Problematics | A trick that works, an equation that doesn’t

This week, we bring to you a party trick with a (slightly) lengthy description and a strange equation of 1s and 0s. Good luck!

How about another one of those party tricks that you can play on an impressionable audience? As we have seen in these columns earlier, it’s fun to work out the mathematical and logical principles that make these tricks work.

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The following trick requires a lengthy description but, from a puzzler’s point of view, it’s one of the best:

#Puzzle 32.1.

From a deck of 52 playing cards, ask a friend to take any number of cards between 1 and 12, and put those in his pocket. You do not know how many he chooses.

From the remainder of the pack, ask him to memorise the card that appears in the same position as the number of cards in his pocket. That is to say, if he has 9 cards in his pocket, he notes the 9th card from the top in the rest of the pack. Again, you don’t know which card that is.

Next, ask him to name any celebrity whose name contains more than 12 letters. Let’s say he names Audrey Hepburn.

“Spell out her name with the cards, one letter at a time,” you say. To demonstrate, you deal the cards one by one, calling out one letter each time, and placing each dealt card over the card that was dealt immediately before it.

Once AUDREY HEPBURN has been spelt out, you pick up the dealt cards and put them back on top of the deck. And on top of these, ask your friend to place the cards from his pocket.

“With this full pack,” you tell him, “spell out the chosen name I way I showed you.”

He does so. “Now,” you continue, “turn over the next card.”

To his surprise, it turns out to be the same card he had memorised earlier.

What makes the trick work every time?

#Puzzle 32.2:

Consider the following equation:

1 + (–1) + 1 + (–1) + 1 + (–1) + 1 + (–1) +… = 1 – 1 + 1 – 1 + 1 – 1 + 1 – 1 +… => 1 + (–1 + 1) + (–1 + 1) + (-1 + 1) +… = (1 – 1) + (1 – 1) + (1 – 1) + (1 – 1) +… => 1 + 0 + 0 + 0 + … = 0 + 0 + 0 + 0 + … => 1 = 0

What’s going on here?

Mailbox: Last week’s solvers:

#Puzzle 31.1 Hello Kabir, The answer is 50 metres. Suppose x is the length of the pool, S₁ is the speed of one swimmer and S₂ is that of the other. The time taken by swimmer 1 to cover (x – 22) metres is the same as the time taken by swimmer 2 to cover 22 metres. Therefore, (x – 22)/S₁ = 22/S₂; and similarly (2x – 16)/S₁ = (x + 16)/S₂ Solving for x by eliminating S₁/S₂, x = 50 — Sanjay Gupta, Delhi

#Puzzle 31.2

Good morning, Kabir ji,

The values in the table provided show the object's relative diameter in comparison to the (Earth's) Moon. The missing value for the Earth, therefore, would be approximately 3.67.

Sun	401	Jupiter	41
Mercury	1.40	Saturn	34.7
Venus	3.50	Uranus	14.0
Earth	—	Neptune	14.4
Mars	1.95	Pluto	.072

— Sandeep Bhateja, Hoshiarpur

Solved both puzzles: Sanjay Gupta (Delhi), Sandeep Bhateja (Hoshiarpur), Anil Kumar Goyal (Delhi), Amardeep Singh (Meerut), Jasvinder Singh (Nabha), Harshit Arora (Delhi), Vasu Handa (Sonipat), Shishir Gupta (Indore), Rahul Agarwal (Bay Area, California), Khushi Verma & Lucky Singh Randhawa (Mandawali, Delhi), Ajay Ashok (Mumbai)

A number of other readers have figured out the ratio of 3.67 for #Puzzle 31.2 but haven’t pointed out the connection with the Moon.

Solved #Puzzle 31.1: Dr Nakkul Makkar (Noida), Dr Sunita Gupta (Delhi), Sunita Gupta (also Delhi), NC Srivastava (Indirapuram), Puneet Vashistha (Delhi), Sunita & Naresh Dhillon (Gurgaon), Dr Mrinalini Parashar (Delhi), Rutu M Save (Mumbai), Rajesh Bansal (Noida), Rajender Parsad Agarwal (Delhi), Bharti Budhiraja (Delhi), Apaala Ghai (Gurgaon), Deviprakash Seksaria (Navi Mumbai), Amar Lal Miglani (Mohali), Rimzim Ahuja (Noida), Simran Pushkarna & Sarthak Pushkarna (Delhi), Ravi Sondhi & Rudra Sondhi (Gurgaon).

Problematics will be back next week. Please send in your replies by Friday noon to problematics@hindustantimes.com