**Question:**
John
was born on Feb 29

^{th}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 29

^{th}2012 = Wednesday => Feb 28^{th}2012 = Tuesday
Feb
28

^{th}2013 = Thursday (because 2012 is a leap year, there will be 2 odd days)
Feb
28

^{th}2014 = Friday, Feb 28^{th}2015 = Saturday, Feb 28^{th}2016 = Sunday, Feb 29^{th}2016 = Monday
Or,
Feb 29

^{th}to Feb 29^{th}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 84

^{th}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 29

^{th}would be a Wednesday.
So,
there are 4 occurrences of birthday falling on Wednesday – 2040, 2068 and 2096,
2108.

**Level of Difficulty - 2**

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