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Problematics | Dice and oranges
In this week’s puzzles, decode a party trick that uses dice, and reconstruct a pyramid using oranges
When describing puzzles related to dice, it is always necessary to mention the configuration at the cost of repetition. If the die is standard, the 6 is always opposite the 1, the 5 is always opposite the 2, and the 4 is always opposite the 3. We are mentioning this for the umpteenth time, and will do so again whenever we have a dice puzzle where opposite faces are relevant.
Dice now, numbers last week, and cards on so many previous occasions. This week’s puzzle involves yet another party trick.
#Puzzle 182.1
A magician hands you three standard dice and turns his back. He asks you to roll the dice, look at the spots on the top faces and treat them as a three-digit number. For example, if you have thrown 3, 5 and 2, he is to treat the throw as 352.
The magician asks you to look at the bottom faces to the three-digit number, and append these spots to get a six-digit number. Since the dice are standard, the bottom faces will show 4, 2 and 5, so you append 425 to get 352425.
The magician asks you to divide the six-digit number by 111, assuring you there will be no remainder. Indeed, 352425/111 gives you 3175 without a remainder. You announce this result to the magician who, his back still turned, tells you the spots on the three faces.
As a puzzler, you want to know how it’s done. The magician tells you that it takes just two steps. First, subtract 7 from the announced number, 3175 – 7 = 3168. Next, divide this result by 9, 3168/9 = 352, which turns out to be your original three-digit number.
You experiment with a random three-digit number. First, your steps. Say the numbers thrown are 1, 6 and 4. The spots on the bottom faces will then be 6, 1 and 3. Take 164, append 613 to get 164613, and divide by 111 to get 164613/111 = 1483. This will be the announced result.
Now, the magician’s steps. Subtracting 7 from 1483 gives 1476, then dividing this by 9 gives 164. Indeed, these are the three numbers originally thrown. So the trick works with the new combination too. In fact, it works with any combination.
How does the trick work? As usual, give a general algebraic explanation.
#Puzzle 182.2
It is possible to form pyramid structures with cannonballs, billiard balls and even round fruits. The base can be either a square (left image) or a triangle (right image, with top layer missing). A child forms a square pyramid with a number of oranges. Her younger brother takes another orange (not part of the formation) and hurls it at the pyramid, bringing down the pyramid. Unperturbed, the elder sister picks up the fallen oranges, adds her brother’s single orange and uses the lot to form a new pyramid. This one is triangular.
Within plausible limits, how many oranges are used in the square and triangular formations?
MAILBOX: LAST WEEK’S SOLVERS
#Puzzle 181.1
Dear Kabir,
Here is how the trick with consecutive numbers works. First, add three three chosen consecutive numbers (x – 1), x and (x + 1) to get 3x. To this, add a multiple of 3 (let this be 3n) to get 3x + 3n = 3(x + n). Multiplying this sum by 67 gives 201(x + n). The initial digits of this product will be 2(x + n) and the last two digits will be (x + n).
Now 3n and (x + n) are known. The magician takes 1/3 of 3n, adds 1 and subtracts the result from (x + n) to get (x + n) – (n + 1) = (x – 1), which is the first number in the consecutive series. This allows him to announce the entire series. The magician also doubles (x + n) to get 2(x + n), which gives him the initial digits of the product.
— Dr Sunita Gupta, New Delhi
As Sr Sunita Gupta notes, the condition that none of the chosen consecutive numbers should exceed 60 is important. Professor Anshul Kumar has also sent a generalised solution for the different situations that will turn up if this limit is exceeded (which I will not go into here). Meanwhile, let us look at Professor Kumar’s solution to the sweet probability puzzle.
#Puzzle 181.2
Hi Kabir,
Put just one Sweeter in one of the jars and the remaining 49 Sweeters along with all the 50 Bitters in the other jar. If a chocolate is randomly chosen from a randomly chosen container, the probability of finding a Sweeter is
(1/2)(1/1) + (1/2)(49/99)
This is 0.747 and nearly equal to 3/4.
— Professor Anshul Kumar, New Delhi
Solved both puzzles: Dr Sunita Gupta (Delhi), Professor Anshul Kumar (Delhi), Dr Nitin Trasi (Sydney), Shri Ram Aggarwal (Delhi), Yadvendra Somra (Sonipat), Vinod Mahajan (Delhi), Ajay Ashok (Delhi)
Solved #Puzzle 181.1: Anil Khanna (Ghaziabad), YK Munjal (Delhi), Shishir Gupta (Indore), Sabornee Jana (Mumbai)
#Puzzle 181.2: Dr Vivek Jain (Baroda)
