days = 1 odd day
2028
– 6 odd days
2032
– 11 odd days = 4 odd days
2036
– 9 odd days = 2 odd days
2040
– 7 odd days = 0 odd days. So, after 28 years he would have a birthday on
Wednesday
The
next birthday on Wednesday would be on 2068 (further 28 years later), the one
after that would be on 2096. His 84th birthday would again be a leap
year.
Now,
there is a twist again, as 2100 is not a leap year. So, he does not have a
birthday in 2100. His next birthday in 2104 would be after 9 odd days since
2096, or 2 odd days since 2096, or on a Thursday.
From
now on the same pattern continues. 2108 would be 2 + 5 odd days later = 7 odd
days later. Or, 2108 Feb 29th would be a Wednesday.
So,
there are 4 occurrences of birthday falling on Wednesday – 2040, 2068 and 2096,
2108.
Level
of Difficulty - 2