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Hindustan Times
5 days ago
- Entertainment
- Hindustan Times
Gangster wives
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. For representational purposes only.(Shutterstock) 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 Solution to the puzzle. 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@
Hindustan Times
30-06-2025
- General
- Hindustan Times
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@
Hindustan Times
23-06-2025
- General
- Hindustan Times
Find the square
I learnt something new and interesting recently. Most puzzle solvers, including you readers of Problematics, probably know this already, but here it is anyway. To check if a number is a perfect square, one way to proceed is to keep adding its digits until you reach a single digit. If it's a square, the ultimate 'digital root' in a single digit must be any one of 1, 4, 7 and 9 — and nothing else. Welcome to Problematics!(Shutterstock) For example, the digital root of 16 is 1 + 6 = 7, that of 81 is 8 + 1 = 9, that of 49 is 4 + 9 = 13 followed by 1 + 3 = 4, and that of 676 is 6 + 7 + 6 = 19 followed by 1 + 9 = 10 and finally 1. There are, of course, numbers that can have digital roots of 1, 4, 7 or 9 without being perfect squares, for example 70 whose root is 7. So the method may not always tell you if a number is a perfect square, but it will definitely help you rule out many possibilities. #Puzzle 148.1 A century ago, the English puzzler Henry Ernest Dudeney presented six numbers to his readers: 4784887 2494651 8595087 1385287 9042451 9406087 Three of them can be added to form a perfect square, Dudeney wrote, inviting his readers to identify them. I managed to find the three numbers using hit and trial, paired with some observations that eliminated one or more possibilities. Then I looked at Dudeney's solution and learnt about digital roots and how the rule could be applied to solve this puzzle. Who knows, there may be other methods too. Use any method — hit and trial, digital roots, or any procedure you may innovate yourself. But do send me the three numbers that satisfy the condition. #Puzzle 148.2 The Village of Music is organising a concert and has invited a musician from the Village of Astronomy. The Village of Astronomy, on the other hand, is inaugurating a telescope and has invited an astronomer from the Village of Music. By sheer coincidence, both events begin at the same time on the same day. The musician from Astronomy, who wants to get some exercise, decides to walk the whole distance and sets off towards Music at 12 noon on the day of the event. The astronomer from Music chooses to cycle the distance and sets off towards Astronomy at 2pm. Both are good puzzlers and can manage their respective speeds precisely. Indeed, each one arrives at his destination at the exact moment that his event starts. This is after the two had met on the way, which happened at 4:05pm. They called out 'HI!' to each other but neither stopped. At what time do the two events start? MAILBOX: LAST WEEK'S SOLVERS #Puzzle 147.1 HT picture Hi Kabir, Based on the conditions that surnames and addresses do not begin with the same letter, and that of four statements only one is true, we can map different scenarios as shown in the table. Statement 1 is not possible. From the other three statements, whichever one is true, only one address can be established for sure, i.e the Poonawala family lives in Surat. — Sabornee Jana, Mumbai *** Hi, The only address that can be definitely established is that the Poonawala family is from Surat. Statement (1) must be false. Statements (2), (3) and (4) give different combinations that do not violate the given conditions. Therefore any two of them can be false. But no matter which one statement is true, one address is common to each combination — the Poonawala family must live in Surat. — Ajay Ashok, Delhi #Puzzle 147.2 Hi Kabir, Each of the ten statements contradicts every other statement. Therefore, at the most one can be true. That is, either exactly one statement is true (this is what statement 9 is) or none of them is true (this is what statement 10 is). However, statement 10 contradicts itself. Therefore, only statement 9 is true and the remaining nine are false. — Professor Anshul Kumar, Delhi Solved both puzzles: Sabornee Jana (Mumbai), Ajay Ashok (Delhi), Professor Anshul Kumar (Delhi), Dr Sunita Gupta (Delhi), Yadvendra Somra (Sonipat), Vinod Mahajan (Delhi), Abhinav Mital (Singapore), Shri Ram Aggarwal (Delhi) Solved Puzzle 147.2: Kanwarjit Singh (Chief Commissioner of Income-tax, retired), Dr Vivek Jain (Baroda) Problematics will be back next week. Please send in your replies by Friday noon to problematics@
Hindustan Times
09-06-2025
- Entertainment
- Hindustan Times
Who is the postman?
One of the most enduring puzzles from the English pioneer Henry Ernest Dudeney (1857-1930) involves six persons sharing three names. That is to say, two persons each share the same name. Clues are given about their profession, address etc, and the solver tries to establish the identity of one or more of the individuals. Later generations of puzzlers have used Dudeney's template to create newer logic puzzles in the same mould. Here in Problematics, I have taken inspiration from several other Dudeney puzzles, but this is my first attempt at using his 6-persons-with-3-names template. Puzzle #146.1 Some names are gender-neutral; they can be given to either boys or girls. Therefore, meet three men called Gagan, Kiran and Suman and three women called (you guessed it) Gagan, Kiran and Suman. They happen to be three married couples, but we are not given the full details of which one is whose spouse. The men are salaried individuals: a paramedic at a hospital, a plumber engaged by a housing society, and a postman in government service. Among the women, one is a painter, one a poet, and one a playwright. The three couples live on the same street. The female Gagan and her paramedic husband live at the north end of the street, and the female Kiran and her plumber husband at the south end. The plumber's best friend, incidentally, is the male Suman. The female Suman, meanwhile, lives in the middle of the street with her postman husband. Although a career in the arts does not bring a fixed income, one can earn a windfall from a single painting exhibition, a successful book of poems, or one hit play. The female Suman, for example, had a very good year, earning three times as much as her husband. The woman with the same name as the paramedic, meanwhile, earned ₹10 lakh for the year. To prevent any misdirection, the year's total income for each of the six individuals was an exact multiple of ₹1 lakh. Puzzle #146.2 SCAN OF CAIRNS (2 words) SONIA AND LIPI (1 word) A MAN'S ACTION (2 words) SO, I'M IN NEPAL (1 word) LONG LEASES (2 words) LANE OWNERS (2 words) Unscramble the above anagrams to get the names of six cities, all in the same country. MAILBOX: LAST WEEK'S SOLVERS #Puzzle 145.1 Dear Mr Kabir, Let the number of duck eggs = N, which means the number of chicken eggs = (100 – N). On the first day, both sets fetch the same price, i.e. (N) (price of a duck egg) = (100 – N) (price of a chicken egg) On the second day, after the baskets are interchanged unintentionally, the given facts can be represented as follows: Price of a duck egg on the second day = 280/N = price of a chicken egg on the first day; Price of a chicken egg on the second day = 630/(100 –N) = price of a duck egg on the first day Substituting these values in the first day's equation, we get (N)[630)/(100 – N)] = (100 – N)[280/N] => 630N² = 280(100 – N)² => 70 x 9N² = 70 x 4(100 – N)² => [3N]² = [2(100 – N])² Taking the positive square roots of both sides, 3N = 2(100 – N) => N = 40. Thus the number of duck eggs is 40, and the number of chicken eggs is 100 – 40 = 60. This gives the first-day price of a duck egg (= second-day price of a chicken egg) as 630/60 = ₹10.50, and the first-day price of a chicken egg (= second-day price of a duck egg) as 280/40 = ₹7. Sale on day #1 = (40 x 10.50) + (60 x 7) = ₹840; and sale on day #2 = 280 + 630 = ₹910. Thus Father should be happier on Day #2. — Shri Ram Aggarwal, New Delhi #Puzzle 145.2 Hello, In the puzzle with playing cards, start by placing the 4s. There are only two options: 4 __ __ __ __ 4 __ __ __ 4 __ __ __ __ 4 __ 4 A 3 A 2 4 3 2 — Biren Parmar, Bay Area, California *** Position: 1 2 3 4 5 6 7 8 Cards: 4 A 3 A 2 4 3 2 Mirror solution Cards: 2 3 4 2 A 3 A 4 — Vinod Mahajan, New Delhi Solved both puzzles: Shri Ram Aggarwal (Delhi), Biren Parmar (Bay Area, California), Vinod Mahajan (Delhi), Abhinav Mital (Singapore), Anil Khanna (Ghaziabad), Dr Sunita Gupta (Delhi), Kanwarjit Singh (Chief Commissioner of Income-tax, retired), Shishir Gupta (Indore). Yadvendra Somra (Sonipat), Professor Anshul Kumar (Delhi), Sabornee Jana (Mumbai), Ajay Ashok (Delhi), Sampath Kumar V (Coimbatore), Nitin A note on #143.1: Two weeks ago, the list of correct solvers for an Einstein puzzle (#143.1) had omitted the name of Vinod Mahajan. He has since explained the way he interpreted some of the clues, and I have subsequently accepted his answer as correct. In future, I will try and make sure that every clue in an Einstein puzzle is worded in a way that there is no possibility of ambiguity in interpreting it.
Hindustan Times
02-06-2025
- Science
- Hindustan Times
Chicken and egg — and duck too
Here at Problematics we usually aim for puzzles that are not the kind you would find in a textbook, but there are exceptions. Some puzzles that can be solved with textbook methods are still interesting because of the way they are packaged or because of their pedigree, with illustrious minds having dwelt on them at some point in history. A prime example of puzzles that are delightful because of both packaging and pedigree are the problems in Bhaskara's Lilavati. While those are widely known, I recently found one that I hadn't come across earlier. It is said to have appeared in a book by the great Euler, and described by the French writer Stendhal before making its way into the writings of the late Russian mathematician Yakov Perelman. To insulate the solution from an internet search, I have added my customary modifications to the version described by Perelman. I have changed the currency to Indian rupees, and tinkered with the prices to bring them within a range that is credible for the story into which I have packaged my adaptation. The story, of course, is entirely my own. #Puzzle 145.1 A family of poultry farmers collects 100 eggs one morning. They are all chicken and duck eggs, the distribution being unequal. Handing the chicken eggs to their son and the duck eggs to their daughter, the farmer parents send them off to the market. The price for each kind is fixed, with the duck eggs being costlier than chicken eggs, as is the case in most places. Each child sells his or her full share of eggs at the respective fixed rates. In the evening, when they compare their earnings, they are thrilled to find that both have made exactly the same amount. I am no farmer, but the internet tells me that hens and ducks lay about one egg daily at the peak of their productive years. It is not surprising, therefore, that the same birds at our farm lay the same number of eggs the following morning. In other words, the family has 100 eggs again, and the unequal distribution of chicken and duck eggs is the same as on the previous day. Mother segregates the produce into a number of baskets, the chicken eggs on one side, the duck eggs on another. Father passes the orders: 'Pick up your respective shares and come back with the same earnings as you did yesterday.' The kids get mixed up, of course (how else would there be a puzzle?) The son picks up the duck eggs by mistake, and the daughter takes the rest. Neither of them notices that his or her count is not the same as on the previous day. At the market, the boy sells the duck eggs at the price for chicken eggs, and his sister sells the chicken eggs at the price for duck eggs. When they compare their earnings in the evening, the boy is alarmed. 'I got only ₹280 today. I don't know how I can explain this to Father,' he says. The girl is equally puzzled about her collection, but pleasantly so. 'I don't know how, but my earnings rose to ₹630 today,' she tells her brother. #Puzzle 145.2 MAILBOX: LAST WEEK'S SOLVERS Hi Kabir, Assuming that the store owner initially bought cat food for 31 cats for N days, or 31N cans. As each cat consumes 1 can/day, the total consumption reduces by 1 can every day. Again, all cans were consumed in one day less than twice the number of days originally planned, or (2N – 1) days. Thus the total number of cans is the sum of an arithmetic progression of (2N – 1) terms starting 31, and with a common difference of –1. The sum of the AP is: [(2N – 1)/2][2*31 + (2N – 1 – 1) (–1)] = 65N – 32 – 2N² Equating the above to 31N and simplifying, we get the equation 2N² – 34N + 32 = 0. The roots of this equation are N = 16 and 1. As 1 day is not viable, N must be 16. So the total number of cans bought initially = 31*16 = 496. And as it took (2N –1) = 31 days to finish the whole stock of food, only 1 cat was left unsold. — Anil Khanna, Ghaziabad *** Hi Kabir, Suppose the cat food was initially ordered for N days. Then, the number of cans ordered = 31N. Also, suppose K is the number of cats remaining unsold when the food stock got exhausted. On any day, the number of cans consumed is the equal to total number of unsold cats. Thus the total cans consumed = 31 + 30 + 29… + (K + 2) + (K + 1) + K = (31 + K)(31 – K + 1)/2 i.e. 31N = (31 + K)(31 – K + 1)/2 For the right-hand side to be a multiple of 31, K has to be 1. This means 31N = 32*31/2, or N = 16. The number of cans = 31 x 16 = 496. The food lasted for 31 days. If we add one more day, we get 32 days which is twice the original period of 16 days. — Professor Anshul Kumar, Delhi From Professor Kumar's approach, it emerges that the puzzle can be solved even without the information about the cans being exhausted in (2N – 1) days. Many readers, however, have used this bit in solving the puzzle. Puzzle #144.2 Hi Kabir, The puzzle about the party trick is fairly simple — you randomly tap on any two animal names for the first and second taps and then tap in the order of length of the animal names — i.e. COW (third tap), LION, HORSE, MONKEY, OSTRICH, ELEPHANT, BUTTERFLY AND RHINOCEROS. Obviously, this trick will get old very soon because your tapping pattern will become predictable to a keen observer. — Abhinav Mital, Singapore Solved both puzzles: Anil Khanna (Ghaziabad), Professor Anshul Kumar (Delhi), Abhinav Mital (Singapore), Kanwarjit Singh (Chief Commissioner of Income-tax, retd), Dr Sunita Gupta (Delhi), Yadvendra Somra (Sonipat), Shishir Gupta (Indore), Ajay Ashok (Delhi), YK Munjal (Delhi), Sampath Kumar V (Coimbatore) Solved #Puzzle 144.1: Vinod Mahajan (Delhi)
