Problematics | Cat food
Food procured by a pet shop lasts longer than expected, with pets being sold regularly. Can you work out the quantity of pet food?
In the old days, any image of a cat lapping up milk from a saucer would mean a very happy image. But with the times doing what Bob Dylan says they are, we are told today that milk is generally not good for cats as most adults among them are lactose intolerant. In another departure from the old days when cats would share whatever humans themselves ate, we now have a thriving industry in pet food, much of it prepared exclusively for cats. Cans and cans of them.
Pet store owners keep lots of cat food, sometimes to sell to pet owners, and sometimes to feed their own cats on. The following puzzle involves a store owner who sells pets rather than pet food, but he still keeps plenty of packaged cat food because that is the only thing his cats eat.
#Puzzle 144.1
Our pet store owner buys a large number of cans of cat food. He has 31 cats at the time of the purchase, and the number of cans he buys is an exact multiple of 31, planned to sustain the cats for a fixed number of days. Each cat consumes one can daily.
The store owner has made his purchase on the assumption that he will still have 31 cats when the stock of food runs out. But given the high demand for pets, the number of cats in the shop keeps declining steadily. That does not mean any of the cans go waste, because any unused can may be given to any unsold cat.
On day #1, every cat consumes its allotted can. On day #2, one of the cats is sold before mealtime; the remaining cats are fed after the sale. On day #3, a second cat is sold before breakfast, and then the remaining cats are fed. By a stroke of coincidence, the store owner keeps selling exactly one cat every morning, and always before the sold cat has begun to consume its daily ration.
The pet buyers are obviously expected to procure their own cat food. As a result, the store owner’s stock of cans lasts longer than expected. The evening the food is finally exhausted, he ponders: “If my stock had ended tomorrow instead of today, it would have lasted exactly twice as many days as the number of days I had originally intended it to last.” He then proceeds to order a fresh stock of cans.
How many cans of cat food were in the original stock, and how many of the 31 cats remain unsold when the store owner makes his observation?
#Puzzle 144.2
The late Martin Gardner contributed this simple party trick to Children’s Digest magazine in 1952. He presented illustrations and spellings of eight creatures to his child audience. These are (in alphabetical order):
BUTTERFLY, COW, ELEPHANT, HORSE, LION, MONKEY, OSTRICH, RHINOCEROS
Ask a child to secretly select any one of the creatures. Tell her to silently spell out the name of the creature while you tap on the eight images with a pencil, one letter to a tap. Let’s say she has chosen elephant. You tap on one image among these eight, and she spells E in her mind. You tap on a second image, and she counts that as L. While she spells the word in the correct sequence of letters, your taps do not follow the alphabetical order of the creatures. When she reaches the T at the end of E-L-E-P-H-A-N-T, the child calls out “Stop!” And bingo, she finds, your pencil has landed precisely on the elephant image.
A second child makes a different choice and similarly spells its name, letter by letter, tap by tap. Once again, when the spelling ends, your tap lands precisely on the chosen creature. And then a third child: different creature, same result.
How does the trick work?
MAILBOX: LAST WEEK’S SOLVERS
#Puzzle 143.1
Dear Kabir,
The solution is as shown in the table. All clues describing positions are taken as being from the audience's perspective. If I take those clues as being from the stage performers’ perspective, then there will be more than a dozen different solutions possible.
— Sampath Kumar V, Coimbatore
#Puzzle 143.2
Dear Kabir Ji,
I could solve the speed-and-distance puzzle with mere logic.
Time taken by the coach’s car to reach the tea stall after leaving the runner
= time taken by the runner to take 29 strides.
The runner himself needs to take a further 116 strides to reach the stall. Total distance covered = 29 + 116 = 145 = (5 x 29) strides
Logically, this means the speed of the car must be 5 times the speed of the runner (who is going at a steady 10 km/hr). Therefore, the speed of the car = 5 x 10 = 50 kph.
— Shri Ram Aggarwal, Palam, New Delhi
Solved both puzzles: Sampath Kumar V (Coimbatore), Shri Ram Aggarwal (Delhi), Abhinav Mital (Singapore), Dr Sunita Gupta (Delhi), Yadvendra Somra (Sonipat), Kanwarjit Singh (Chief Commissioner of Income-tax, retired), Professor Anshul Kumar (Delhi), Anil Khanna (Ghaziabad), Shishir Gupta (Indore), Ajay Ashok (Delhi)
Solved #Puzzle 143.1: Dr Vivek Jain (Baroda)
Solved #Puzzle 143.2: Vinod Mahajan (Delhi), YK Munjal (Delhi), Aishwarya Rajarathinam (Coimbatore)
