Problematics | Cards on the table
Here is yet another trick with a deck of 52 that you can perform on a gullible audience. As usual, let us examine the mathematics that makes the trick work.
Over the last two years and more, card tricks have provided us with the fodder for many of our puzzles. This will continue, because cards are fun.

In the works of the English puzzler Henry Ernest Dudeney (1857-1930), I came across a trick that I had not noticed before. In keeping with the established traditions of Problematics, of course, I have packaged the puzzle differently. Dudeney tells us how the trick is performed, and wants us to determine why it works. In the following adaptation, packaged in my own story written exclusively for Problematics readers, you need to determine both how and why.
#Puzzle 121.1
A card trickster is showing me… well, a card trick. He hands me a deck of 52 cards, which I check to confirm that each suit, indeed, has the usual 13 cards.
“Consider each ace to be 1, and each jack, queen and king to be 10,” he tells me. “Now lay the cards into a number of piles as instructed. First, turn over the top card.”
It turns out to be a 5. “Pile up more cards on this one by one, counting them as 6, 7, 8 and so until you reach 12, the number at which you must stop.”
On top of the 5 I lay down 7 more cards as instructed, counting: “6, 7, 8, 9, 10, 11, 12.”
“That’s your first pile,” the trickster tells me. “Now start your second pile by turning over the top card from the 44 cards remaining in your hand.”
This turns out to be a Q, which is to be considered 10. I lay down two more cards on top of it, counting: “11, 12.”
We continue in this way, making new piles until I have a small handful of cards remaining. When I look at the top card, it turns out to be a 4, but I do not have enough cards to make a pile counting up to 12. I point that out to him.
“That means you now stop making piles. How many piles and how many cards remaining?”
“As you can see,” I reply, “I have made 6 piles and have 5 cards remaining in my hand.”
“Look at the cards at the bottoms of the 6 piles, with which you began the piles. Their sum is 31,” the trickster says.
I scoff at him: “You saw me make all the piles. You must have added up the bottom cards as they emerged.”
“All right then, I shall turn my back while you go through the same process again,” the trickster says. “Remember, when you reach a stage when you can no longer count up to 12 from the card turned over, you stop, and the cards remaining in your hand may be called unused cards.”
I do as he instructs. Finally I tell him: “5 piles and 3 unused cards.”
“Then the sum of the bottom cards must be 16,” he says, correctly.
“How do you do it?” I wonder.
“Simple,” he says, in a manner as clichéd one would expect. “I perform two mathematical operations on the number of piles, and then to this result I introduce the number of unused cards by means of a third mathematical operation.”
What operations does he perform, and why does his trick work?
#Puzzle 121.2
Here is yet another puzzle that involves people who either always speak the truth or always lie. This time, there is no one who alternates between truth and falsehood; each of the three people you meet must be either consistently truthful or a consistent liar.
You: Are you truthful or a liar?
Person #1: (incoherent)
You: What did he say?
Person #2: He said he is a liar.
Person #3: Person #2 is a liar.
How many truthful persons and liars can you identify?
MAILBOX: LAST WEEK’S SOLVERS
#Puzzle 120.1
As expected, different readers used different methods to solve the puzzle of ages. Some also used hit and trial. Among the methods described, I found the following one the simplest.
Dear Kabir,
Let the ages of the mathematician’s mother, son, brother and daughter be a, b, c and d respectively. Given,
(a + b) + (a – b) + ab + a/b = 243; and
(c + d) + (c – d) + cd + c/d = 243
From the first equation, we get
2a + ab + a/b =243
=> a/b(2b + b² + 1) = 3⁵
=> (a/b) (1 + b)² = (27)*(3²) or (3)*(9²)
Therefore, either a/b = 27 and (1 + b) = 3, or a/b = 3 and (1 + b) = 9. The first option gives b = 2 and a = 54. The second gives b = 8 and a = 24 (which are actually d = 8 and c = 24).
Hence the ages are — mathematician's mother: 54, brother: 24, daughter: 8, son: 2.
— Yadvendra Somra, Sonipat
#Puzzle 120.2
Hi Kabir,
The following bird names came in handy in solving the puzzle: COOT, GUAN, HAWK, IBIS, LARK, RAIL, ROOK, TEAL and TERN. The required words are: B(RAIL)ING/B(ROOK)ING, B(ROOK)LET, H(IBIS)CUS, I(GUAN)ID, MA(LARK)EY, S(COOT)ERS/S(TEAL)ERS, S(COOT)ING/S(TEAL)ING/S(TERN)ING, S(TERN)EST, T(RAIL)ERS, T(OMA)HAWK.
— Professor Anshul Kumar, Delhi
While Anshul Kumar has given more options than anyone else, we still do not have 10 unique birds. If we take B(RAIL)ING as the first word, the same bird reappears in T(RAIL)ERS. If we take B(ROOK)ING, the same bird reappears in B(ROOK)LET. The bird I had in mind was LOOM, giving B(LOOM)ING.
It was not a condition, of course, that the 10 birds must be unique. The following lists include those who have correctly sent the names of 9 or 10 birds.
Solved both puzzles: Yadvendra Somra (Sonipat), Professor Anshul Kumar (Delhi), Sanjay Gupta (Delhi), Anil Khanna (Ghaziabad), Ajay Ashok (Delhi), Shishir Gupta (Indore), Kanwarjit Singh (Chief Commissioner of Income-tax, retired)
Solved #Puzzle 120.1: Sanjay S (Coimbatore), Amarpreet (Delhi), Dr Sunita Gupta (Delhi)
Problematics will be back next week. Please send in your replies by Friday noon to problematics@hindustantimes.com.
