30-06-2025
Groping in the dark
What follows below should have happened much sooner in a column that is now nearly three years old. Puzzle columns that work on reader interaction should have readers contribute not only solutions but an occasional puzzle too. It has been a long wait for Problematics, but we have finally arrived there. Representational image.(Shutterstock)
This week's main puzzle comes from Kanwarjit Singh, a retired chief commissioner of income-tax based in Delhi, who writes to Problematics regularly with solutions. Now he has sent a challenging logic puzzle that may appear impossible at first, but turns out quite easy once the answer is spelt out. I do not know who originally created this puzzle, but I have made significant modifications just in case the answer is on the internet somewhere. The principle on which the puzzle is based, of course, remains the same as in the version Kanwarjit Singh sent to me.
#Puzzle 149.1
Normal playing cards have their rank and suit on one side, and some kind of design printed on the reverse. Now imagine a special deck in which both sides of any card have a rank and suit on them. For example, one cad may have the 7 of diamonds on one side and the 8 of clubs on the other.
If such a special deck existed, the standard deck of 52 could have been printed on 26 cards. Not in this case, however. This deck has more cards than 26, more than 52 even, although the total number is unspecified. This is because some faces repeat themselves on different cards. For example, the 4 of hearts may appear on two cards, and the king of hearts may appear on three. One thing is common across the deck, though. On every card, one side is black (spades or clubs) and the other is red (hearts or diamonds).
Late one evening, when it is very dark outside, a master puzzler throws you a challenge. He spreads out the entire special deck on a table and invites you to divide it into two piles so that the number of red cards face up in one pile is equal to the number of red cards face up in the other pile.
What if the visible red faces are an odd number, you ask. The puzzler clears your doubts: 'Just flip any card or cards to make the number of red faces even.'
You decide that it would be a waste of time to count all the cards. You count the red faces alone, and find to your relief that the total is an even number: 30 of them, spread out across the table. Ah, you tell yourself, all I need to do is to make two piles with 15 red-upwards cards in each. But before you can proceed, the puzzler switches off the light.
You cannot see the cards, but you can feel them. Every card, of course, feels the same, as does its red side and its black side. In the dark you can no longer locate the red cards you had counted. You can still count any number of cards, but without knowing which ones are red face up and which ones black face up. Can you meet the challenge in the dark?
#Puzzle 149.2
So far, 976 individuals have won a Nobel prize, five of them twice. This count does not include organisations. The five double winners are well-known: Marie Curie for physics and chemistry, John Barding for physics, and Linus Pauling, Frederick Sanger and Karl Barry Sharpless for chemistry.
The chances of any individual chosen at random winning one Nobel during their lifetime are extremely low: about 1 in 8 billion if you take that as the world's population. On the other hand, winning a second Nobel should be more probable if you go by the following logic. Assume that there are 100 single laureates alive at any given time. The probability of any of them becoming a double laureate should be 1 in 100, which is much better than 1 in 8 billion. Agree or disagree?
MAILBOX: LAST WEEK'S SOLVERS
#Puzzle 148.1
Hi Kabir,
The three required numbers are 2494651, 1385287 and 9406087, which add up to 13286025, a perfect square (= 3645²).
Adding the digits of the given six numbers repeatedly until one digit for each remains, the 'digital roots' are 1, 4, 6, 7, 7, 7 respectively. Given that the digital root of a perfect square must be 1, 4, 7 or 9, the next step is to add the roots of any three numbers, taking one triplet at a time, and finding the digital roots of these sums. We get only one combination of 7, 4, 7 with a digital root of 9. All other combinations give values other than 1, 4, 7, 9 values. The numbers leading to a digital root of 9 are the ones mentioned above (2494651, 1385287 and 9406087).
The fun part of solving your puzzles is that we learn something new also, like Dudeney's digital roots last week. I am going to teach my grandchildren about these roots to identify a perfect square.
— Anil Khanna, Ghaziabad
While most readers have solved this puzzle using digital roots, some including Professor Anshul Kumar and Yadvendra Somra have additionally relied on the knowledge that the last digit of a perfect square must be 0, 1, 4, 6 or 9.
#Puzzle 148.2
Hello Kabir,
I initially found the second puzzle complicated, then realised that no, it's not. The two events start at 7pm.
The journeys of the two travellers are depicted in the illustration. Let the distance between the two villages be 'd'. Where the walker and the cyclist meet, let that point be at a distance 'a' from the walker's village and (d – a) from the cyclist's village.
The walker covers distance 'a' in 245 minutes from 12:00 to 4:05, his speed being a/245. To travel the remaining distance (d – a), he needs another 245(d – a)/a minutes.
The cyclist covers distance (d – a) in 125 minutes from 2:00 to 4:05, his speed being (d – a)/125. To travel the remaining distance 'a', he needs another 125a/(d – a) minutes.
Since the two complete the remaining distance in the same time,
245(d – a)/a = 125a/(d – a)
=> [(d – a)/a]² = 125/245 = 25/49 = (5/7)²
=> (d – a)/a = 5/7
This is the ratio between the distances from the meeting point to the two villages.
To travel 7, the walker needed 245 minutes. To cover 5, he will require 5/7 x 245 = 175 minutes.
To travel 5, the cyclist needed 125 minutes. To cover 7, he will require 7/5 x 125 = 175 minutes.
Time of reaching the respective events = 4:05 + 175 minutes = 7pm
— Dr Sunita Gupta, Delhi
Solved both puzzles: Dr Sunita Gupta (Delhi), Professor Anshul Kumar (Delhi), Yadvendra Somra (Sonipat), Vinod Mahajan (Delhi), Shishir Gupta (Indore), Ajay Ashok (Delhi)
Solved #Puzzle 148.1: Anil Khanna (Ghaziabad), Sabornee Jana (Mumbai), Shri Ram Aggarwal (Delhi) Problematics will be back next week. Please send in your replies by Friday noon to problematics@
