One of the many interesting aspects about time measurement is the concept of leap years. We have 365 days in non-leap years because that is the closest integer to the period of one orbit of the Earth. But since the actual orbital period is closer to 365.25 days, that makes 4 x 365.25 = 1461 days in 4 years instead of 4 x 365 = 1460, which makes it necessary to have a 366-day calendar (with February 29) about once every four years.
It would have been fine to have exactly one leap year in four years if the orbital period were exactly 365.25 days (365 days 6 hours), but it is not really so: the true period is 365 days, 5 hours, 48 minutes, 46 seconds. The correction required is 4 x 5:48:46 = 23:15:04 hours, so 1 leap day in 4 years is an overcorrection averaging to about 11 seconds per year.
After leap years were first introduced in 46 BC, it led to an accumulation of the error starting, and it took centuries before a solution was devised. Some historical details you may find interesting are discussed in an article I wrote during the leap month of February 2024.
The rules for leap years are now established: we need 97 of them in a span of 400 years. So we have one leap year every 4 years, but then remove 3 of them across 400 years. The last year of a century ends in 00 and is therefore a multiple, but it is not a leap year unless it is a multiple of 400 (eg 2000). Across 4 centuries, gives us three multiples of 4 that are non-leap years. In the last 400 years, the three non-leap multiples of 4 have been 1700, 1800 and 1900.
{{/usCountry}}The rules for leap years are now established: we need 97 of them in a span of 400 years. So we have one leap year every 4 years, but then remove 3 of them across 400 years. The last year of a century ends in 00 and is therefore a multiple, but it is not a leap year unless it is a multiple of 400 (eg 2000). Across 4 centuries, gives us three multiples of 4 that are non-leap years. In the last 400 years, the three non-leap multiples of 4 have been 1700, 1800 and 1900.
{{/usCountry}}#Puzzle 167.2
The first day of the 21st century (January 1, 2001) was a Monday, and the first day of the 22nd century (January 1, 2101) will be a Saturday. There are certain days of the week that can never be the first day of any century (January 1, xx01).
Which are these days, and why can they never come at the beginning of a century?
#Puzzle 167.2
If you throw three dice, we know there are 216 possible permutations that they can fall in, but with only 16 possible totals on the top face, ranging from 3 to 18. A stall at a gambling fair allows visitors to bet on any two numbers as the possible total. If the thrown total matches any of the chosen two, the customer wins; otherwise the stall-keeper collects the amount betted.
A couple visiting the fair try their luck. The husband chooses 7 and 13. He will win if the dice thrown show, for example, (2, 4, 1) or (6, 3, 4) among various possible combinations. His wife chooses two different numbers, neither of which is 7 or 13. Her probability of winning, however, is exactly the same as her husband’s chances.
What pair of numbers has the wife chosen?
MAILBOX: LAST WEEK’S SOLVERS
#Puzzle 166.1
The figure above and the solution below are from two different solvers, as credited. I have changed one of the variables in Professor Anshul’s drawing so that it matches Amarpreet’s notations.
Hi Kabir,
The amusement park is 20 km from Hotel B and 36 km from the two families’ home.
The lake is 9 km from A and 16 km from B. But A & B are 25 km apart on the same road (given). Therefore, the lake must be on road AB.
In triangle ALP, 9² + 12² = 15², therefore ALP is a right-angled triangle at L. This means BLP too must be a right-angled triangle and PB = √(12² + 16²) = √400 = 20. So Hotel B is 20 km from the amusement park.
In triangle APB, 15² + 20² = 25², therefore APB is a right-angled triangle at P. Therefore, APC too must be a right-angled triangle and CP = √(39² – 15²) = √1296 = 36. So the amusement park is 36 km from home.
— Amarpreet, New Delhi
#Puzzle 166.2
Dear Kabir,
The Scottish novelist in the anagram (OUR BEST NOVELIST, SENOR) is ROBERT LOUIS STEVENSON. The puzzle was worded in such a way that the solution was very easy to guess or search. I just searched for "Scottish novelist" in Google; a list of about 50 Scottish novelists was the result.
— Y K Munjal, Delhi
Solved both puzzles: Amarpreet (Delhi), Professor Anshul Kumar (Delhi), YK Munjal (Delhi), Yadvendra Somra (Sonipat), Dr Sunita Gupta (Delhi), Kanwarjit Singh (Chief Commissioner of Income-tax, retired), Shri Ram Aggarwal (Delhi), Anil Khanna (Ghaziabad), Vinod Mahajan (Delhi), Sabornee Jana (Mumbai), Ajay Ashok (Delhi), Shishir Gupta (Indore)
Solved #Puzzle 166.2: Dr Jeffrey R Geist (Columbus, Ohio), Nitin Trasi (Sydney)