Problematics | Multiplying with playing cards
Here is an old party trick in which playing cards magically reveal the result of a multiplication exercise. What is the secret behind it?
Neither of this week’s puzzles is very difficult. The second, in fact, is a sitter, although it may appear impossible to solve at first glance. The first puzzle too should be easy to get but it requires a little thought.
It is based on a party trick invented in 1942 by Lloyd Jones of California (my source: Martin Gardner). Although the trick is not difficult to see through, I thought readers would enjoy exploring the way it works, and so turned it into a puzzle. It is the kind in which all the fun comes in the process of solving it.
#Puzzle 109.1
A magician hands you six cards, all red: 2, 3, 4, 5, 6. He shows you six black cards in his hand: A, 2, 4, 5, 7, 8. I mention these in increasing order of rank, with the Ace counting as 1, but this is not the order in which the magician holds the cards. In fact, he has prearranged the cards without making it obvious.
The magician cuts his set of six cards and lays them out on the table, face up, in the following order:
A, 4, 2, 8, 5, 7
This is very much deliberate, the exact order the magician wants. After showing you a different order, he has cut the set at the precise card to bring the cards to the above arrangement, forming the number 142857.
The magician now asks you to select any one of your six cards and place it face up below your six cards. Suppose you select the 5 from among your six cards.
“Take the six-digit number formed by my six black cards and multiply it by the number represented by your single card,” the magician tells you.
While you bring out your calculator, the magician picks up the six black cards and cuts them, seemingly at random (but at a deliberately chosen card, of course). He now lays out the six cards again, but face down this time.
Meanwhile, you have completed the multiplication and announce it: 142857 x 5 = 714285.
Now, the magician delivers the punch. He turns the black cards face up again, and you find, to your surprise:
7, A, 4, 2, 8, 5
As you can see, this order is easily achieved with a single cut to the previous face-up sequence (A, 4, 2, 8, 5, 7). The trick will work whether you choose to multiply 142857 with 2, 3, 4, 5, or 6 (i.e. with any of the numbers represented by the red cards). The magician will show you the right product each time.Neither of this week’s puzzles is very difficult. The second, in fact, is a sitter, although it may appear impossible to solve at first glance. The first puzzle too should be easy to get but it requires a little thought.
It is based on a party trick invented in 1942 by Lloyd Jones of California (my source: Martin Gardner). Although the trick is not difficult to see through, I thought readers would enjoy exploring the way it works, and so turned it into a puzzle. It is the kind in which all the fun comes in the process of solving it.
#Puzzle 109.1
A magician hands you six cards, all red: 2, 3, 4, 5, 6. He shows you six black cards in his hand: A, 2, 4, 5, 7, 8. I mention these in increasing order of rank, with the Ace counting as 1, but this is not the order in which the magician holds the cards. In fact, he has prearranged the cards without making it obvious.
The magician cuts his set of six cards and lays them out on the table, face up, in the following order:
A, 4, 2, 8, 5, 7
This is very much deliberate, the exact order the magician wants. After showing you a different order, he has cut the set at the precise card to bring the cards to the above arrangement, forming the number 142857.
The magician now asks you to select any one of your six cards and place it face up below your six cards. Suppose you select the 5 from among your six cards.
“Take the six-digit number formed by my six black cards and multiply it by the number represented by your single card,” the magician tells you.
While you bring out your calculator, the magician picks up the six black cards and cuts them, seemingly at random (but at a deliberately chosen card, of course). He now lays out the six cards again, but face down this time.
Meanwhile, you have completed the multiplication and announce it: 142857 x 5 = 714285.
Now, the magician delivers the punch. He turns the black cards face up again, and you find, to your surprise:
7, A, 4, 2, 8, 5
As you can see, this order is easily achieved with a single cut to the previous face-up sequence (A, 4, 2, 8, 5, 7). The trick will work whether you choose to multiply 142857 with 2, 3, 4, 5, or 6 (i.e. with any of the numbers represented by the red cards). The magician will show you the right product each time.
How does the trick work?
#Puzzle 109.2
There are five apples in a basket and you want to distribute them among five girls so that each girl gets an equal number of apples. One apple must, however, remain in the basket.
How do you manage this?
MAILBOX: LAST WEEK’S SOLVERS
#Puzzle 108.1
Hello Kabir,
This one is full of nostalgia. When I was a kid, living on what was then known as Wellsley Road, my dad had a wire contraption of a bread toaster. It took two slices, but did only one side each at a time. One had to open the lids and flip the slices each time. Your puzzle reminded me of that era, over 50 years ago.
Of that ancient toaster, let’s call one side X and the other Y. Let the toasts be numbered 1 through 3. For toasting 3 slices for 3 minutes per side, the most efficient time possible is: (3 toasts x 2 sides each x 3min each side)/2 sides of the toaster = 9min. The time table is as shown in the table.
— Sanjay Gupta, Delhi
#Puzzle 108.1 (b)
Hi Kabir
Total toasting time required =2*2 + 3*2 + 4*2 = 18 minutes. If we ensure toasting two sides together as above, the minimum time required will be18/2 = 9 minutes. By toasting the slices as shown in the table, we can achieve this.
— Yadvendra Somra, Sonipat
#Puzzle 108.2
Hi Kabir,
The diagonally opposite squares at the corners of a chessboard are of the same colour, either white or black. Say we remove the white corner squares. Now, 30 white and 32 black squares remain on the board. The cardboard strip measuring 2 x 1 covers 1 white square and 1 black square on the chessboard. So, 30 cardboard strips can cover 30 white and 30 black squares, leaving 2 black squares, which is impossible to cover with 1 cardboard strip, as 1 strip can cover only 1 white square and 1 black square.
— Shishir Gupta, Indore
Solved both puzzles: Sanjay Gupta (Delhi), Yadvendra Somra (Sonipat), Shishir Gupta (Indore), Dr Vivek Jain (Baroda), Dr Sunita Gupta (Delhi), Professor Anshul Kumar (Delhi)
Also solved both (almost): Anil Khanna (Ghaziabad), Sabornee Jana (Mumbai), Shri Ram Aggarwal (Delhi), Sanjay S (Coimbatore), Ajay Ashok (Delhi), YK Munjal (Delhi), Raghunathan Ravindran (Coimbatore), Aditya Krishnan (Coimbatore), Kanwarjit Singh (Chief Commissioner of Income-Tax, retired)
Those named in the second list solved the first part of the toaster puzzle correctly but described procedures that use 10 minutes (instead of 9) for the second part. Aditya Krishnan rightly said 9 minutes but did not describe the procedure. All these readers also solved the chessboard puzzle.
Problematics will be back next week. Please send in your replies by Friday noon to problematics@hindustantimes.com.
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