Problematics | Crorepati at 100
Celebrate the 100th week of Problematics with a special puzzle centred on Kaun Banega Crorepati and a sitter based on Wordle
In 2001, we set out to chase a Bombay dream, only to return disappointed. Having received an invitation to participate in a ‘couples special’ episode of Kaun Banega Crorepati, my wife and I sat by the buzzer but were not quick enough in the ‘fastest fingers first’; we did not qualify for the hot seat. Our consolation was meeting Amitabh Bachchan in person.
KBC is a game of skill and luck, and also probability. More than a year ago (Puzzle #41.1, June 2023), while presenting my variation of a classic probability puzzle, I had mentioned that I had another variation of the same puzzle based on a game show. That show is KBC, of course, and I had kept the puzzle on hold for a special occasion. And no occasion could be more special than the 100th consecutive week of Problematics.
Thank you for the continued feedback that has made it possible for the column to keep running.
#Puzzle 100.1
At the time we went to KBC, there were only three lifelines in the game. One of these was 50:50, which works like this. The host presents you with four options as usual. If you opt for 50:50 (you are allowed to do that only once in the game), two of the four options will disappear from the screen. Of the two remaining two options, one is right and the other wrong. If you have no idea which option is correct, you can choose either one and your chances of being correct are 50%. Hence the name 50:50.
There are times, however, when your chances are not 50% even after going for the 50:50 lifeline. Suppose you have already recognised one or two incorrect answers before going for the lifeline? Let's look at such possibilities.
Scenario #1. Out of the original four options, you recognise one wrong answer but don’t know which of the other three options is correct. If you pick one option at random, your chances of being correct are 1/3 (33.3%). Instead of taking a chance now, you go for 50:50, so that only two options remain.
Have your chances improved from the previous 1/3?
Scenario #2. Out of the original four options, you can recognise two as incorrect, but don’t know which one of the other two options is correct. If you gamble on any one, your chances of getting it right are 50%. Instead, you opt for 50:50, so that only two options remain.
Do your chances improve from the previous 50%?
Scenario 3. You are clueless about all four options. You decide to gamble on any one, so it’s 25% in your favour. Mind you, you have not opted for 50:50. The host, however, suddenly decides to remove two wrong answers. This is after you have made your choice. Now, two options remain on the screen, including the one you have already chosen. You still don’t know which one is correct. The host gives you a choice: either stick to your original choice, or switch to the other remaining option.
Should you stay put or switch?
#Puzzle 100.2
To recap the rules of Wordle, the hidden word is always 5 letters long. If your test word gets a green cell, that letter is in the same position in the hidden word. If you get yellow, that letter occurs in a different position in the hidden word. A grey cell means that the letter in that cell does not appear in the hidden word at all.
In the illustrated game, a weak player exhausts all six test words without uncovering the hidden word.
What is the word?
MAILBOX: LAST WEEK’S OPTIONS
#Puzzle 99.1
Hi Kabir,
The puzzles were quite simple. Hope you like the following method.
Number of matchboxes = x, soap bars = y, milk packets = z
Total no of objects = x + y + z = 100
Cost-wise, x + 10y + 50z = 500
From these two equations, we get 9y + 49z = 400
Both y and z have to be integers, and z can have values between 1 to 8 only. The only number that satisfies both these two conditions is z = 1.
Which means y = 39 and x = 60. So, the woman bought 1 packet of milk, 39 bars of soap and 60 matchboxes.
— Dr Sunita Gupta, Delhi
***
Dear Kabir
Let the number of matchboxes, bars of detergent soap and milk packets be x, y, z respectively.
x + 10y + 5z = 500 (money) and x + y + z = 100 (count)
Subtracting, 49z + 9y = 400
Or, y = (400 – 49z)/9
So (400 – 49z) needs to be a multiple of 9. Now,
400 – 49z = 396 + 4 – 54z + 5z = (396 – 54z) + 4 + 5z
Since (396 – 54z) is a multiple of 9, it follows that (4 + 5z) too must be a multiple of 9.
4 + 5z = 9 when z = 1
4 + 5z = 54 when z = 10
4 + 5z = 99 when z = 19
And so on.
But z cannot be 10 or more, because if z = 10, then 10 x ₹50 = ₹500 is taken care of by the milk alone, when it is given that she also bought matchboxes and soap. So, the only solution is z = 1, y = 39, x = 60.
— Sampath Kumar V, Coimbatore
(Many solvers have used the same method as Dr Sunita Gupta’s, or something similar. A few have used hit and trial. Sampath’s method, on the other hand, is the traditional way to solve a Diophantine equation. Thank you to everyone who has shared their method; the variety makes it interesting. — KF)
#Puzzle 99.2
Hi Kabir,
For the anagrams of movie titles, the lack of information on whether we are referring to Hollywood films or Indian films made it a little complex. I went with a gut feeling and started with Hollywood, and hit upon the films easily. Actually, one film title is good enough, because with that one knows who is the superstar you are referring to, and the rest follows.
The superstar is Clint Eastwood. The films are:
Dirty Harry (anagram of TRY DRY HAIR)
Sudden Impact (anagram of DUSTIN CAMPED)
True Crime (anagram of RECRUIT ME)
Pale Rider (anagram of LIPREADER)
Looking forward to the 100th edition of Problematics. Huge congratulations to you for reaching that milestone.
— Akshay Bakhai, Mumbai
Solved both puzzles: Dr Sunita Gupta (Delhi), Sampath Kumar V (Coimbatore), Akshay Bakhai (Mumbai), Raghunathan Ravindranathan (Coimbatore), Harshit Arora (IIT Delhi), Professor Anshul Kumar (Delhi), Ajay Ashok (Mumbai), Shishir Gupta (Indore), Sanjay S (Coimbatore), Sundarraj C (Bengaluru), Aditya Krishnan (Coimbatore), Jaikumar Inder Bhatia & Disha Bhatia (Ulhasnagar, Thane), YK Munjal & friends (Delhi), Hari Velaayutham
Solved #Puzzle 99.1: Yadvendra Somra (Sonipat), Anil Khanna (Ghaziabad)