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Maths Week: Tuesday puzzle

Fancy another mathematics challenge? (And get the answer to yesterday’s puzzle.)

Image: Alamy Stock Photo

MATHS WEEK STARTED on Saturday and, as is our annual tradition, we’re setting our readers some puzzles. Give them a go!

Day 4

“Everything should be made as simple as possible, but no simpler.” – Albert Einstein

So, when a problem looks daunting and complicated, we should try to simplify it. A simpler version of the problem might actually give us the answer – or more often give us insights that help guide us to the ultimate solution.

For example: In a knockout tournament, there are 100 players with one player advancing after each match. How many matches will there be?

This problem looks quite difficult, or tedious if you want to count through all the matches.

Why don’t we simplify it and consider 2 players? Then try 3 and then 4 and see what happens. For 2 players we require 1 match, for 3 players, 2 matches. For 4 and 5 we will get 3 and 4 matches. A pattern emerges that we need one match less than the number of players. For a 100 players we can be confident about claiming 99 matches are needed.

(You might also have the insight that every match creates a loser that exits the tournament. Losers number one less than the total number in the tournament.)

It is a little more complicated for a round robin or a league, where each player plays every other player. So today’s questions: 

How many matches would be needed in a league of 8 players, where each player plays every other player once?

What if there were 100 players?

Come back tomorrow for the answer to today’s puzzle.

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Puzzles compiled for The Journal by Eoin Gill of Maths Week Ireland / Waterford Institute of Technology.

Monday’s puzzle: The answer 

1) 3 socks. When you pull two socks out you might get a pair or you might get one of each. The next sock would have to match one of them (as there are only two colours).

2) Yes we can be certain. The least number of people a mathematician can shake hands with is zero and the maximum is 99. That’s 100 options and with 100 mathematicians; it seems like every mathematician could shake hands with a different number of people. However, if one mathematician shakes hands with nobody else, another mathematician
can’t shake hands with 99. Therefore, there are actually 99 possibilities (pigeon holes) and 100 mathematicians (pigeons). So, we can be certain that at least two mathematicians shook hands with the same number of people.

3) 10. We have three colour balls – imagine we have three pigeon holes, one for each colour. As we pull out the balls we put them in the appropriate pigeon hole. The most we could withdraw without getting 4 in a pigeon hole is 3 in each pigeon hole. Then the next ball drawn will have to go into one of the pigeon holes, so that is 3×3+1 = 10.

4) 17. Here we have 4 pigeon holes (one for each suit) and we’re looking to be certain of
getting 5 in one. The most we could have without 5 in one hole is 4 in each pigeon hole. That is 4 x 4 = 16 and the next card assigned will guarantee 5 cards in one pigeon hole.

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