Problematics | Magic squares in a calendar
Two party tricks, one with dates on a calendar and the other with playing cards. What makes them work?
I am not sure if tricks with numbers, cards and dice are played at parties any longer, but they will always be part of Problematics. Many of these tricks can be quite impressive, mysterious even, if they are played with a certain amount of flair and if the ‘magician’ has a gift for what we call showmanship. However, as we have seen in previous episodes of Problematics, it is often a very simple underlying principle that makes these tricks work.
The following trick is one of those. I found it in the writings of Martin Gardner, who attributes it to Tom Sellers.
#Puzzle 105.1
Should we go back to the smart alec who plays tricks on you at every party? Why not.
“Get me that calendar hanging on the wall,” the smart alec tells you, and you do that.
Smart alec: Circle any 3 x 3 square containing nine dates. Do not choose a section where any square is blank. I won’t be looking while you do this, but let everyone else see the 9 dates you have selected. I am turning my back now.
You: Done. The circled 3x3 square contains nine dates, no blank squares.
Smart alec: What is the smallest date inside the circle?
You: It is ___
Smart alec: The sum of the nine numbers is ___
The section circled in the illustration is just an example. The smart alec will be able to tell you the sum of the nine dates in any 3x3 square (no blanks) that you select.
How does he do it, and what is the arithmetic going on here?
#Puzzle 105.2
A simpler trick now, this time with playing cards. The smart alec holds a set of 13 cards, all face down. He transfers the top card to the bottom, calling out “A”. The next card goes to the bottom, with a call of “C”. And then the third card, with a call of “E”. The smart alec now calls out “Ace!” and turns up the fourth card. Bingo: It is indeed an ace. The card is now removed from the set.
The so-called magician continues the same procedure with the remaining 12 cards. “T, W, O,” he calls out as he transfers cards one by one from the top to the bottom. He turns up the next card, which is indeed a 2, and which is now removed. Then “T, H, R, E, E,” after which the next card is a 3, and so on. The smart alec exhausts all 13 cards this way, right up to J-A-C-K and Q-U-E-E-N. When just one card remains, he needlessly calls out “K, I, N, G” before turning it up — a king, of course.
Yes, he ordered the 13 cards in advance. What is that order?
MAILBOX: LAST WEEK’S SOLVERS
#Puzzle 104.1
Hi Kabir,
I made a table of all combinations in which the four batsmen’s unequal and nonzero scores add up to less than 18. We know that solving the puzzle depends on knowing whether Garfield scored 1 or more than 1. In the table, there is only one product (120) that is common between the combinations in which Garfield’s score is 1 and the combinations in which Garfield’s score in more than 1. Now, if Garfield scored 1, there are two possible combinations for a product of 120: (1, 3, 5, 8) and (1, 4, 5, 6). In that case, the solver would still not be able to determine the answer. But if Garfield scored more than 1, there is only one combination remaining: (2, 3, 4, 5). Since the solver cracked the puzzle after knowing whether Garfield scored 1 or more than 1, this must be the right combination of scores. So, Garfield scored 2, Viv scored 3, Don scored 4, and Sunny scored 5.
— Dr Sunita Gupta, Delhi
#Puzzle 104.2
The algebraic puzzle (x/3 + x²/2 + x³/6 = 1), unfortunately, did not appear the way I had planned it to be. I first called for integer solutions to the equation, after which I received a couple of replies. Then it struck me that the puzzle was not what I had meant it to be, and I called for non-integer solutions. More solutions arrived, but the puzzle was still different from the intended form.
I will revive the puzzle at a later date, framing it exactly as it was meant to be. For now, I am crediting all correct answers to either of the two versions published.
Dear Kabir,
x/3 + x²/2 + x³/6 = 1
=> x³ + 3x² + 2x – 6 = 0; which can be factorised to:
(x – 1) * (x² + 4x + 6) = 0
Integer solution means x – 1 = 0 or x = 1. Non-integer solution means (x² + 4x + 6) = 0, or
(x + 2)² + 2 = 0 => (x + 2)² = –2
This will have imaginary roots. So, no real solutions exist other than x = 1, which is an integer solution.
— Sampath Kumar V, Coimbatore
Solved both puzzles: Dr Sunita Gupta (Delhi), Sampath Kumar V (IIT Delhi), Harshit Arora (IIT Delhi)
Solved #Puzzle 104.2: Anil Khanna (Ghaziabad), Akshay Bakhai (Mumbai), Kanwarjit Singh (Chief Commissioner of Income-Tax, retired), Aditya Krishnan (Coimbatore), Yadvendra Somra (Sonipat), Sanjay Gupta (Delhi), Professor Anshul Kumar (Delhi), Shri Ram Aggarwal (Delhi), YK Munjal (Delhi), Ajay Ashok (Delhi), Shishir Gupta (Indore)