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The roots of a card trick
The roots of a card trick

Hindustan Times

time14-07-2025

  • Entertainment
  • Hindustan Times

The roots of a card trick

By a happy coincidence, just weeks after we discussed digital roots and how they can help in determining if a number is a perfect square, I came across the concept again. To recap, the digital root is the result of repeatedly adding the digits of a number until you are left with a single digit. For example, the digital root of 2697 is obtained by adding 2 + 6 + 9 + 7 = 24, then 2 + 4 = 6. Welcome to Problematics! (Shutterstock) The late American puzzler and author Martin Gardner describes a card trick based on digital roots. The trick was invented by Stewart James, a magician from Ontario. It is very easy to see why it works, but deducing that should be fun. You may also enjoy playing the trick on an audience. #Puzzle 151.1 From a standard deck of 52, a magician selects and separates nine cards, and orders them in the following sequence: ace (top)-2-3-4-5-6-7-8-9 (bottom). He shows the audience these cards, telling them the ace represents 1. Now holding them face down, the magician pretends to cut the nine cards randomly, but what he actually does is bring the bottom three cards to the top. The order, which is not shown to the audience, is now 7 (top)-8-9-1-2-3-4-5-6 (bottom). The magician now places the pile of nine cards and the remainder of the deck side by side, and invites a member of the audience on stage. 'I will transfer cards, one by one, from the pile of nine to the main deck. You may select any card being transferred, calling out 'Stop!', at which stage I shall stop the transfer and display the selected card,' he tells the spectator. Just to make sure there is no misunderstanding, let's take an illustrative example. The magician starts to transfer cards from the pile of nine to the main deck of 43. Let us say he has transferred four cards (7-8-9-1) when the spectator says 'Stop!' on the fifth card. The magician turns up the fifth card, which is 2. He places the 2 face-up on the remainder of the pile (the cards that have not been transferred). The magician invites two more spectators. One of them is asked to cut the deck (the set not containing the face-up card) into two new piles, randomly. Let us say the spectator cuts the deck (which now has 47 cards after the transfer) into piles of 18 and 29 cards. The third spectator is asked to count the cards in each pile, to add these digits until she is left with a single digit. She adds as follows: 1 + 8 = 9, and 2 + 9 = 11, then 1 + 1 = 2. She then adds these two roots to get 9 + 2 = 11, then 1 + 1 = 2.. Bingo! The result matches the face-up card selected by the first spectator. To reiterate, the above is only an illustrative example. No matter which card the first spectator selects, and no matter how the second spectator splits the deck into two piles, the prescribed mathematical steps will always lead to a digital root that is the same as the selected card. What is the mathematical principle that makes the trick work? #Puzzle 151.2 An ace, a king, a queen and a jack are taken from a standard deck of cards. Each of them belongs to a different suit, so that all four suits are represented. They are now laid side by side in a row. Your clues: The king is to the right of the club. The club is to the right of the ace. The spade is to the left of the ace. The heart is to the left of the jack. The diamond is to the right of the jack. Name the four cards in order from left to right. MAILBOX: LAST WEEK'S SOLVERS #Puzzle 150.1 Dear Kabir, The first puzzle last week seemed tough at first, but I was able to solve it easily after I wrote down the prime factors of 19800, grouped these factors in all possible combinations and rearranged the corresponding letters. The two possible names of Bloke's wife are: DIVYA and VIDYA. — Y K Munjal, Delhi #Puzzle 150.2 Hi Kabir, The two given statements are paradoxical — each statement contradicts itself. — Professor Anshul Kumar, Copenhagen *** Another example of an invalid statement is: "I am a liar", which if true, would nullify me being a liar, and if false, would indicate that I am lying. — Sampath Kumar V, Coimbatore Apologies to Yadvendra Somra for omitting his name last week. He had correctly solved Puzzle 149.1 Solved both puzzles: Y K Munjal (Delhi), Professor Anshul Kumar (Copenhagen), Sampath Kumar V (Coimbatore), Yadvendra Somra (Sonipat), Kanwarjit Singh (Chief Commissioner of Income-tax, retired), Dr Sunita Gupta (Delhi), Shri Ram Aggarwal (Delhi), Abhinav Mital (Singapore), Ajay Ashok (Delhi), Sabornee Jana (Mumbai), Shishir Gupta (Indore), Aditya Krishnan (NMIMS Mumbai) Solved #Puzzle 150.1: Anil Khanna (Ghaziabad) Solved #Puzzle 150.2: Dr Vivek Jain (Baroda) Problematics will be back next week. Please send in your replies by Friday noon to problematics@

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