Problematics | More tricks to play on impressionable friends

Welcome to week 23 of Problematics! Here are this week's puzzles, admittedly not my own.

Neither of this week’s puzzles is my own. One has been in circulation for such a long time that I have no idea where it originated from. The other one, the first of the two below, comes from Martin Gardner’s Mathematical Magic Show.

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#Puzzle 23.1.

#Puzzle 23.1:

With your back turned, ask a friend to lay out three coins so that they are not all heads up or all tails up. Without looking, specify any one coin and ask your friend to flip it over. Carry on this way, flipping one coin at each move, until you have all three heads or all three tails. What is the best strategy to use, and what is the minimum number of moves necessary to reach the goal?

Gardner’s book goes on to a more advanced version. Give instructions as before, but this time you want all three coins to be heads up. Again, what is the minimum number of moves required?

#Puzzle 23.2.

#Puzzle 23.2:

You have three glasses, one of them upside down. Flipping any two glasses at a time, one move is enough to get all three glasses upside down. Can you make three such moves and still get all three glasses upside down?

Mailbox: Last week’s solvers:

#Puzzle 22.1:

The puzzle about cats and dogs, in hindsight, could have been made more difficult by withholding some information. Kudos to Gopal Menon for a solution that is better than my own:

Dog	Cat
Name and Price	Name and Price
Akita ---------- ₹15432	British Shorthair ---------- ₹30864
Collie ---------- ₹15320	Siamese ---------- ₹15320
Dalmatian ---------- ₹15376	Manx ---------- ₹7688

You have provided excess information for this puzzle. The total cost of the three dogs need not be given. It can be determined.

Let the first letter in the name of each animal denote its cost. Then, a = d + 56 and c = d – 56.

Let x, y and z be three distinct numbers {0.5, 1, 2}, not necessarily in that order. Then, the costs of the cats are xa, yc and zd.

a + c + d + xa + yc + zd = 100000

=> (d + 56) + (d – 56) + d + x(d + 56) + y(d – 56) + zd = 100000.

=> (3 + x + y +z)d + 56(x – y) = 100000

=> 6.5d + 56(x – y) = 100000, since x + y + z = 3.5.

=> d = [200000 – 112(x – y)] ÷ 13.

(x – y) can be 0.5, 1, 1.5, (–0.5), (–1) or (–1.5). Now, for these values, check whether [200000 – 112(x – y)] is divisible by 13.

Only x – y = 1 satisfies this requirement. From this, we get d = 15376 and, a = 15432, c = 15320.

x – y = 1 implies that x = 2, y = 1, z = 0.5.

Using the equation for the total cost, we get, b = 30864, m = 7688 and s = 15320.

We can then determine the combinations in the enclosures.

— Gopal Menon, Mumbai.

#Puzzle 22.2:

Hello Sir,

The longest word formed by anagramming the letters of DICTIONARY would be INDICATORY (10 letters).

— Yojit Manral, Faridabad

Solved both puzzles: Gopal Menon (Mumbai), Yojit Manral (Faridabad), Gulraj Singh Negi (Ludhiana), Akshit Goel, Anil Kumar Goyal (Delhi), Dhiren Gupta (DPS Indirapuram), Vasu Handa (Sonipat), Gaganjot Kaur (Mohali), Sunita & Naresh Dhillon (Gurgaon), Shaivya Shivangi (Gurgaon), Rohit Khanna (Noida), Maninder Singh Dang, Himanshu Jain (Gurgaon), Harshit Arora (Delhi), Raghav Khapre (Shri Ram School, Moulsari), Varsha Jain (Mumbai), Rakesh Behari (Delhi), Anushka Rai (Delhi University), SR Aggarwal (Delhi), Amardeep Singh (Meerut), Vinod Mahajan (Delhi), Mayobhav Pathak (Gurgaon), Meenakshi Gupta, Sandeep Ohlan (Rohtak), Sumanyu Aggarwal (Delhi), Dr Savita (Sonipat), Kanishka Kataria (Delhi), Simran Pushkarna (Delhi), Jasvinder Singh (Nabha), Avanti Kashikar (Mumbai), Nishant Iyengar (Faridabad), Rina Kumari (Noida Extension), Sahil Chawla (Delhi), Musarrat Rai Handa (Faridabad), Shishir Gupta (Indore), Agastya Malhotra (Noida), Lucky Singh Randhawa (Mandavali, Delhi), Rahul Agarwal (Bay Area, California), Mukesh Arora (Gurgaon), Dr Roona Poddar (Delhi University), Juneja Satishwar (Delhi)

Solved #Puzzle 22.1: Mahesh Mundhra (Indirapuram), Saksham Bhatnagar (Delhi), Dr Nakul Makkar (Noida), Prabhjot Singh (Delhi)

Solved #Puzzle 22.2: Manish Jain (Delhi), Ahmad Adil, Dr Sana Parveen (Delhi), Sudesh Dogra (Delhi), Arun Kumar Gupta (Noida Extension), Philomina PS (Delhi), Ravi Sondhi & Rudra Sondhi (Gurgaon), Jaikumar Inder Bhatia (Ulhasnagar, Thane)

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