If there is one place that can successfully make you buy what you may not need, it is your child’s school. Selling uniforms and books is one thing, but many schools also force parents to buy shoes of chosen brands which could be procured elsewhere, not to mention stationery items ranging from fancy pens and novelty erasers to custom packages including the pen, the eraser and sundry other items.
Having crossed that stage of parenting years ago, I no longer have any reason to complain. So let us look at such a situation clinically, as puzzlers do. Many items and many children mean there are always interesting numbers on offer.
#Puzzle 201.1
A profit-minded school administrator is calculating and exacting. His school sells custom stationery packages at ₹100. It gets all the items from a wholesaler, who assembles each set free and charges only for the individual items, which come to ₹49.36 per set.
A fresh session is starting and the administrator sends a worker to the wholesaler to buy sets for a certain class (different classes are given packages). The worker pays in cash and the wholesaler makes out a bill. After the worker delivers the goods, the administrator accidentally spills some ink on the bill. This should not have been a bother at an age when records are stored electronically, but the administrator is the grumpy kind.
Part of the bill is legible, and part smudged in ink. Here is how it looks:
Received ₹XXX7.28 from YYYYYY YIGHER SECYYYYYY SCHYYL for XX stationery sets @ ₹49.36 each. Each X stands for a digit, and each Y for a letter. The letters are of no consequence to the puzzle.
How many stationery sets were sold, and what was the bill amount?
#Puzzle 201.2
Your World Cup puzzle for this week involves five anagrams:
GOOD DAM IN AREA
DULL MERGER
FRANK SEES CUP
GO GET BEERS
SHY ELVINA
Unscramble the above to get the names of five deceased football legends, with each name consisting of two words.
MAILBOX: LAST WEEK’S SOLVERS
#Puzzle 200.1
Hi,
From a starting position of 1, 3, 5 and 7 matches in four rows, the second player should make the following moves in the two given situations:
(a) When the first player removes the entire row of 7 matches, the second player should remove 3 matches from the row of 5, leaving rows of 1, 3, 2
(b) When the first player removes 4 matches from the row of 5, the second player should remove 4 matches from the row of 7, leaving rows of 1, 3, 1, 3
In both cases, the second player should ensure that, after the numbers in the rows are written in three-digit binary form, the total number of 1s in each binary column is even. This restores a losing position for the first player.
(a) In the first situation, after the first player removes the entire row of 7, the rows become 1, 3, 5, which in binary form are:
001
011
101
Column totals: 113 (every column has an odd number of 1s).
The second player needs to make one move that restores all column totals to even. If the second player reduces the row of 5 matches to 2, the rows will become 1, 3, 2, which in binary form are
001
011
010
Check: all column totals (022) are even. This is the unique optimal move.
(b) In the second situation, after the first player removes 4 matches from the row of 5, the rows become 1, 3, 1, 7, which in binary form are:
001
011
001
111
Column totals: 124 (one column has an odd number of 1s, two have an even number).
The second player again needs one move to make all column totals even. This is achieved by reducing the row of 7 to 3. This now leaves rows of 1, 3, 1, 3 matches, which in binary form are:
001
011
001
011
Check: all column totals (024) are even. This is again the unique optimal move.
— Ajay Ashok, New Delhi
The question asked last week was only about the second player’s immediate move after the first player’s opening. The strategy Ajay Ashok has described above is the way to go at that stage, and also in subsequent steps, but not until the end of the game. In the beginning, make sure that you leave your opponent with an even number of 1s in every binary column. Then there comes a stage when you change your strategy. When every remaining row has just one match, forget the binary rule. Simply leave your opponent an odd number of single-object rows. There are two such situations: 1 remaining match in any row, or 3 remaining matches in 3 different rows. Conversely, 2 rows with 1 match each gives an even number of 1s in all binary columns, yet your opponent is not in a losing situation when they face this. Yadvendra Somra sums all this up in binary language:
Dear Kabir,
In a nutshell, when the total number of 1s in each binary column is 0 or 2 or 4, the second player has placed the first player in a losing situation. The only exception is 002.
Conversely, when the total number of 1s in any binary column is 1 or 3, the second player has not placed the first player is a losing situation. The only two exceptions are 003 and 001.
— Yadvendra Somra, Sonipat
#Puzzle 200.2
Hello Kabir,
The match results are:
Mexico v South Korea: 0–0
Mexico v Czech Republic: 2–0 (Mexico won)
Mexico v South Africa: 2–0 (Mexico won)
South Korea v Czech Republic: 0–0
South Korea v South Africa: 2–1 (South Korea won)
Czech Republic v South Africa: 1–1
— Sunita Gupta, New Delhi
***
These results match the final standings in Group 1 of the 1966 FIFA World Cup, which consisted of Uruguay, hosts England, France and Mexico.
— Vinod Mahajan, New Delhi
The actual results in 1966 were (England 0 Uruguay 0), (England 2, Mexico 0), (England 2 France 0),
(Uruguay 0 Mexico 0), (Uruguay 2 France 1) and (Mexico 1 France 1). Mexico is the common team between the real 1966 group and the fictional group in the puzzle, finishing third 60 years ago and topping the group in Problematics.
Apart from Vinod Mahajan, the only other reader to identify the 1966 group is Ajay Ashok who, however, miscalculates a couple of the scores. The lists below acknowledge solvers of #200.2 only if they have given the correct goal score for each match.
Solved both puzzles: Yadvendra Somra (Sonipat), Dr Sunita Gupta (Delhi), Vinod Mahajan (Delhi), Sabornee Jana (Mumbai)
Solved #Puzzle 200.1: Ajay Ashok (Delhi), Professor Anshul Kumar (Delhi)
Solved #Puzzle 200.2: Shishir Gupta (Indore), Shri Ram Aggarwal (Delhi)
