10th Putnam 1950

Problem B3

Leap years have 366 days; other years have 365 days. Year n > 0 is a leap year iff (1) 4 divides n, but 100 does not divide n, or (2) 400 divides n. n is chosen at random from the natural numbers. Show that the probability that December 25 in year n is a Wednesday is not 1/7.



365 = 52.7 + 1, so in non-leap years Christmas day is a day later in the week than the previous year. In leap years it is two days later. Now consider a period of 400 years. There are 100 multiples of 4, 3 or which are multiples of 100 but not 400, so there are 97 leap years. Thus after 400 years the day has advanced 303·1 + 97·2 = 497 days or exactly 41 weeks. In other words after 400 years Christmas is on the same day of the week. This cycle then repeats indefinitely. But 400 is not a multiple of 7, so each day cannot occur with exactly probability 1/7.

If we look in more detail at the 400 years we find that Su, Tu, Fr each occur 58 times, We and Th each occur 57 times, and Mo and Sa each occur 56 times (compared with the expected 57 1/7).



10th Putnam 1950

© John Scholes
5 Mar 2002