Play the magician

Hindustan Times

05-05-2025

Exactly 20 episodes ago, readers solved a card puzzle that I had found, as I mentioned at the time, among the works of the English puzzler Henry Dudeney. I felt it necessary to mention that just in case this week's card puzzle sounds familiar.
Indeed, there are similarities. In this puzzle, as in the last one, you are asked to count to 10 while you place a series of new cards on an exposed card. On both occasions, you count the court cards as 10. Rest assured, though, that the puzzle is quite different. Although you may (or may not) use the same kind of mathematical reasoning to solve it, that one came from Dudeney while the following one comes from the writings of Martin Gardner:
#Puzzle 141.1
A magician invites you to take part in a card trick he wants to play on you. He hands you a deck of 52 cards and asks you to shuffle thoroughly. You do so and hand the deck back to him.
The magician deals 12 cards from the top, one at a time, face down. From this pile, he asks you to select any four cards at random and place them side by side on the table. 'Place each card face up,' he says. You randomly turn up the 4 of clubs, 7 of diamonds, jack of hearts and ace of spades, and place them in a row. These are just illustrative examples; the trick will work with any four cards.
The magician picks up the remaining eight cards and restores them to the deck. Your observant eye notes that he has placed these cards at the bottom of the deck, not the top. This must be significant, you guess.
The magician hands the deck to me. 'Deal more cards face down on the top of each face-up card. Count to 10 while you deal, starting with the number immediately higher than the pip value of the face-up card. For example, on top of the 4 of clubs, you count 5, 6, 7, 8, 9, 10, and stop when you reach 10,' he instructs you.
But what about the jack of hearts, you want to know. 'Count all court cards (jacks, queens and kings) as 10, so nothing needs to be added to this jack,' he replies.
So you deal cards on the 4 of clubs (counting '5, 6, 7, 8, 9, 10'), then on the 7 of diamonds ('8, 9, 10') and finally on the ace of spades ('2, 3, 4, 5, 6, 7, 8, 9, 10'). You deal nothing atop the jack of hearts, which is already assumed to have a pip value of 10.
'Add the numbers on the face-up cards. Court cards are 10, as previously stated,' the magician says, and you add: '4 + 7 + 10 + 1 = 22.'
'What is the 22nd card in the deck you hold?' the magician teases you, then answers his own question: 'It's the six of diamonds. Check and see.'
You check by dealing the cards from the deck and counting: '1, 2, 3, 4… 21.' You then turn the 22nd card face upwards. Indeed, it is the six of diamonds.
You make an informed guess: 'This appears to be a mathematical trick. After I had shuffled the deck in the very beginning and handed it back, you quickly looked at one card in a certain position, knowing that the subsequent operations would bring this card to the exact position derived from the sum of the pip values.'
The magician agrees. 'You are right, but can you identify which card I silently observed?'
#Puzzle 141.2
'My turn now,' you say, offering to play a trick on the magician. After he agrees, you give him your instructions:
(1) Write down the year of your birth.
(2) Write down your age in years on your birthday in 2024.
(3) Write down the year of your wedding.
(4) Write down the number of years you had been married when you celebrated your anniversary in 2024.
(5) Write down the number of people in this room.
'Don't show me any of these five numbers,' you continue. 'Add them all up, but don't tell me the sum either. Just tell me when you're done.'
'Done,' says the magician.
'Your total is 4061,' you announce.
MAILBOX: LAST WEEK'S SOLVERS
#Puzzle 140.1
Hi Kabir.
The solution to the wages puzzle is shown in the table. Let us take s, w₁, w₂ and w₃ as the years of experience of the supervisor and workers #1, #2 and #3 at t = 0 (15 years ago). At t = 5 (10 years ago), let the experience of the new worker #4 be w₄. And at t = 15 (today), let the experience of the new worker #5 be w₅. From the equations that are given and/or formed (see table), we can equate (4) and (5) to get:
w₃ = 2w₅ – 15
Now 0 < w₅ <10, and w₃ > 0., for which we need w₅ = 8 or 9. But from equation (1), we know s is an even number. So, from (3), w₅ must be odd. In other words, we need w₅ to be 9 and not 8. From the value w₅ = 9, we can derive s = 24, w₃ = 3, w₄ = 2 , w₁ = 6 and w₂ = 3. The completed years of experience will be (s + 15 = 39), (w₁ + 15 = 21), (w₂ + 15 = 18), (w₃ + 15 = 18), (w₄ + 10 = 12), and (w₅ = 9).
— Sabornee Jana, Mumbai
***
Worker 2's cryptic remark at the end (in response to worker 3's remark that worker #1 and worker #2 together earn as much as the sum of the other three workers' salaries) means that he could say the same about worker #1 and worker #3 together earning the same as the other three. Workers #2 and #3 earn the same amount.
— Kanwarjit Singh, Chief Commissioner of Income-tax, retired
#Puzzle 140.2
Hi Kabir,
It could be that the two sisters, although born to the same parents on the same day of the same month in the same year, are not twins because they are two among triplets or two among quadruplets etc.
— Dr Sunita Gupta, Delhi
***
Here's how it works. They are not 'twins' because this word means exactly two babies born from the same pregnancy. If there were three (or more) babies born to the same mother on the same day, the term would be triplets, quadruplets etc. So if two of the triplets are celebrating their fifth birthday, they are sisters, same age, born on the same day — but not twins, because they are part of a larger multiple birth.
— Vinod Mahajan, Delhi
Solved both puzzles: Kanwarjit Singh (Chief Commissioner of Income-tax, retired), Dr Sunita Gupta (Delhi), Sampath Kumar V (Coimbatore)
Solved #Puzzle 140.1: Sabornee Jana (Mumbai), Professor Anshul Kumar (Delhi), Yadvendra Somra (Sonipat)
Solved #Puzzle 140.2: Vinod Mahajan (Delhi), Aishwarya Rajarathinam (Coimbatore), Sanjay Gupta (Delhi), Ajay Ashok (Delhi)