There is a lot about party tricks that make them irresistible. They are easy to play and hold appeal for everyone, those who play them as well as those they are played on. Not everyone who knows how to play a mathematical trick, however, knows why it works; they can get by knowing what they need to do to make it work. This makes them an infinite source for puzzles, as we have seen in these columns before; the solver can have fun determining the principles a given trick is based on.
The following trick may be familiar; many may have played it on friends at school.
Invite a friend to write any number of four digits but less than 9000. Suppose she writes 3854. Below that, you write another number, seemingly at random: 6145.
Ask the friend to write another number below these two and suppose she chooses 2946 this time. You add a fourth number now, which again appears to be random: 7053.
Turning your back, you ask your friend to write down a fifth number, promising her that this will be the last. She writes down, say, 1563.
You explain that you had turned your back so that you wouldn’t be doing mental calculations while she was writing the fifth number. Turning back to face her, you look at the sheet and immediately announce the sum of the five numbers: 21561.
You demonstrate the trick a few more times, taking turns at writing a number until you have five of them. Each time, you correctly announce the sum after one glance, without spending time on actual calculations.
It is obvious that the numbers you placed in between were not chosen at random.
But what is the secret behind determining the sum, and what makes it work?
There will be very few people who do not know the rules of Wordle, but to reiterate them all the same: a green square means the right letter in the right place, yellow means the right letter is wrongly placed, and grey means the letter does not appear in the hidden word at all.
What is the hidden word in the game shown?
Mailbox: Last week’s solvers
Every time the gambler wins a bet, the number of his coins gets multiplied by 1½ and every time he loses, the number gets multiplied by ½. Therefore, with 3 wins and 3 losses, the number of coins gets multiplied by (1½)³ X (½)³, irrespective of the order of wins and losses. Thus, the number of coins at the end will always be 64 x (1½)³ x (½)³ = 64 x 27/64 = 27, which amounts to a loss of 64 – 27 = 37 coins.
— Professor Anshul Kumar, Delhi
We set up some basic equations. The distance run (D) is the same for both. Assume the slower runner’s speed is V. Then the champion’s speed is 2.5V. The time taken by the slower one is T. Then, the champion’s time is (T – 6).
Distance = VT = 2.5V (T –6).
Solving for T, we get T = 10 minutes.
Then, the champion’s time = 4 minutes.
— Rohit Khanna, Noida
Solved both puzzles: Prof Anshul Kumar (Delhi), Rohit Khanna (Noida), Amardeep Singh (Meerut), Dr Sunita Gupta (Delhi), Yadvendra Somra (Sonipat), Ajay Ashok (Mumbai), Shri Ram Aggarwal (Delhi), Group Captain RK Shrivastava (retired; Delhi), Vivek Aggarwal (Bangalore)
Solved #Puzzle 65.2: Hanit Kaur (Chandigarh), Anil Khanna (Ghaziabad), Dr Nakul Makkar (Noida), Sandeep Asthana
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