Problematics | Marital status
A bachelor meets three women and wants to know who is married and who is not, for whatever reason. Can you help him?
Consider the following two statements: (A) If X is a physicist then Y is a real estate agent; (B) If X is a physicist then Y is not a real estate agent. Is it possible for both statements to be true?

The answer is yes, if you consider that X is not a physicist. In that case, it does not matter which clauses are dependent on X being a physicist. Assuming that both statements are true, we know for sure that X is not a physicist because the two dependent clauses are mutually contradictory. But we cannot determine whether Y is a property dealer or not, unless we are given additional information.
In the following puzzle, you may find statements that appear to be mutually contradictory, but will find that they are not actually so if you approach them the right way.
#Puzzle 153.1
At a party, a bachelor meets three young women and strikes up a conversation. He finds that each one has a name with the same initial as the state she belongs to, but that is not the information he is looking for. Men being men, he is more curious about their marital status.
The women decide to tease the young man. Rather than spell out who is married and who is not, they let out little bits of information (all true) one by one:
(1) If the woman from Chhattisgarh is married then either the woman from Assam or Bonnie is married but not both
(2) If neither the Assamese woman nor the Chhattisgarhi woman is married then Bonnie must be married
(3) If Apsara is married then the woman from Bihar and Chitra are both unmarried
(4) The woman from Bihar is unmarried if both Apsara and Chitra are unmarried
Who is married and who is not?
#Puzzle 153.2
(0) (1) (2) (3) (4) (5) (6) (7) (8) (9)
The above represents a 10-digit number but the digits are not as shown. Instead, the numbers in the brackets signify the number of each digit in the number. To make that clear, in the bracket with 0, enter the number of times 0 appears in the final number. In the bracket with 1, enter the number of 1s in the final number. In the bracket with 2, enter the number of 2s in the final number. Finally, remove all brackets.
What is the number?
MAILBOX: LAST WEEK’S SOLVERS
#Puzzle 152.1
Hello Kabir,
I learnt something new from this puzzle. The amount earned should be a square, be even, and have an odd number before the units digit. That is to say, the amount should be 10n + c, where c < 10 and n is odd. I checked all squares up to 40² and found that if n is odd, then c is always 6. This means one vendor has got ₹4 extra, so he will send the other vendor ₹2 via UPI.
— Dr Sunita Gupta, Delhi
Sanjay Gupta, who looks at all squares up to 100², shows that the only squares with an odd number in the tens place are 36, 196, 256, 676, 1156, 1296, 1936, 2116, 2916, 3136, 4096, 4356, 5476, 5776, 7056, and 7396 — all ending in 6. This will be true no matter how far you go.
A few readers have solved this correctly up to the point that the initial difference in the two vendors’ shares is ₹4, but then made the mistake of concluding that ₹4 is also the amount transferred via UPI (missing the point that the transfer must be half of that). Nevertheless, I am counting those as correct along with those who have correctly concluded their solution at ₹2.
#Puzzle 152.2
Kabir, Hi
At times one tends to overthink the solution when, all the time, the simple one is staring you in the face. The solution in this case is clearly in the wordings of the guarantee. The poor parrot is stone deaf and the shop manager has pulled a fast one on the unsuspecting customer.
— Sanjay Gupta, Delhi
Solved both puzzles: Dr Sunita Gupta (Delhi), Sanjay Gupta (Delhi), Kanwarjit Singh (Chief Commissioner of Income-tax, retired), Ajay Ashok (Delhi), Shishir Gupta (Indore), YK Munjal (Delhi), Shri Ram Aggarwal (Delhi), Yadvendra Somra (Sonipat)
Solved Puzzle 152.1: Anil Khanna (Ghaziabad), Vinod Mahajan (Delhi), Sabornee Jana (Mumbai)
Solved Puzzle 152.2: Aditya Krishnan (NMIMS Mumbai)