Latest news with #Problematics
Hindustan Times
4 days ago
- 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)
Hindustan Times
12-05-2025
- Entertainment
- Hindustan Times
World Cups with and before Virat Kohli
Those of us who have lived through both of India's ODI World Cup victories will remember the joy of younger cricket fans who hadn't been born yet in 1983 but were old enough to celebrate the victory in 2011. Some of us bragged to our younger colleagues that we had watched both triumphs live, even though that only meant television for most of us. But there are others who lived through both victories and showed no reaction, because sport does not enthuse them the way it enthuses others. This week's puzzle includes one such bore. Coincidentally, after I created the puzzle, the last remaining active member of the 2011 team has announced his retirement from Test cricket. Virat Kohli remains available for ODI selection, of course, but his announcement somehow makes the following puzzle seem topical. #Puzzle 142.1 We go back a few years into the recent past, when a neighbourhood in Mumbai is celebrating India's World Cup victory at the Wankhede Stadium. A young pair of twins have just returned from watching the match and are chattering away about how the match might have been won earlier if Batter X hadn't lost his wicket playing that rash shot, or if Batter Y had played a different shot on a certain delivery that would have got him a boundary instead of a single. The noise upsets their grumpy old neighbour, who has little interest in cricket and has resented the student twins ever since they moved in next door from London, where their parents are working. 'What is there fuss so much over a cricket match? India won another World Cup 28 years ago when I was close to your age, and my generation never made such a noise,' he yells at the two students. 'Hey uncle, we have just watched the match on the ground. You probably watched both matches on TV, which may explain your lack of enthusiasm,' the youngsters shoot back. 'My generation doesn't waste time watching cricket,' Grumpy says. 'Who are you kidding?' the brother says. 'Our grandmother was at Lord's during the previous victory in 1983, and celebrated just like we are today. Coincidentally, India won that World Cup on our grandmother's birthday, and it's our birthday today when India have won it again. Granny made it a point to celebrate each birthday simultaneously with the anniversary of the Lord's victory, and so shall we." 'How old is your grandmother now?' 'Sadly, she did not have a very long life and died on her birthday many years ago. This was less than three months before we were born, so we never met her. Coincidentally, her age at death was exactly 1/38 of the year of her birth,' the sister replies. #Puzzle 142.2 This is an old puzzle and it is highly likely that many Problematics readers have come across it before. Nevertheless, it is such a classic that it had to appear in this column sooner or later. A hot tap fills a bathtub in 5 minutes, and a cold tap fills it in 4 minutes. This is when each tap is turned on singly. If you turn off both taps and pull out the plug, a full bath empties in 10 minutes. One day, a user finds that the bath is taking unusually long to fill even though both taps are on. The poor fellow has not realised that he has forgotten to plug the bottom, which is draining out part of the water coming in. Kabir, The practical answer to the magician's card trick is quite straightforward; the magician observed the bottom-most card after the deck was shuffled. It is unlikely he could have seen any other card. The mathematically derived answer is also the bottom-most card. After dealing the first 12 cards, 40 remain in the pile. When the 8 cards are stacked at the bottom, then the original bottom card is #40 out of 48. Now, irrespective of what cards are face up, since we are topping up till 10, the sum of (the four face-up values) and (the number of top-up cards) will always be equal to 40. So if X number of cards dealt out on top of face-up cards, then the number of cards remaining in the deck would be (48 – X) with the original bottom card at position (40 – X) which is the sum of the pip values of the face-up cards. — Abhinav Mittal, Singapore #Puzzle 141.2 Hi Kabir, In the mathematical trick, the sum of the first two numbers is always 2024, and the same is the sum of the next two numbers. This means that the sum of the first four numbers is 2024 + 2024 = 4048. Therefore, the sum of all the 5 numbers = 4048 + the number of people in the room. In this case, since the sum is 4061, the number of people in the room = 4061 –4048 = 13. — Professor Anshul Kumar, Delhi Solved both puzzles: Abhinav Mittal (Singapore), Professor Anshul Kumar (Delhi), Kanwarjit Singh (Chief Commissioner of Income-tax, retired), Anil Khanna (Ghaziabad), Yadvendra Somra (Sonipat), Dr Sunita Gupta (Delhi), Shishir Gupta (Indore), Ajay Ashok (Delhi) Solved #Puzzle 141.2: Vinod Mahajan (Delhi), Y K Munjal (Delhi)
Hindustan Times
21-04-2025
- Science
- Hindustan Times
Walking on an escalator
Suppose you walk up a escalator that is itself moving up. Do you count more steps or fewer compared to moving up a stationary staircase of the same length? What happens if you walk down while the escalator moves up, or vice versa? How does the number of steps you walk change as you vary your walking speed, or the number of steps you cover with each stride? Puzzles with escalator steps offer a number of variations. As far as I can remember, they have been used only once in Problematics, and that was in the early days of this column in its current form in HT, close to two-and-a-half years ago. So here's a selection for you, picked up from various published sources and adapted to Problematics-like situations. #Puzzle 139.1 #139.1A: Professor Pandit walks up a small escalator that is also moving up. Walking one step at a time, she takes 16 steps to reach the top. As an experiment, she walks downstairs on a regular staircase. She then takes the upward-moving escalator again. This time, she walks up two steps at a time, and reaches the top with 12 steps. She takes the staircase down again, but unfortunately cannot continue the experiment because a power failure results in the escalator shutting down. #139.1B: In a mall not far away, Professor Pandit's colleague Lecturer Lalaji is conducting his own experiment on another escalator, this one moving downward. Like his colleague, Lecturer Lalaji begins by walking in the same direction as the escalator. Walking slowly downward, he reaches the bottom in 50 steps. The next step of his experiment is different, though. From the bottom, he runs up the down-moving escalator. His upward speed now is 5 times his previous downward speed, but he still maintains the discipline of travelling only one step at a time. To accelerate so much without increasing his stride length, indeed, tells us volumes about Lecturer Lalaji's fitness, even though he is not getting any younger. But to come back to more relevant details, he takes 125 steps to reach the top. He would have loved to continue his experiment, but the manager of the mall has observed him on CCTV. 'These professors!' the manager mutters, and switches off the escalator, much to Lecturer Lalaji's disappointment. #139.1C: In yet another location, two students are conducting an escalator experiment assigned by their teachers Professor Pandit and Lecturer Lalaji. Young Ms Ahmed and her classmate Mr Basu are both walking up an upward escalator. In the time that Mr Basu takes one step, Ms Ahmed takes three. She counts 150 steps, he counts 75. 'How many steps will we see if the escalator's switched off?' the young man wonders. 'That's easily calculated,' his classmate says. #Puzzle 139.2 The two professors we met on the above escalators, incidentally, are both family persons and have two children each. Each child is either a boy or a girl, and all four are friends, we are told. MAILBOX: LAST WEEK'S SOLVERS #Puzzle 138.1 There are several ways that you can avoid relying on longitudes and still explain why any traveller, like Phileas Fogg, will see the sun rise 80 times if they go around the Earth in 79 days, travelling eastward. It was a pleasure going through the various approaches that readers took: Hi Kabir, Here is a simple way to explain why a traveller going around the globe in 79 days, travelling eastward, will see the sun rise 80 times. The sun's angular speed with respect to the Earth is 1 rotation per day, westwards, and the traveller's angular speed with respect to the Earth is 1 rotation per 79 days or 1/79 rotations per day, eastwards. Since the two are rotating in opposite directions, their relative angular speed with respect to each other is the sum of their individual speeds = (1 + 1/79) rotations per day = 80/79 rotations per day. Therefore, in the 79 days that the traveller takes for one round, he will see the sun going around 79 x (80/79) = 80 times. — Professor Anshul Kumar, Delhi *** One rotation of the Earth involves one sunrise. When a traveller is also moving in the easterly direction, he adds one rotation by coming back to the starting point. Hence one extra sunrise. Anybody travelling east and covering the distance in N days will see N + 1 sunrises. — Kanwarjit Singh, Chief Commissioner of Income-tax, retired *** Think of an experiment where you are on a merry-go-round, and your friend is standing outside and watching you. You are running in the same direction as the platform, say clockwise. Now if you move a distance of x and the platform travels a distance of y, the effective distance travelled is (x + y). This gives an analogy of our Problematics puzzle. The traveller and the Earth are rotating in the same direction. So for an outside observer (sun), the total distance covered by the traveller is the sum of the distances by the traveller and the Earth = (1 round by the traveller) + (79 rounds by Earth in 79 days) = 80 rounds in total. — Sampath Kumar V, Coimbatore *** Imagine you're on a 12-hour analog clock, and each full revolution is a day. If you move one hour on the clock face each day, you'll complete the circle in 12 days, but you'll have seen the hour hand 13 times. Apply the same idea to Fogg's travel. — Vinod Mahajan, Delhi [In Vinod Mahajan's analogy, the Earth is presumed to be stationary while the sun rotates around it. The clock face represents the stationary Earth. The sun (hour hand) and the traveller (you on the clock) must move in opposite directions for this analogy to work. — KF] #Puzzle 138.2 Hi Kabir, Other than 3-4-5 (all three sides single digits), the only Pythagoras triple with two integer sides of single digits is 6-8-10. To prove that the solution is unique, we can check all possible combinations with two or all three sides being single digits. This shows that (3, 4, 5) and (6, 8, 10) are unique. — Shishir Gupta, Indore To make it clear, a rigorous proof does exist. My sincere thanks to Professor Anshul Kumar and Sampath Kumar V, who have satisfied my curiosity by sending two different proofs to show that the solution is unique. These are, however, too lengthy to present in this column. As Shishir Gupta and a couple of other readers have pointed out, merely examining at the finite number of possibilities is as good a method as any. Solved both puzzles: Professor Anshul Kumar (Delhi), Kanwarjit Singh (Chief Commissioner of Income-tax, retired), Sampath Kumar V (Coimbatore), Vinod Mahajan (Delhi), Shishir Gupta (Indore), Dr Sunita Gupta (Delhi) Yadvendra Somra (Sonipat), Dr Sunita Gupta (Delhi), Shri Ram Aggarwal (Delhi), Aishwarya Rajarathinam (Coimbatore), Ajay Ashok (Delhi)
