Problematics | A magic square, but not the kind everyone knows
This week, we have a one-of-its-kind magic square and a math problem. Good luck!
Magic squares are just that, magic, but they may be too familiar for puzzle solvers like you (and setters like me) who are looking for originality. There are variations, of course, and I shall try and bring them to you from time to time, but not today.
What you see below is not a variation of the traditional magic square in which all rows and columns and both diagonals add up to the same number. It is a magic square of its own kind, its properties entirely different.
#Puzzle 34.1
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The one labelled “original square” has, as you can see, 25 numbers including two negative integers. Start by selecting any of the 25 numbers and deleting all other numbers in its row and in its column. To illustrate with an example, let us say you select 13 in row 4, column 2. So, delete all other numbers in row 4 and column 2, eliminating them from the rest of this exercise.
Storing 13 in your kitty, select another number from what remains, i.e. from among those that have not yet been selected or eliminated. Let’s say that this time you select –2 (row 2, column 4) and consequently delete all other numbers in row 2 and column 4. Add the number –2 to your kitty, which already has the number 13.
Carry on this way, three more times. On the fifth round, there will be only one number remaining, which you select. In this illustrative round, the numbers chosen after 13 and –2 are 42, 15 and 32 (the only number remaining at the end).
Add the five chosen numbers, and you get 13 – 2 + 42 + 15 + 32 = 100.
Start from scratch, selecting any five numbers and eliminating other numbers, following the same rules as discussed above. Add the five selected numbers. No matter which numbers you choose, you will find that the sum is 100 again. Each time.
What is the mathematics behind the magic?
| #Puzzle 34.2: |
a² – b² = x³ a³ – b³ = y² I know that some readers will run this through a computer or a mobile app, but still: If a, b, x, and y are all positive integers, what are the smallest possible values for a and b? |
Mailbox: Last week’s solvers:
Hi Kabir, If I am the Amateur and given the first chance to remove matches as suggested by you, I will remove 2 on first move, hence leaving 31 for the Ace Player. In the first move, if Ace Player removed x number, I will remove (5 – x) in my next move leaving 26 for Ace Player. If I repeat the strategy, Ace Player will be left with 21, 16, 11, 6 and finally 1. Hence, Ace Player will lose the game contrary to what you want. For Ace Player to win, he has to move first and follow the strategy suggested by me. — Dr G L Arora, Delhi [Like Dr Arora, Akshay Bakhai of Mumbai too has argued (correctly) that the Ace Player cannot win every time. The only way the Amateur may win is by picking 2 matches in the beginning. But even if he does, he can still lose: as many others readers have observed, the Ace can wait for subsequent rounds when the Amateur fails to pick up (5 – x) matches (that’s why he is an amateur).] |
Hi Kabir, The couple reached home 20 minutes earlier when the husband walked a bit. Because the wife started at her usual time and travels the same distance each way, she must have saved 10 minutes each way. This means she must have picked her husband 10 minutes earlier, i.e, at 7:50 pm. This means the husband must have walked for 50 minutes. — Rahul Agarwal, Bay Area, California |
| Solved both puzzles: Rahul Agarwal (Bay Area, California), Dr G L Arora (Delhi), Akshay Bakhai (Mumbai), Puneet Vashistha (Delhi), Amardeep Singh (Meerut), Anil Kumar Goyal (Delhi), Harshit Arora (Delhi), Amar Lal Miglani (Mohali), Sunita and Naresh Dhillon (Gurgaon), Jasvinder Singh (Nabha), Nishil Rathod |
| Solved #Puzzle 33.1: Aditya Mittal (Gurgaon), Sumit Sharma |
| Solved #Puzzle 33.2: Dr Anurag Mishra (Maulana Azad Medical College), Vikas Goel (Gurgaon), Naresh Kumar Khurana (Sonipat), Dr Sunita Gupta (Delhi), Dr Nakul Makkar (Noida), Shishir Gupta (Indore). Madhuri Patwardhan (Thane), Nipun Bamania (Mumbai), Mayobhav Pathak (Gurgaon) |
Problematics will be back next week. Please send in your replies before Friday noon to problematics@hindustantimes.com
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