There was a time when we had coins of denominations 1, 2 and 3 paise. Coins of denominations 5 and 10 paise lasted longer than 1, 2 and 3, and somewhere in between we had 20-paisa coins in orange circles and later in dull grey hexagons. Coins of 25 paise and 50 paise remained in circulation much longer than any of the above, although we rarely see them now. It is the 25p coins we will concern ourselves with today.
Another thing that has gone out of vogue, and rightly so, is gambling fairs. They were probably illegal even in the 1970s when I first saw them, but I remember them thriving until at least the 1990s. School fetes offered a middle path: the visitor would pay money to play a game but the prize would be in the form of a toy or any other item but not a cash prize. That way, the school student hosting the fair would earn money but a winning player would get something more than his or her money’s worth.
The following puzzle describes an unapologetically gambling game at a fair rather than a disguised one at a school fete. The game was common in many gambling fairs. The odds are, of course, stacked heavily against the visitor playing the game.
#Puzzle 169.1
{{/usCountry}}#Puzzle 169.1
{{/usCountry}}In a stall at a gambling fair, a table stands at a distance from the edge where visitors stand. It is covered by a tablecloth with a square pattern, each square measuring 3.5cm x 3.5cm. You play by tossing a 25-paisa coin at the table. If you have no 25-paisa coins, the stall will be happy to provide you with change for your 1-rupee coin.
Your object is to toss the coin so that it lands completely within any of the squares on the tablecloth. “Completely” means that no part of the coin touches any edge of any square. If you manage that, you get back your coin and win another coin from the stall-keeper. If your coin touches the edge(s) of any one or more squares, you lose the coin. If your coin does not land on the table at all and falls off, you are allowed to toss it again.
What is the probability that you will win? Take a 25-paisa coin as being 2cm in diameter.
#Puzzle 169.2
aᵇ x cᵃ = abca
In the above equation, a, b and c represent three distinct digits, and abca is a four-digit number with the digits in that order.
Replace the letters with digits to make the equation correct.
MAILBOX: LAST WEEK’S SOLVERS
#Puzzle 168.1
Nice puzzle — the five scores are 54, 56, 58, 59, 62. Quick derivation: arrange the scores in increasing order as a₁, a₂, a₃, a₄, a₅. The smallest pair-sum 110 must be (a₁ + a₂), and the second smallest sum 112 is (a₁ + a₃), so (a₃ = a₂ + 2). The largest pair 121 must be (a₄ + a₅) and the next largest 120 is (a₃ + a₅), which leads to (a₄ = a₃ + 1) and (a₂ + a₅) = 118. Checking which of the remaining given pair-sums fits (a₂ + a₃ = a₂ + a₂ + 2 = 2a₂ + 2) gives a₂ = 56. From that we get a₁ = 54, a₃ = 58, a₄ = 59, a₅ = 62. The ten pair-sums match the listed values.
— Vinod Mahajan, Delhi
#Puzzle 168.2
Many possible solutions exist to the “non-magic square” puzzle. The above six are a selection from combinations sent by readers. A couple of readers have sent grids in which one sum or the other repeats itself; these have obviously not been included.
Solved both puzzles: Vinod Mahajan (Delhi), Professor Anshul Kumar (Delhi), Kanwarjit Singh (Chief Commissioner of Income-tax, retired), Shishir Gupta (Indore), Sabornee Jana (Mumbai), Shri Ram Aggarwal (Palam, Delhi), YK Munjal (Delhi), Yadvendra Somra (Sonipat), Dr Sunita Gupta (Delhi)
Solved #Puzzle 168.1: Ajay Ashok (Delhi)