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Problematics | Roll of the dice
This week, a puzzle based on a dice trick first published in 1937, and one with liquid containers that is centuries older.
We have said it before, and we will say it again, for puzzles involving dice must always begin with this disclaimer. The standard for dice is that the face with one spot is opposite the face with six spots, 2 is opposite 5, and 3 is opposite 4. In other words, the spots on any two opposite should add up to 7. Not all dice manufacturers follow this, but that is the standard. Party tricks involving dice have to use the standard kind. Such as the following one, contributed by Frank Dodd of New York to The Jinx magazine in 1937.
#Puzzle 157.1
A magician keeps his left hand inside his pocket, secretly holding a certain number of matches. With his right hand, he gives three standard dice to a spectator and turns his back. “Roll them around your fingers until you are satisfied that the faces are now random. Then place the three dice one on top of the other.”
After the spectator has done so, the magician asks her to add the spots on the touching faces of the top two dice. In other words, she adds the bottom spots on the topmost die to the top spots of the second die.
Next, he asks her to add the spots on the touching faces of the lower two die, i.e. the bottom spots on the second die and the top spots on the lowermost die.
Finally, the magician asks the spectator to add those two totals, and then add the spots on the bottom face of the lowermost die. “Don’t tell me the grand total,” he advises her.
The magician now turns back to face the spectator. He looks at the spots on the top face of the topmost die, which are visible to all. He now drops a certain number of matches from his hidden left fist into his pocket. Finally, he takes out his hand and scatters the remaining matches on the table, with a dramatic flourish. Count them, he invites the spectator. She does that. Yes, the number of matches is exactly the grand total she had achieved!
How is the trick done?
#Puzzle 157.2
This is a puzzle that mathematics writer Ian Stewart attributes to Italian mathematician Tartaglia of the 1500s. You have three containers of capacities 8, 5 and 3 litres respectively. You have no other measuring instrument. The 8-litre container is full of water; the other two are empty. By repeatedly transferring the contents of one container into another, you need to end up with 4 litres each in the larger two containers. Assume you are skilled in transferring liquid without spilling a drop.
In how many steps can you do it?
MAILBOX: LAST WEEK’S SOVERS
#Puzzle 156.1
Hello Kabir,
The wife had ₹6331 in 63 notes and 31 coins. The husband had ₹3163 in 31 notes and 63 coins. The wife spent ₹5 leaving her with ₹6326, which is twice the amount her husband had. Back in the 1980s, ₹9494 was a lot of money to be carrying around. And they went to see a movie for ₹2.50 a ticket? Real cheapskates!
— Sanjay Gupta, New Delhi
Cheapskates, yes, especially if you consider that ₹9494 was the minimum they must have been carrying. The solution would have been unique if I had specified an upper limit of ₹9494, but I missed that. It was only after receiving the following explanation from Professor Anshul Kumar that I realised there could be an infinite number of solutions, starting ₹9494 and going higher:
Hi Kabir,
Suppose the wife is carrying A ₹100 notes and B ₹1 coins, so she has a total amount of ₹(100A + B) while her husband is carrying is ₹ (100B + A). Then we can say that,
100A + B = 2(100B + A) + 5
=> 98A – 199B = 5
=> 98(A – 2B) = 3B + 5
The right hand side must be a multiple of 98. We can, therefore, write -
3B + 5 = 98n, where n = A – 2B, a positive integer
=> 3B = 98n - 5
=> 3B = 93n + 5(n - 1)
n – 1 must be a multiple of 3. Let it be 3p, where p is a non-negative integer. Then,
3B = 93(3p + 1) + 15p
=> B = 31(3p + 1) + 5p = 98p + 31 and A = 2B + n = 196p + 62 + 3p + 1 = 199p + 63
For different values of p, we get different solutions.
p = 0 => A = 63, B = 31
p = 1 => A = 262, B = 129
p = 2 => A = 461, B = 227
p = 3 => A = 660, B = 325
and so on.
— Professor Anshul Kumar, New Delhi
# Puzzle 156.2
Hi Kabir,
The puzzle should have stated “names of the states to which the professors belong” instead of “names of the professors”. It can be inferred that the woman professor is not from Punjab, that she is not the mathematician, and the mathematician (male) cannot be from Manipur. Since the second statement is by a professor from Bihar, he cannot be teaching botany. Therefore, he teaches physics. As one of the male professors belongs to Punjab, this must be the one who made the third statement. Therefore, (1) the woman professor belongs to Manipur and teaches botany; (2) the other male professor is from Bihar and teaches physics; (2) the other male professor is from Punjab and teaches mathematics.
— Yadvendra Somra, Sonipat
Solved both puzzles: Sanjay Gupta (Delhi), Professor Anshul Kumar (Delhi), Yadvendra Somra (Sonipat), Dr Sunita Gupta (Delhi), Anil Khanna (Ghaziabad), Vinod Mahajan (Delhi), Dr Vivek Jain (Baroda), Shishir Gupta (Indore), Ajay Ashok (Delhi), Kanwarjit Singh (Chief Commissioner of Income-tax, retired), Shri Ram Aggarwal (Delhi)
Solved #Puzzle 156.2: YK Munjal (Delhi)
