»  Solutions to puzzles in my National Review Online Diary

  October 2010


In my October diary I posed the following brain-teaser.

According to the London Sunday Times October 31 edition:  "Burma's military dictators have turned to soothsayers, astrologers and numerologists to help secure victory in national elections next Sunday [i.e. November 7]. Citizens were bewildered when a new national flag was hoisted on Oct. 21 at 3 p.m. — a highly auspicious moment as the digits of the day, the time and the year all add up to three, and together equal the lucky number nine."

Digits of the day (2+1 = 3), the time (3 = 3), and the year (2+0+1+0 = 3) … yep, it all works out.

OK, here's the puzzle:  What's the probability that the particular hour passing over us satisfies the "Burmese condition"?

Let me be more explicit. We'll consider only years in the range from a.d. 2001 to a.d. 3000. That's 1,000 years even. Multiply by 365, that's 365,000 days. Add 250 for the years divisible by four (2004, 2008, 2012, … 3000), it's 365,250 days. Subtract for the 8 century years when you don't get a leap day on the Gregorian calendar (bypassing 2400 and 2800, when you do), and I think the final total is 365,242 days. Multiply by 24 and you have 8,765,808 hours.

How many of those hours — and then, what proportion — satisfy the Burmese condition? Use a 12-hour clock, since that seems to be the one favored by the Burmese junta.



Way too easy. The hour condition is satisfied four times a day: at 12am, 3am, 12pm, and 3pm. The day condition is satisfied on the 3rd, 12th, and 21st of every month, and on the 30th of every month except February, for a total 47 times a year. The year condition is satisfied by only four years in the range specified: 2001, 2010, 2100, and 3000. Solution: 4 × 47 × 4 = 752. The ratio is about 1 in 11,657 overall, though of course much higher in years satisfying the year condition: almost 1 in 47 for this year.