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Problematics | Calculating her name
This week, use arithmetic to work out the name of a bloke’s wife and try a logic puzzle that sounds illogical.
This week marks a milestone with the 150th consecutive episode of Problematics. The puzzles are intended to be special, with the first one (my own creation) having been saved up for such an occasion. Neither puzzle is particularly difficult, but both will hopefully be fun to solve.
#Puzzle 150.1
There is a bloke whose name is so long and so difficult to pronounce that everyone, including his wife, calls him “Bloke”. The good thing about the couple is that both of them are fond of puzzles, particularly cryptograms.
Bloke: Have you noticed something common between our names?
Wife: How can I? I cannot even pronounce your name.
Bloke: Not my real name, but the name everybody knows me by.
Wife: OK, Bloke. What’s common between that name and mine?
Bloke: Take each letter of BLOKE, replace it with its position in the alphabet, and multiply those values.
Wife: Let me see. B = 2, L = 12, O = 15, K = 11, E = 5, so the product is 19800.
Bloke: Now work out the product from the letters of your name.
Wife: Ah, 19800 again. Wonderful coincidence, isn’t it?
The wife’s name, like Bloke’s nickname, has five letters. No letter repeats itself in her name, and no letter in her name appears in BLOKE.
You could note that CHOKE has the same product as BLOKE, but that cannot be a woman’s name. I am not sure how many other words will satisfy these conditions. So let’s narrow it down. The wife has a common Indian name. Its letters can be anagrammed into two distinct names, simply by swapping the first and third letters. This hopefully makes the answer a unique pair; either of the two names will count as correct.
What is the name of Bloke’s wife?
#Puzzle 150.2
The second puzzle is not my own creation. I have found it different publications, so there’s no one to credit:
Statement A: Statement B is true
Statement B: Statement A is false
Is something wrong somewhere?
MAILBOX: LAST WEEK’S ANSWERS
#Puzzle 149.1
The total number of red faces visible is 30, known from the initial count before the lights were turned off. To meet the challenge in the dark, follow these steps.
Arbitrarily select and separate 30 cards from the deck to form Pile P. This can be done by feeling the cards in the dark. The remaining cards form Pile Q.
Let the number of red-face-up cards in Pile P be R. Then the number of red-face-up cards in Pile Q is (30 – R).
Flip all cards in Pile P. This changes all faces: a red face up now becomes black, and vice versa. The number of red-face-up cards in Pile P, therefore, becomes (30 – R).
In other words, the number of red-up cards visible in Pile P will equal the number of red-up cards in Pile Q, regardless of the initial distribution of red and black faces.
The strategy relies only on knowing the number of visible red faces (30) and does not require knowledge of the total number of cards, the specific cards, or their colours, making it feasible in the dark.
— Vinod Mahajan, Delhi
#Puzzle 149.2
Dear Kabir,
Disagree with the reasoning given in the puzzle. The probability of any of them becoming a double laureate is equal to (probability of getting the first Nobel) * (probability of getting second Nobel) = (1/8 billion)*(1/8 billion). Since the second award is independent of the first award, it is the product of each award's probability.
— Sampath Kumar V, Coimbatore
Last week’s puzzles may have been among the most difficult ever set in Problematics, with fewer correct answers received than ever before. My thanks to every reader who attempted them in spite of not being sure about or satisfied with their own answers.
Among those who solved the cards puzzle, Dr Vivek Jain insists that the trick will work only with a total of 60 cards. However, as Vinod Mahajan and Sampath Kumar V point out, it will work with any number of cards.
For the probability puzzle, a number of readers have argued that the probability of winning a second Nobel is 100 in 8 billion. I think that would have been the correct answer if the question had been: “What is the probability that one of the Nobel prizes will go to someone who had also won an earlier Nobel?” But the statement of the puzzle was: “winning a second Nobel should be more probable”. For the probability of winning two Nobel prizes, Sampath’s answer is correct.
Solved both puzzles: Sampath Kumar V (Coimbatore)
Solved Puzzle #149.1: Vinod Mahajan (Delhi), Dr Vivek Jain (Baroda)
Problematics will be back next week. Please send in your replies by Friday noon to problematics@hindustantimes.com.
