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?

1. 3
2. 4
3. 5
4. 1

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 birthday – 6 odd days
2032 birthday – 11 odd days = 4 odd days
2036 birthday – 9 odd days = 2 odd days
2040 birthday – 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 29^th would be a Wednesday.

So, there are 4 occurrences of birthday falling on Wednesday – 2040, 2068 and 2096, 2108.

Answer Choice (B)