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Problematics | Women batting, two by two
This week, a cricket-themed puzzle following the World Cup, and a magic square that isn’t.
India’s triumph in the ODI World Cup came just over a week ago, recent enough to theme a puzzle on women’s cricket. To be honest, the puzzle below (adapted from Henry Dudeney’s infinite resources) would work equally well for men’s cricket, but let’s not fuss over trivial matters. Let’s just note that we are looking at the scores of women batters — or rather pairs of women — and see how they break up.
#Puzzle 168.1
Ahead of national selection, 22 women players are training in a camp. Apart from regular and improvised training exercises, the camp includes a practice match with the 22 players split into two teams of 11 each. The batters score runs and the bowlers take wickets. The coach is happy that all the top five batters have scored well, but is disappointed that none has got a century, for which he credits his bowlers.
The coach notes, however, that no pair of batters has a combined score of less than 100. Being something of a puzzler, he uses pen and paper to combine the five batters into all 10 possible pairs and adds their scores: 110, 112, 113, 114, 115, 116, 117, 118, 120, 121.
What are the five individual scores?
#Puzzle 168.2
The illustration shows a magic square. We all know how these works: the distinct numbers in all three rows, all three columns, and both diagonals add up to the same total. In this example, using the distinct numbers 1 to 9, the common total is 15.
Can you rearrange the same 9 numbers in a way that the sums obtained from the three rows, three columns and two diagonals are all different? (If you find more than one solution, send any one.)
MAILBOX: LAST WEEK’S SOLVERS
#Puzzle 167.1
Dear Kabir,
In a century not having any year divisible by 400, there are 24 leap years and therefore (365 x 76) + (366 x 24) = 36524 days. This is 5217 and 5 days. Therefore as 01.01.2001 was a Monday (1st day of week), 01.01.2101 will be a Saturday (1 + 5 = 6th day of week). Proceeding further, 01.01.2201 will be 6 + 5 = 11, or 11 – 7 = 4th day of the week i.e. Thursday. And then 01.01.2301 will be 4 + 5 – 7 = 2nd day of the week, i.e. Tuesday. Now 2400 will be a leap year because it is divisible by 400. In this century there will be one day extra i.e. 36525 days, or 5217 weeks and 6 days. Therefore 01.01.2401 will be 2 + 6 – 7 = 1st day of the week, i.e. Monday again. Hereafter the cycle repeats itself for the next four centuries and so on.
Hence, we see that Monday, Saturday, Thursday and Tuesday are getting repeated in this order as the first day of a century. So, we conclude that the century will never begin on a Wednesday, Friday or Sunday.
— Yadvendra Somra, Sonipat
***
Hi Kabir,
I noticed that even 97 days as leap days in a span of 400 Years is an overcorrection of 2 hours 53 minutes 20 seconds, or approximately 3 hours. So a correction of one day is required in a span of 24/3 × 400= 3200 years.
— Shishir Gupta, Indore
(Indeed, no correction can ever be perfect. But 1 day in 3200 years is probably insignificant. At least the makers of our calendar appear to have thought so — Kabir)
#Puzzle 167.2
Hi Kabir,
The probability of getting a particular total after three throws of dice is proportional to the number of permutations resulting in that total. Note that for every permutation (x, y, z) of three dice adding up to [x + y + z], there is a permutation (7 – x, 7 – y, 7 – z) that adds up to [21 – (x + y + z)]. This means that the number of permutations resulting in a total of 7 is equal to the number of permutations resulting in a total of 21 – 7 = 14, and the number of permutations resulting in a total of 13 is equal to the number of permutations resulting in a total of 21 – 13 = 8. Therefore, we can conclude that the wife has chosen the numbers 8 and 14. On a funny note: although the wife has chosen to be “one up” over her husband, the probability of her winning is the same!
— Professor Anshul Kumar, New Delhi
Solved both puzzles: Yadvendra Somra (Sonipat), Shishir Gupta (Indore), Professor Anshul Kumar (Delhi), Kanwarjit Singh (Chief Commissioner of Income-tax, retired), Vinod Mahajan (Delhi), Sabornee Jana (Mumbai), Dr Sunita Gupta (Delhi), Amarpreet (Delhi), YK Munjal (Delhi)
Solved Puzzle #167.2: Nitin Trasi (Sydney)
Problematics will be back next week. Please send in your replies by Friday noon to problematics@hindustantimes.com.
