Keith Austin, Author at New �Ӱ�ԭ��

Enigma 877 : Fantastic football

Keith Austin — Fri, 31 May 1996 23:00:00 +0000

OUR league consists of four teams: Albion, City, United and Victoria. Each
plays each of the other teams once in a season and the six matches are played on
six consecutive Saturdays. Penalty shoot-outs are used to ensure there are no
draws. Before the first match and after each match a league table is drawn up
with the teams ordered by the number of points they have and teams with the same
points in alphabetical order.

For each match the winner receives the number of points equal to their
position in the league immediately before the match. If the winner was equal on
points with one or more other teams then it receives the average of the
positions of these teams, including itself. For example, suppose City wins the
first and third matches and Victoria the second. Before the first match the
table is:

1-A-0, 2-C-0, 3-U-0, 4-V-0 (position-team-points), so City receives (1 + 2 +
3 + 4)/4 = 2.5 points. Before the second match the table is 1-C-2.5, 2-A-0,
3-U-0, 4-V-0, so Victoria receives (2 + 3 + 4)/3 = 3 points. Before the third
match the table is 1-V-3, 2-C-2.5, 3-A-0, 4-U-0, so City receives 2 points.

What is the largest total number of points a team can receive in a
season?
What is the largest possible total number of points for all four teams at
the end of the season?
What is the smallest possible total number of points for all four teams at
the end of the season?

A £10 book token will be awarded to the sender of the first correct
answer opened on Thursday, 4 July. The Editor’s decision is final. Please send
entries to Enigma 877, New �Ӱ�ԭ��, King’s Reach Tower, Stamford Street,
London SE99 0BB. The winner of Enigma 871 was William Swindell of Kenley,
Surrey.

Answer to Enigma 871

Pandigitals

17820.

Enigma 872 : All move round

Keith Austin — Fri, 26 Apr 1996 23:00:00 +0000

THE LAYOUT. Take 99 plates and lay them on the ground in a large circle. Take
99 coins and put one on each plate; they must not be all showing heads and they
must not be all showing tails.

THE MOVE. Take each coin that is heads side up and move it, in a clockwise
direction, to the next plate; put it down with the same side still facing up. Do
the same with the coins facing tails up, except that you move each one
anticlockwise.

Repeat the Move, starting with the position reached after the first Move.
Continue in this way, making Move after Move.

Q1. Is it possible that immediately after the first Move

there are no empty plates?

Q2. Is it possible that immediately after the second Move

there are no empty plates?

Q3. Is it possible that immediately after the third Move there are no empty
plates?

Q4. Is it possible, for some starting Layout of heads and tails, that there
is always at least one empty plate however many Moves you make, 1 or 2 or 3
�ǰ�…?

A £10 book token will be awarded to the sender of the first correct
answer opened on Thursday 30th May. The Editor’s decision is final. Please send
entries to Enigma 873 New �Ӱ�ԭ��, King’s Reach Tower, Stamford Street, London
SE99 0BB. The winner of Enigma 867 was Steve Hughes of Saffron Walden

Answer to Enigma 866

Own goals galore

Attack beat Brave 1-0 and 2-0

ENIGMA 864

Keith Austin — Sat, 02 Mar 1996 00:00:00 +0000

I HAVE just bought a new electronic clock and I have to put in the fraction of a second which elapses between updates of the clock’s display. For example if I put in 8/13 then the clock’s display will update at the following times after I start the clock:

Equation

The clock’s display actually gives the time to the nearest second and so in this case it shows, in turn:

1,1,2,2,3,4,4,5,5, …

I can put in any fraction I choose, provided that the denominator is odd.

For each of the following sequences, say whether or not I can put in a fraction so that the clock’s sequence of displays begins with the given sequence:

(a) 1,2,3,4,5,5,6,6 .

(b) 1,2,3,4,4,5,6,7,8,8 .

(d) 1,2,3,4,4,5,6,7,8,9,10,11,12,13,14.

A £10 book token will be awarded to the sender of the first correct answer opened on Thursday 4th April. The Editor’s decision is final. Please send entries to Enigma 864, New �Ӱ�ԭ��, King’s Reach Tower, Stamford Street, London SE99 0BB. The winner of Enigma 858 was John Hobley of Como, Western Australia.

Answer to Enigma 858

A matter of degree

156° and 62°F .

Weight for it!

Keith Austin — Sat, 03 Feb 1996 00:00:00 +0000

I USED to run a village shop and I had a pair of scales and a set of 10
weights. Each weight was a whole number of ounces and a total weight was less
than 1024 ounces. I found that I could put one or more weights on each of the
two pans of my scales so that they balanced.

One day I lost a weight. By chance I was able to replace it with a spare;
unfortunately it was not the same weight as the one I lost. However, now I
could put one or more weights on each of the two pans of my scales so that
they balanced; in fact, I could arrange for each side to weigh 6 ounces.
Unfortunately I could not weigh 13 ounces, although I could weigh 3
ounces.

Q1. What was my set of weights after the replacement?

Q2. Suppose I was back with my original set of weights. Could I have
selected one and replaced it by some other whole-number-of-ounces weight, of
my own choosing, so that the total weight was still less than 1024 ounces and
I still could not put one or more weights on each of the two pans of my scales
so that they balanced?

Q3. As Q1 but now I can replace six weights instead of just one. The new
set of weights must not be the same as my original set.

A £10 book token will be awarded to the sender of the first correct
answer opened on Thursday 7 March 1996. the Editor’s decision is final. Please
send entries to Enigma 860, New �Ӱ�ԭ��, King’s Reach Tower, Stamford
Street, London SE99 0BB. The winner of Enigma 853 was J. A. Grant of Guiseley
in West Yorks.

Answer to Enigma 853

Franglais

4302568

First you see it, then you don’t

Keith Austin — Sat, 20 Jan 1996 00:00:00 +0000

THE way mathematics is taught in Britain is worrying a lot of people right now. It was the subject of a recent report (Tackling the Mathematics Problem) by three learned bodies, the London Mathematical Society, the Institute of Mathematics and its Applications, and the Royal Statistical Society. The report particularly expressed unease about “mathematical proof”, which was also the concern of teachers from schools and institutes of higher education when they met in London in December. So what is so special about proof? Why does its teaching appear to leave so many dissatisfied?

A good place to start is with what is probably the most significant word in mathematics: “all”. More or less any statement a mathematician wants to prove will contain the word “all”, either explicitly or implicitly. For example, in the early primary years we might want to prove the statement: “If we take any even number between 1 and 11 and add 2 to it then the result is an even number”. I shall call this Statement 1, or S₁ for short.

Now at this stage in the journey, proofs are straightforward. We make a list of all the numbers referred to in S₁: 2, 4, 6, 8, 10, and check each number in the list in turn to see that adding 2 to it gives an even number.

It is when we move to the next stage of our work that the problems set in. Suppose we have to prove the statement: “If we take any even number between 1 and 10 000 001 and add 2 to it then the result is an even number” (S₂). But clearly it is not practical to write out the list of all the numbers referred to in S₂

It is at this point that one of the key ideas of mathematics is brought into play – that the same argument can be used to check out each of the different numbers on the list. For example, we could give the following argument: “We take an even number between 1 and 10 000 001. Let us call it a. Because it is even it is 2 × some other number, b, say. Now, if we add 2 to a then we get 2 × b + 2. But this equals 2 × (b + 1), which is 2 × a number and so is even.”

At this point we have good news and bad news. The good news is that we have covered 5 million different numbers by an argument that is between five and six lines long. The bad news is that anything that is that powerful is liable to overwhelm us or at least have the appearance of the magical. The move from simply checking a list to using this mystical power can destroy our faith in the rational basis of mathematics.

But worse is to come. The proof did not refer to the list that we are checking and so in future the list will be forgotten and the proof will simply become a ritual unhooked from reality. Where the ritual leads and the rules that determine how it proceeds become the dominant features. All thoughts of a proof being an argument about something real have long since vanished.

Now I have to admit that a number of pupils do see mathematical proof as a rational argument about reality. However, for the great majority the above description is probably closer to their experience. This picture of mathematical proof as magical and uncoupled from reality is the darkest and most frightening view of mathematics. So the teacher of mathematical proof is involved in a battle, not just for the minds of the schoolchildren and students but also for their souls, a battle between the rational and the irrational.

Somewhere between S₁ and S₂ mathematical proof has stopped being an argument about reality and has become a mystery. So perhaps we should discard proofs like that of S₂. But those who use mathematics creatively, such as the scientists and engineers who use it to develop theories, need mathematical proof to give security to their conclusions. Is there any reason why others should bother with mathematical proof?

In fact, if we tackle this question and the question of making proof rational together then we come up with an answer. In both cases, the essential point is to keep proof grounded in reality. Provided the teacher ensures this, the pupils can see proof as being a rational argument about reality. What is more, the pupil learns to appreciate that the techniques and skills of proof can be useful in their daily lives, since they are powerful arguments about the real world.

So how can we keep proof grounded in reality? Statement S₁ provides a suitable starting point but the development from there to S₂ needs to be thought through very carefully. It needs to proceed by a series of steps, none of which must break the tie between proof and reality, and yet which must involve an increasing level of sophistication in the arguments.

There have been many calls for action on education in mathematics, including the one from the report of the learned bodies, Tackling the Mathematics Problem. Perhaps one appropriate response to these calls would be an investigation into the guiding of learners from S₁ to S₂, from simple proofs to sophisticated proofs, in such a way that learners and proofs remain firmly grounded in reality.

ENIGMA 854: Colourful Christmas

Keith Austin — Sat, 23 Dec 1995 00:00:00 +0000

AS Christmas approaches and the winter weather takes hold, so two nuns bring food and clothing to the refugee camp that has been established near to their hospital. There are 27 huts in the refugee camp, ranged in a circle and numbered 1, 2, 3, …, 26, 27 as we go round the circle, with 27 also being next to 1. In order to brighten the camp up for Christmas, each hut has a coloured flag in front of it, red or blue or yellow. The rule for the camp is that if two huts are next to each other in the circle then their flags must be of different colours. Each morning the flags are put out as follows:

RBYRBYRBYRBYRBYRBYRBYRBYRBY.

That is to say, 1=R, 2=B, 3=Y, …, 27=Y.

The children in the camp have a supply of flags of all three colours. They choose a hut and go and change its flag, ensuring that they keep the camp rule. They then choose another hut and go and change its flag, still keeping the camp rule. They carry on in this way, choosing one hut after another.

It was late on Christmas Eve when the nuns completed their task and set off for the hospital. When they looked back at the refugee camp they saw it was bathed in moonlight shining through a gap in the clouds, while the surrounding area was in darkness. The younger nun remarked on the contrast and the older agreed and continued, “but it does help us see where our priorities lie”.

On various days the children try to achieve certain patterns of flags. Which of the following are they able to produce?

(i) RBYRBYRBYRBYRBYRBYRBYRBYRBY.

(ii) RYRBYRBYRBYRBYRBYRBYRBYRBY.

(iii) RBYRBYRBYRBYRBYBRYBRYRBYRBY.

ENIGMA 850: Swapping the letters

Keith Austin — Sat, 25 Nov 1995 00:00:00 +0000

I have 26 boxes numbered 1 to 26 and 26 cards labelled A to Z. I start with A in box 1, B in box 2 … Z in box 26 and I want to end with A in box 2, B in box 3 … Y in box 26, Z in box 1.

I am allowed to make a series of swaps, where a swap consists of choosing any two boxes and swapping over the two cards in them. So, can I achieve my aim with a series of: 1) 27 swaps; 2) 26 swaps; 3) 25 swaps; 4) 23 swaps? (Four answers.)

A £10 book token will be awarded to the sender of the the first correct answer opened on Thursday 28 December. The Editor’s decision is final. Please send entries to Enigma 850, New �Ӱ�ԭ��, King’s Reach Tower, Stamford Street, London SE99 0BB. The winner of Enigma 844 was Åke Nord of Solna, Sweden.

Answer to Enigma 844

Add what?

ADDITIONS = 277616985

Enigma 846: Winners on the left

Keith Austin — Sat, 28 Oct 1995 00:00:00 +0000

THE 100 towns, Oneton, Twoton … Hundredton, have just finished a tennis tournament in which every town played every other town once and there were no draws. In my report on the tournament I listed the towns so,

Oneton, Twoton, … Hundredton.

Then I decided I would like my list to reflect the results as follows: I would call a pair of towns X and Y a bad pair if they occur next to each other in my list, with X on the left, and Y beat X. I would then try to make a list that had no bad pairs.

Question 1: Suppose I find a bad pair in my list and I swap the two towns in the pair over. Do I necessarily reduce the number of bad pairs in my list?

I now carry out the following process:

(i) I start with my initial list.

(ii) I look for a bad pair in my list and if I am able to find one I swap the two towns over.

(iii) I take the new list I have made and do (ii).

(iv) I repeat (iii) over and over, each time doing (ii) on the list I created by the previous doing of (ii).

Question 2: Is it certain that if I carry on the process long enough I will reach a list which contains no bad pairs?

A £10 book token will be awarded to the sender of the firt correct answer opened on Thursday 30 November. The editor’s decision is final. Please send entries to Enigma 846, New �Ӱ�ԭ��, King’s Reach Tower, Stamford Street, London SE99 0BB. The winner of Enigma 840 was Peter Mabey of Harlow, Essex.

Answer to Enigma 840

Four graduates

7558 miles

ENIGMA: Cover photo

Keith Austin — Fri, 29 Sep 1995 23:00:00 +0000

LET me explain what you can see in the picture. I drew the large rectangle ABCD and then I took a photograph of it. I placed the photograph, that is to say the small rectangle ABCD, on the large rectangle as you see in the picture. Note that the sides of the photograph are parallel to the sides of the large rectangle.

What I want to know is whether there are any points of the large rectangle which are exactly covered by their photographic image. For example, the point A is not covered by its photographic image A.

Are there any points of the large rectangle which are exactly covered by their photographic image?
If there, then make a copy of the figure and carefully mark on it all such points.
If I placed the photograph elsewhere within the large rectangle, not necessarily with the sides parallel, which of the following situations are possible:

• No point is exactly covered by its photographic image;
• One point is exactly covered by its photographic image;
• More than one point is exactly covered by its photographic image.

A £10 book token will be awarded to the sender of the first correct answer opened on Thursday 2 November. The Editor’s decision is final. Please send entries to Enigma 842, New �Ӱ�ԭ��, King’s Reach Tower, Stamford Street, London SE99 0BB. The winner of Enigma 836 was Andrew Plant of Lymington, Hampshire.

Answer to Enigma 836

Who buys the drinks?

His initial was I (Ian?)

Enigma: No. 837

Keith Austin — Fri, 25 Aug 1995 23:00:00 +0000

The quick brown …

We arrange the 26 letters of the alphabet in a row as follows:

THEQUICKBROWNFXJMPDVLAZYGS (*)

Now take any letter, say P, and find the longest chains in (*) in alphabetical order ending with P. We find some of length 5, for example EIKMP, but none any longer. Next we find the longest chains in (*) in reverse alphabetical order starting with P. We find some of length 3, for example PLG, but none any longer. We say P has alphabetical length, α=5, and reverse alphabetical length, ω=3.

Question 1: I have chosen a letter which comes before P in the alphabet and to the left of P in (*). Can you say for certain my letter has α less than 5?

Question 2: I have arranged the 26 letters in a new row (**). Can you say for certain that if you choose a letter in (**) and I choose a different letter in (**) then they will have different α’s or different ω’s?

Now I want you to write on a piece of paper, a list of 25 possibilities for α and ω so [α=1 ω=1], [α=1 ω=2], …, [α=1 ω=5], [α=2 ω=1], …, [α=2 ω=5], [α=3 ω=1], …, [α=5 ω=5].

Next I want you to take each letter in (*) and work out α and ω for it and mark it on the list, for example you will write P against [α=5 ω=3]. Unfortunately, some letters wil have a combination of α and ω that is not on the list, for instance X has α=6 and ω=4.

Question 3: Can you arrange the 26 letters of the alphabet in a row so that every letter has a combination of α and ω that is in the list? If your answer is “yes” then give such a row.

Question 4: Can you be certain that you will be able to find in my row (**) a chain of 6 letters that are in either alphabetical or reverse alphabetical order?

A £10 book token will be awarded to the sender of the first correct answer opened on Thursday 28 September. The Editor’s decision is final. Please send entries to Enigma 837, New �Ӱ�ԭ��, King’s Reach Tower, Stamford Street, London SE99 0BB. The winner of Enigma 831 was Andrew Plant of Lymington, Hampshire.

Answer to Enigma 831

Pot Black

(1) 14 and 13; (2) Whyte.