Problematics | Shopping basket
Given the price of each item, how many matchboxes, bars of detergent soap and packets of milk must you buy for the total spending to be exactly ₹500?
There have been at least two occasions in Problematics when I have presented puzzles that could be solved with Diophantine equations, only to find readers using their own ingenious methods to arrive at the solution. These are equations with two or more variables, with integer solutions, and the most common kind is a single equation in two variables.
The following puzzle involves three variables and can be solved by reducing the equation(s) to a form where only one set of integer solutions is possible. From experience, however, I can anticipate readers using their own methods, which is part of the fun of writing Problematics.
#Puzzle 99.1
A woman shopping for household items spends exactly ₹500 on three kinds of items. This does not necessarily mean that she bought nothing else, but the only items we are concerned about are matchboxes, detergent soap and packets of milk. The prices are: ₹1 for a matchbox, ₹10 for a bar of soap and ₹50 for a packet of milk. Put together, the woman buys exactly 100 of these three items.
How many of each item does she buy? Do share the methods you use to arrive at the solution.
#Puzzle 99.2
These are anagrams of the titles of four movies, each featuring the same superstar. It should be easy solving them.
MAILBOX: LAST WEEK’S SOLVERS
#Puzzle 98.1
Hi Kabir,
Let us refer to five persons as P1, P2, P3, P4 and P5. First let us figure out whether the truthful outnumber the liars or the other way round.
I. Assume that the truthful outnumber the liars. This means that the statements of P2, P3 and P4 are false and that of P5 is true. But this means that there are at least 3 liars, contradicting our assumption. Therefore, this assumption does not hold.
II. Assume that the liars outnumber the truthful. This means the statement of P4 is true and that of P5 is false. With P4 identified as truthful, there can be either 3 liars or 4 liars.
IIa. Assume that there are 4 liars. This means that P1, P2 and P3 must also be liars in addition to P5 to make up the number. The statements of P1, P2 and P3 being false means: (i) P1 is not truthful; (ii) Either the number of truthful is not 2 or P2 is not included among the truthful; (iii) The number of liars is not 3. All these three inferences are consistent with the assumption. Therefore, the assumption is valid.
Conclusion: P1, P2, P3 and P5 are liars. P4 is truthful.
IIb. Assume that there are 3 liars. This implies that P3 is also truthful along with P4. For the number of liars to be 3, P1 and P2 must also be liars. The statements of P1 and P2 being false means: (i) P1 is not truthful; (ii) Either the number of truthful is not 2 or P2 is not included among the truthful. Both these inferences are consistent with the assumption. Therefore, the assumption is valid.
Conclusion: P1, P2 and P5 are liars. P3 and P4 are truthful.
Thus, there as two feasible solutions, as concluded above.
— Professor Anshul Kumar, Delhi
#Puzzle 98.2
Hi Kabir,
Say, the numbers are x, y and z on the tops of the first, second and third dice. After the calculations step by step, the number becomes 2x + 5, then 10x + 25, then 10x + 25 + y, then 100x + 250 + 10y, and finally 100x + 250 + 10y + z.
If the magician subtracts 250 from the number reported by the spectator, the number becomes 100x + 10y + z. It will be a three-digit number as x, y and z are all from 1 to 6 only. Hence, the number will be "xyz". Thus the numbers of all three dice are obvious to the magician now.
— Yadvendra Somra, Sonipat
On this, Akshay Bakhai has a question: “I am a bit surprised that you did not word the puzzle to have a round figure (multiple of hundred) in place of 250. Subtracting 250 is not that easy. A round figure such as 200 or 300, or even better if it was 1000, would have made it faster and safer for the magician to guess the numbers on the 3 dice.”
The answer is that subtracting 200, 300 or 1,000 would have made it easier for the audience to guess how the trick works. Let us assume that the magician’s arithmetic skills are sound enough (as they usually are) to subtract 250 from a three-digit number in his head.
Meanwhile, the observation that I had made about a puzzle in a book was correct. Thank you to all readers who have reconfirmed my views.
Notes about #Puzzle 98.1
Readers have offered two possible answers to #Puzzle 98.1 Originally, I intended the puzzle to have only one solution, that the fourth person speaks the truth while the other four are liars. But Professor Anshul Kumar (see mail above) and a couple of other readers have provided the alternative solution: that the first, second and fifth are liars while the third and fourth are truthful.
This calls for a brief discussion. The second alternative springs from the second person’s statement: “Only two [of us are truthful], including myself.” As Professor Kumar has shown, one part of his statement (“including myself”) can be false while the other part (“two truthful persons”) can be true. Strictly speaking, if a person is a liar, you would expect both parts of his statement to be false.
The ambiguity could have been avoided if I had kept his statement simple: “Two,” instead of “Two, including myself.” Again, because I added that unnecessary embellishment, we got these insights that added an alternative way of looking at the puzzle. Setting puzzles can be a learning experience for the setter too.
The following lists include the names of those who have proposed any one or both alternative solutions.
Solved both puzzles: Professor Anshul Kumar (Delhi), Yadvendra Somra (Sonipat), Akshay Bakhai (Mumbai), Sampath Kumar V (Coimbatore), Kanwarjit Singh (Chief Commissioner of Income-Tax, retired), Dr Sunita Gupta (Delhi), Sanjay S (Coimbatore), Ajay Ashok (Mumbai), YK Munjal (Delhi), Harshit Arora (IIT Delhi), Raghunathan Ravindranathan (Coimbatore)
Solved #Puzzle 98.1: Shishir Gupta (Indore), Dr Vivek Jain (Baroda)
Solved #Puzzle 98.2: Anil Khanna (Ghaziabad)
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