Problematics | Gangster wives
Three gangsters and their wives indulge in different pursuits. Who does what?
Readers of Problematics might feel we are dealing with Einstein puzzles less frequently these days, but the fact is that I have been working on many more of them than I used to. An Einstein puzzle appears online in an interactive format in the Games section of the HT website every week. The way one solves those is different; users anonymously click on the correct options without needing to write back.
It is true, however, that Problematics has not had an Einstein puzzle for several weeks now. It’s good when so many users go online to solve my puzzles, but nothing can beat the satisfaction a creator derives from the old-fashioned writeback format that we follow in Problematics. The Einstein puzzle below is on the easier side, but do be sure to send in your answers.
#Puzzle 155.1
Here is some more information on Mrs Anand, Mrs Bora and Mrs Chopra, whose children we counted last week. Each of them is married to a gangster. Their husbands are called Corleone, Scarface and Big Boy, in no particular order. One of them deals in arms, one in diamonds, and one in bootleg liquor.
Their wives work too, or so they think. The connections their husbands enjoy allow them to indulge in whatever pursuit they wish to. One of them sings at functions attended only by gangster friends of her husband. One writes books published privately, finding no readers, and one acts in B-movies financed by her husband.
1. Corleone is Mrs Chopra’s brother.
2. Scarface’s wife is the actress.
3. The so-called author is married to the bootlegger.
4. Mrs Anand’s husband smuggles diamonds.
5. Mrs Bora calls herself a singer.
Who is what, and who is married to whom?
#Puzzle 155.2
Note the numbers 512 and 4913. Both are perfect cubes (512 = 8³, and 4913 = 17³). An unusual property shared by the two numbers is that their cube roots equal the sum of their respective digits. That is to say, 5 + 1 + 2 = 8 is the cube root of 512, and 4 + 9 + 1 + 3 = 17 is the cube root of 17.
Can you find any other numbers with the same property?
MAILBOX: LAST WEEK’S SOLVERS
#Puzzle 154.1
Hello Kabir,
In the puzzle about children, the speaker must have been either Mrs Anand or Mrs Chopra because she mentioned Mrs Bora as her friend. The number of children is less than 25 but must be unique in the sense that it can immediately inform us of the number of children the speaker has. So the speaker’s individual count should uniquely appear more than once for the same total. After examining all combinations, I found that only a total of 19 satisfies this condition. There are two combinations for (A, B, C) when the total is 19 children: (6, 9, 4) and (6, 5, 8). In both cases, Mrs A has 6 children, so she must be the speaker. No such combination satisfying the condition exists when Mrs C is the speaker. Since Mrs B has the fewest children, the correct combination is (6, 5, 8). So you met Mrs Anand who has 6 children (2 girls and 4 boys). Mrs Bora has 5 children (1 girl and 4 boys) and Mrs Chopra has 8 (2 girls and 6 boys).
— Dr Sunita Gupta, New Delhi
#Puzzle 154.1
Hi Kabir,
Here is an attempt to explain the speed-and-distance puzzle without using algebra. Let us imagine that all three persons stop cycling at the exact moment when the mother reaches the destination. Then the daughter (15 kph) will have cycled for 1 hour more than she actually did, overshooting the destination by 15 km. The son (10 kph) will have cycled for 1 hour less than he actually did and fallen short by 10 km. This would mean the daughter has covered 15 + 10 = 25 km more compared to the son. Their relative speed being 15 – 10 = 5 kph, the time taken would be 25/5 = 5 hr. This is the time taken by the mother in the imaginary scenario and also in the actual scenario.
Now back to the actual scenario. The time taken by the daughter is 5 – 1 = 4 hr. Therefore, the distance travelled is 15 x 4 = 60 km. This distance is travelled by the mother in 5 hours, so her speed is 60/5 = 12 kph.
— Professor Anshul Kumar, New Delhi
Notes on # 154.1
For the first puzzle of last week, some readers have given the total number of children as 23, but that cannot be true. Various combinations for (A, B, C) give a total of 23: (6, 9, 8), (6, 13, 4), (6, 5, 12) and (12, 7, 4). Remember, the first statement (grand total) alone was enough to establish the number of children the speaker has. But whether the speaker was Mrs A or Mrs C, the first statement in isolation will not allow you to determine which of the above combinations is true. If the speaker is Mrs A, you don’t immediately know if she has 6 children or 12. If Mrs C is speaking, you don’t immediately know if she has 8, 4 or 12 children.
One reader has given a total of 24, which does not work either. Different combinations such as (3, 13, 8), (9, 11, 4) and (9, 7, 8) mean that the first statement alone is not enough. If the speaker is Mrs A, she might have 3 or 9 children. If Mrs C is speaking, she might have 8 or 4.
With 19 children, there are only two combinations, (6, 9, 4) and (6, 5, 8). In both cases, Mrs A has 6 children, meaning she is the speaker. Only a handful of readers have given 19 as the unique answer. Some readers have given 19 as one of two possible answers, but I have counted them as correct too.
Solved both puzzles: Dr Sunita Gupta (Delhi), Professor Anshul Kumar (Delhi), YK Munjal (Delhi), Yadvendra Somra (Sonipat), Kanwarjit Singh (Chief Commissioner of Income-tax, retired), Shri Ram Aggarwal (Delhi)
Solved #Puzzle 154.2: Vinod Mahajan (New Delhi), Shishir Gupta (Indore), Ajay Ashok (Delhi)
Problematics will be back next week. Please send in your replies by Friday noon to problematics@hindustantimes.com.
