MATHS WEEK STARTED on Saturday and, as is our annual tradition, we’re setting our readers some puzzles. Give them a go!
We had a slight technical hitch with Wednesday’s puzzle so we’re giving you a bumper double edition tonight.
Day 6 Puzzle One
Working backwards is a useful strategy in many cases.
In fact, it can be the only way to solve a problem when we know where we want to end up but aren’t sure how to get there.
Children learn quite young that this is a useful strategy doing mazes – start at the end and work backwards.
The following puzzle is adapted from legendary puzzler Henry Dudeney’s Amusements in Mathematics (1917).
Three people, whose names were Adams, Baker and Carter, played a card game.
They had agreed that whenever a player won a game, that player should double the money of each of the other players — that is, give the players just as much money as they had already.
They played three games, and, strange to say, each won a game in turn, in the order in which their names are given above. But a more curious coincidence is this — that when they had finished play, each of the three had exactly the same amount — €8.
The puzzle is to find out how much money each had when they sat down to play.
In this puzzle, we know the end state but we don’t know anything about the starting conditions.
Therefore, we’ll have to work backwards. Another useful strategy is to draw a table.
We have the conditions after game 3 – what must the conditions be after game 2 etc?
Give it a try and see how you get on.
Day 6 Puzzle Two
In problem solving, it’s often a useful strategy to examine the penultimate step.
This might be the last step before the solution, or commonly the last step before you realise you have gone wrong. If you realise you’ve gone astray it might be a good idea to back up and try something different.
We’re going to play a game of Nim, which is a mathematical strategy game where players take away pieces from a gathering of objects.
The form we want you to play now is for two players and uses 18 pieces laid out on a table (matchsticks, counters, coins, sweets, can be used).
A move consists of removing between 1 and 4 pieces. Players take their moves in turn and the player left with the very last piece on the table loses.
There’s a strategy where you can always win if you go first. Working backward is the key here.
Can you use your problem solving skills to figure out the winning strategy? How many pieces should you leave on your second last go to be sure of winning?
In this way can you identify all the intermediate places you should aim for at each go. It might be useful to use the ‘simplify’ strategy – start with a lot less than 18 pieces and build up. If you can’t find pieces to use, drawing a picture will be useful.
For further exercise, you could change the rules of the game – start with a different number of pieces, allow a different range of pieces to be taken away, have the person who takes the last piece become the winner. Determine the winning strategy in your variation.
Come back tomorrow for the answer to today’s puzzles.
Puzzles compiled for The Journal by Eoin Gill of Maths Week Ireland / Waterford Institute of Technology.
Tuesday’s puzzle: The answer
1) 28 matches are needed for the league of 8 players.
How we got that answer: Simplifying the problem is a useful strategy here. One player will require no matches. Two players will require one match. If another player joins them (3) that match will still go ahead, but each player will play the new player making 3 matches.
Adding another player will still have those 3 matches but each player will play the new player – 6 matches in total. So adding a new player adds the previous number of players to the previous total of matches.
2) 4,950 matches would be needed with 100 teams
You might further notice from the above exploration that for 3 players there is 2 + 1 = 3 matches
For 4 players there is 3+2+1 = 6 matches
For 5 players there is 4 + 3 + 2 +1 =10 matches
So for 100 matches there would be 99+98+97….+1 matches.
There is a quick way of adding these by pairing up numbers -
(99+1) + (98+2) + (97+3) ……(51+49) + 50 (there is nothing to pair with 50)
That is 49×100 + 50 = 4,950
A more sophisticated and versatile approach would be to derive a formula from your observations. The formula n(n-1)/2 (where n is the number of players involved) allows us to calculate the number of matches.
If n=3 we need 3(3-1)/2 = 3 matches (as above).
If n=100 we need 100(100-9)/2 = 4,950 (as above)
The formula can be applied to other situations, such as a number, n, of people that all shake hands with each other. Or the number of teams, n, in a football league (if teams play home and away, you multiply the answer by 2).
If we wanted to understand how people are connected together on the internet or social media, we could use a similar approach. As we model complex real world applications, we need to derive more complex formula. We can programme computers to run these complex formulae on masses of data to give confident predictions of the future. That is the power of maths.
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