The turnout this month was slightly lower than usual, but many of our regulars were there and we had the chance to have some good discussions and really get into things.
We kicked off with a starter puzzle: Find the smallest (non-zero!) number which is: the sum of 5 consecutive integers; the sum of two consecutive odd numbers; and the sum of three consecutive even numbers. Some people got the answer by trial and error, and others used algebra, but everyone got there within a few minutes – including a couple of people on Twitter who felt the need to give away the answer to anyone reading the @MathsJam Twitter feed. Mean!
Another simple starter went as follows: four cards are placed face down, and each has a number on the back. When three of the cards are turned over, the sum is either 14, 16, 19 or 20 depending on which three cards are turned. What are the four cards? Everyone copied my diagram, even though I’d just drawn four rectangles and labelled them A B C and D, which started a requirement for me to draw diagrams accompanying every puzzle, no matter how unnecessary. A couple of people got different values for B, even though it wasn’t specified which total corresponded to which set of cards, and a small fight broke out.
Last month, one of our simple starter puzzles was to find a four-digit number, which when quadrupled, gives the same digits in the reverse order. Since then, Paul had checked using a computer that no other four-digit numbers than the solution we got (2178) have this property. In doing so, he discovered that 21978 also does this, as does 219978, and in fact any number of 9s can be inserted in the middle and the number will still reverse its digits when quadrupled. We all tried to find a way to prove this fact, and we got to some really nice (slightly hand-waving) proofs, involving an errant 3 bouncing along the top of a row of 9s. Try it!
We briefly discussed the issue raised in my tweet, ‘Do you call yourself a mathematician and why/why not?’ Many viewpoints were raised, and some of the chatter has found its way into my article, which can be found here.
A puzzle posed by Glasgow MathsJam had a thousand barrels of wine, one of which is tainted with a slow-acting poison, and ten wine tasters. The poison takes thirty days to kill you. In thirty-one days, can they determine which barrel is poisoned? If you’d like to try to solve this puzzle, look away now, but we came to the following: you get each of the wine tasters to taste different subsets of the wine, as follows: the first taster tastes odd-numbered barrels (1, 3, 5, …); the second taster tastes two consecutive barrels in every four, starting at barrels 2 & 3, (2, 3, 6, 7, …) and so on, so that each taster will taste a barrel if and only if the binary expansion of the number on the barrel contains a 1 in the position he is standing in the line. In this way, each of the 100 barrels will have been drunk by a distinct subset of the wine tasters. Then you line the wine tasters up on day 30, from right to left, and – this is the genius bit – look at this line from above, e.g. in a helicopter. Then, especially if the wine tasters have bald patches, you can look down and see a 0 in any place a man is standing, formed by his round head, and a 1 wherever a man is lying dead having fallen over. This will then spell out the binary expansion of the number of the poisoned barrel. You can’t say Manchester MathsJam isn’t aesthetically-minded. The puzzle also prompted Andrew to draw a Donkey Kong-themed cartoon.
After discussing ordering of bridge hands last month, Ben brought another cards puzzle. An old card trick, performed by two magicians, has five random cards being drawn from a deck. One magician passes the other magician four of those cards, one at a time, and then the other can name the fifth card without seeing it. A nice explanation of how the trick works is given here. Ben’s puzzle was, given a bridge hand of 13 cards, can you devise a similar strategy, using as few of the cards as possible, to allow your accomplice to identify the remaining cards? We looked at a few different ideas, and Paul hit on the following: in a bridge hand, you’ll always have at least one suit containing four or more cards (by the Pigeonhole Principle), and so you could construct a strategy to allow your partner to identify those four plus one from the suit above, or five the same if you have a five-card suit. Can you work out what the theoretical strategy might be?
We finished the evening by playing Perudo, a dice game which is similar to that played by characters in the Pirates of the Caribbean films. It involves making a bet as to how many dice showing a particular number are on the table (covered by each player’s cup), and whoever bids the highest gets to see if they are right, and lose a dice if they are wrong. It’s quite good fun, and requires a bit of memory and a bit of probabilistic reasoning. We also noticed that the images of dice on the box are not only all identical, but they’re also not standard dice – 3 and 4 are on adjacent faces.
It also turns out I’m not very good at Perudo.