A number of puzzles that have appeared in Problematics over the last three-and-a-half years have involved a magician playing tricks on a gullible audience, sometimes with cards, sometimes with matches, and sometimes simply with numbers. In most cases, it is not difficult to figure out how a given trick works.
You may agree, however, that most such tricks provide a lot of fun. For the magician, the fun lies in packaging his chosen mathematical principles into a trick that engages/enthralls his audience. For puzzlers, of course, the fun lies in going backwards and finding out how it’s done. I think readers will enjoy this unusual little trick I came across very recently.
#Puzzle 181.1
Our magician asks you to think of any three consecutive positive integers, none greater than 60. He then asks you to announce any multiple of 3. You choose 24.
“Add the three consecutive numbers and 24,” the magician tells you, “but don’t give me the result.”
You do the addition. The magician gives further instructions: “Multiply the sum by 67 and tell me the last two digits of the product.”
“That’s 51,” you announce.
The magician’s answer is immediate: “Your chosen numbers are 42, 43 and 44. The product you got is 10251, from which you told me the last two digits.”
You wonder how it’s done, so the magician tells you. “When you announced your chosen multiple of 24, I divided it by 3 and added 1, i.e. 24/3 + 1 = 8 + 1 = 9. That was step #1,” he explains.
“In step#2, I took the last two digits of the product that you told me, which was 51, and then subtracted the earlier result of 9. That gave me 51 – 9 = 42, and this was the first of the three consecutive numbers.”
“But how did you get the entire product so fast?” you wonder.
“That was even simpler,” the magician says. “I simply multiplied 51 by 2 and got 102, which are the digits preceding 51 in the product of 10251.”
You wonder if it’s a fluke, so you check with another set of consecutive numbers, 28, 29 and 30. This time, you choose 54 as the multiple of 3. Adding the four numbers gives you 141. When you multiply that by 67, you get 9447. “If I was playing this trick on a friend, he or she would have announced the last two digits, which are 47.”
You now try the magician’s steps. The multiple of 3, which is `54, would have been announced by your friend. “Let me see. I need to divide that by 3 and add 1 to the quotient, which gives 54/3 + 1 = 19. When my friend gives me the last two digits, 47, I subtract 47 – 19 = 28. And indeed,” you observe, “that is the first of the three consecutive numbers, which gives me 28, 29 and 30.”
“Well done. Now try deducing the product,” the magician says.
“I double 47, which gives me 94, and indeed these are the preceding digits in the product of 9441,” you note, with a sense of wonder.
Why does the trick work every time? Explain in general algebraic terms rather than with more examples.
#Puzzle 181.2
A man likes a chocolate brand called Sweeter, while his wife prefers the Bitter brand. They have 50 Sweeter and 50 Bitter chocolates, which they want to place in two containers. “Assume that the lights go out and I have to pick a chocolate, in the dark, from either jar at random. Divide the 100 chocolates between the two jars in a way that gives me the highest possible probability of getting a Sweeter.”
Help his wife find the ideal distribution.
#Puzzle 180.1
In last week’s betting puzzle, it turns out, I provided more information than was necessary. Thanks to Yadvendra Somra for showing how it could have been solved even if one piece of data had not been provided:
Dear Kabir,
Say, the amounts bet on France, Argentina, Croatia and Morocco are F, A, C and M respectively. The bettor’s returns will be 3F = 4A = 9C = 12M. It is given that M = ₹300. So 3F = 4A = 9C = 12M = 3600, which gives F = ₹1200, A = ₹900, C = ₹400. The total money borrowed is F + A + C + M = ₹2800.
Note that the profit of ₹800, which was given in the puzzle, has not been used in the above solution. The profit can be derived from subtracting 3600 – 2800 = ₹800. This shows that M = ₹300 (given) and the profit of ₹800 are dependent on each other. The puzzle could have given either of these, rather than both.
Suppose the puzzle had given the required profit of ₹800, but had not given the amount bet on Morocco.
3F = 4A = 9C = 12M = E (say)
This means E has to be multiple of the LCM of 3, 4 9 and 12, which is 36. For E = 36, we have M = 3, C = 4, A = 9 and F = 12. The profit = E – (F + A + C + M) = 36 – (12 + 9 + 4 + 3) = 36 – 28 = 8. So, for the profit to be 8, M should be 3. Therefore for profit to be ₹800, M should be (3/8) x 800 = ₹300. Therefore F = ₹1200, A = ₹900, and C = ₹400. Amount borrowed = total investment = ₹2800.
— Yadvendra Somra, Sonipat
#Puzzle 180.2
Hi Kabir,
Among nightingales, it is actually the male bird that majorly sings. In his poem, Keats addresses the bird as "dryad" meaning a wood nymph, which is a female celestial being. Hence Keats is referring to the female bird. The same fallacy holds when we give the sobriquet of nightingale to female personalities such as Sarojini Naidu or Lata Mangeshkar for their eloquence/singing talent. It would be more appropriate if Rabindranath Tagore/Kishore Kumar were referred to as “Nightingale of India” in that light.
— Sabornee Jana, Mumbai
Some readers have pointed out other discrepancies, such as nightingales preferring dense undergrowth rather than high perches in beech trees as described by Keats, and "nightingale of India" being a misnomer because the bird is not native to India. The most glaring issue, however, was about male and female. The list below acknowledges solvers who have pointed out that it's the male that sings.
Solved both puzzles: Yadvendra Somra (Sonipat), Sabornee Jana (Mumbai), Dr Sunita Gupta (Delhi), Shishir Gupta (Indore)
Solved #Puzzle 180.1: Vinod Mahajan (Delhi), Kanwarjit Singh (Chief Commissioner of Income-tax, retired), Ajay Ashok (Delhi), Shri Ram Aggarwal (Delhi), Y K Munjal (Delhi)