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Programming Enigma Puzzles

23 May 2016

Posted by on **From New Scientist #2322/#2323, 22nd/29th December 2001**

Each year there are eight presents in the sack, their values whole numbers of pounds from 1 to 8, to be distributed to the children Amber, Ben, Christine, and Dick. The presents are drawn from the sack at random, one at a time. Each is given to the child who has already received the lowest total value; if more than one qualify, then to whichever of them whose name comes first in alphabetic order.

Last year, Amber received a final total value of £8, Ben £5, Christine £11, Dick £12.

Q1:What were the values of the first and the seventh presents drawn?I recorded the values in pounds for the previous five years:

1995: A 6, B 13, C 10, D 7

1996: A 10, B 6, C 14, D 6

1997: A 14, B 10, C 5, D 7

1998: A 9, B 6, C 14, D 6

1999: A 13, B 6, C 5, D 12Unfortunately I must have made some mistakes because for some of those years the values could not have been achieved.

Q2:Which years were correctly written down?

Thanks to Hugh Casement for providing the transcript of this puzzle.

[enigma1166]

20 May 2016

Posted by on **From New Scientist #1494, 6th February 1986** [link]

Six girls were sitting equally-spaced around a circular table. Each of them was either honest or consistently told lies.

Anne said: “The person on my left is more honest than me”. None of the other girls would say the same.

Barbara said: “The person on my right is not less honest than me”. None of the other girls was able to say the same.

Christine said that the girl opposite her was of different honesty from her, and Doreen said that the majority of the girls were honest. Eileen said that the people either side of her were honest, and Fiona said that she was between two people of equal honesty.

Fill in the seating plan ticking those who are honest.

[enigma345]

16 May 2016

Posted by on **From New Scientist #2322/#2323, 22nd/29th December 2001**

With the help of my niece and nephew I have designed my own Christmas Card this year. We started with a large rectangular piece of card and then near to the top left-hand corner we drew a red dot and a gold dot representing two baubles. Then I wanted to draw a line through at least one of the baubles to create a triangle in that corner of the card.

Linus suggested we draw the line through the red bauble in the direction which made the line as short as possible. Tynia said to draw the line through the gold bauble in the direction which made the area of the triangle formed as small as possible. As a compromise we decided to draw the line through both baubles.

Luckily this line satisfied both Linus’s and Tynia’s requirement. It was 20 cm long and the triangle formed had an area of 80 cm².

How far apart were the two baubles?

[enigma1167]

13 May 2016

Posted by on **From New Scientist #1493, 30th January 1986** [link]

England, Scotland, Wales, Ireland and France took part in the Rugby union five-nations championship, each country playing each other country once. Two match points were awarded for a win and one match point for a draw, and the total number of match points gained by each country was: England 7, Scotland 6, Wales 4, Ireland 2, France 1.

No two matches had the same score and no country scored more than six points in any match; but each country scored exactly the same number of points in the championship as each other country, and Wales also had this number of points scored against them.

Those not familiar with the point scoring system of rugby union merely need to know that it is not possible for a side to score one, two or five points in a match.

What were the results and scores of Wales’s matches in the championship?

[enigma344]

10 May 2016

Posted by on **From New Scientist #2324, 5th January 2002**

I have found a four-digit number such that it is impossible to factorise the numbers formed by its first digit or last digit or first two digits or middle two digits or last two digits or first three digits or last three digits or all four digits. In other words all those eight numbers are prime except that either or both of the single-digit numbers may be unity.

Harry and Tom have also each found such a four-digit number. The four-digit numbers that we have found are all different; but Harry’s number uses the same digits as Tom’s number, though in a different order.

Which four digit number have I found?

This completes the archive of puzzles from 2002.

[enigma1168]

6 May 2016

Posted by on **From New Scientist #1492, 23rd January 1986** [link]

Silent Mews, the cul-de-sac where I live, is dull and uniform. Every house on this side of the road has one exactly corresponding to it on the other side, and vice versa. The only exciting thing about it is that the houses are numbered

boustrophedon. That is, starting at number 1 and visiting each house in numerical order takes you up one side of the street, along the end of the mews and back along the other side.To lend my house individuality, I have bought brass numbers for my front door (all the other houses have plastic). Each digit costs the same, so it would cost 222 times what I paid for my door number to replace everyone else’s door numbers with brass. Not only does my property have a special number, but my number has a special property. Rearranging its digits produces exactly four numbers. One of these is my number and one is the number of another house in the mews.

If more houses were to be added on after the last house on the mews, it it would be possible for the set of digits in my number to be the only set which appeared in a total of exactly four house numbers, two of which were in the old part of the mews, and two which appeared in the newly extended part.

You may wonder whether, besides the houses on the two long sides of the mews, there are also houses at the blind end of the mews, but the answer to this should be obvious, as is the answer to this puzzle, namely: what is the number of the house directly opposite mine?

I am going to be away next week, but I will try and keep to the posting schedule if possible.

[enigma343]

2 May 2016

Posted by on **From New Scientist #2325, 12th January 2002**

George lives not far from the Crucible Theatre, where the annual World Snooker Championship has been held for the past 25 years. Each year 32 players compete in a straightforward knockout tournament to determine who is the champion.

George, and his brother Fred, have entered a less prestigious tournament at their local pub. Since they cannot be certain that the number of entrants will be a power of two, lots will be drawn to see who is awarded byes in the first round. The tournament will then continue as a knockout, also drawn at random.

While waiting for the draw to be made, George calculated the chance that he will at some stage have to play against his brother, assuming all the players are of equal ability. It is exactly one in ten.

How many players entered the tournament?

[enigma1169]

29 April 2016

Posted by on **From New Scientist #1491, 16th January 1986** [link]

A full set of dominoes has been arranged in a 7×8 block. Please mark in the boundaries between the dominoes. There is only one answer.

[enigma342]

25 April 2016

Posted by on **From New Scientist #2326, 19th January 2002**

I have a rectangular piece of paper which I have folded twice and then unfolded again. The first fold was as shown:

In that figure the area of each of the three triangles is a perfect square number of cm², and each area is less than 100 [cm²].

The second fold was through X and parallel to the first. When unfolded the two creases divided the rectangle into three regions, namely two triangles and a pentagon. The area of each was again a perfect square number of cm².

What is the area of the rectangle?

[enigma1170]

22 April 2016

Posted by on **From New Scientist #1490, 9th January 1986** [link]

Four football teams, A, B, C and D are to play each other once. After some — or perhaps all — of the matches have been played, a table giving some details of matches played, won, lost and so on, looks like this:

(2 points are given for a win and 1 point to each side in a draw).

Find the score in each match.

[enigma341]

18 April 2016

Posted by on **From New Scientist #2327, 26th January 2002**

I drew an irregular hexagon. I also drew three straight lines, each of which ran from one angle of the hexagon to the next angle but two. Since the lines all intersected at the same point the hexagon was divided into six triangles. At the point of intersection each triangle had an angle of 60°.

All sides of all the triangles were an integral number (less than 50) of centimetres in length, and the six sides that met at the point of intersection (each one common to two triangles) all had different lengths.

What in centimetres was the perimeter of the hexagon?

[enigma1171]

15 April 2016

Posted by on **From New Scientist #1489, 2nd January 1986** [link]

Your task this week is to find the six three-figure numbers making up this cross-figure:

Across:1. When added to 4 across this give an odd total.

4. My middle digit equals the number of “1”s used altogether in the 3×3 answer.

5. A perfect cube.

Down:2. A factor of the difference between 3 down and 1 down.

3. A perfect square.Although you have not been given the clue for 1 down, I can tell you that the 3×3 answer is an

magic square. In other words, the sums of the digits in each row of three, each column of three and each diagonal of three are all different.unFind the 3×3 answer.

[enigma340]

11 April 2016

Posted by on **From New Scientist #2328, 2nd February 2002**

George’s local council celebrated National Plant a Tree Year by planting four saplings in a public park. Unfortunately, a few nights later, vandals dug up one of the trees and cleared the ground leaving no evidence of where the tree had been. George has been asked to help.

The Mayor explained that the straight-line between the six pairs of trees had been carefully measured so that there were only two

differentdistances. With three trees remaining where they had been planted, he asked it George could work out the original position of the fourth.“No,” George replied, “I have identified four possible positions and I am still thinking”.

How many possible positions were there?

[enigma1172]

8 April 2016

Posted by on **From New Scientist #1487/#1488, 19th/26th December 1985** [link]

In HAPPY CHRISTMAS if you replace each letter consistently by a digit you get a five-figure number for HAPPY but, more interestingly, CHRISTMAS has the property that:

C ends in 1 or is divisible by 1(!);

CH ends in 2 or is divisible by 2;

CHR ends in 3 or is divisible by 3;

CHRI ends in 4 or is divisible by 4;

CHRIS ends in 5 or is divisible by 5;

CHRIST ends in 6 or is divisible by 6;

CHRISTM ends in 7 or is divisible by 7;

CHRISTMA ends in 8 or is divisible by 8;

CHRISTMAS ends in 9 or is divisible by 9.Find the numerical value of CHRISTMAS.

This completes the archive of *Enigma* puzzles from 1985. All *Enigmas* from #1 (Feb 1979) to #339 (Dec 1985) [7 years] along with those from #1173 (Feb 2002) to #1780 (Dec 2013) [12 years] are now available on the site. So, almost 19 years worth of puzzles are now available, and there’s only 16 years of puzzles left to put on the site!

[enigma339b] [enigma339]

4 April 2016

Posted by on **From New Scientist #2329, 9th February 2002** [link]

The months MAY, JUN, JUL and AUG are indicated in the two multiplications shown where all the digits have been replaced by capital letters and asterisks. In these multiplications, and in the clues that follow, different capital letters stand for different digits but the same capital letter always stands for the same digit while the asterisks can be any digit.

Typically, the months JAN and MAR are both exactly divisible by 31, and of course you can also assume that any YEAR is exactly divisible by 12 or even 52.

What is the product of each multiplication?

[enigma1173]

1 April 2016

Posted by on **From New Scientist #1487/#1488, 19th/26th December 1985** [link]

During last Christmas’s intellectual activity — backgammon, ludo, and so on — Pam complained about the unfairness of using a pair of ordinary dice.

“With

pair,” she said, “you can only throw a total from 2 up to 12. But I should like to be able to throw any whole number from 2 up tothismore than that. And withmuchpair I am much more likely to throw some totals that others — I get 7, for instance, six times as often as I get 12. What I want is a pair of dice which will throw every possible total with equal probability. And finally, there’s a 4 on each of these dice, and I object to square numbers. I don’t mind 1 — I don’t think of 1 as really square — but I don’t like 4 and I would equally object to 9 or 16 or 25 or 36.”thisSo, I have designed a special pair of dice for Pam’s Christmas present this year, which, I am glad to say, meets her wishes entirely. They are six-sided dice of the ordinary shape, with a positive whole number on each face, and they are equally likely to throw any total from 2 to 37 inclusive.

What are the numbers on the faces of each die, please?

There are now 950 *Enigma* puzzles available on the site.

[enigma339a] [enigma339]

28 March 2016

Posted by on **From New Scientist #2330, 16th February 2002** [link]

An example of a letter sum is:

ABC + DCEF = AEFB.

That is to say word + word = word, where each word is a string of letters of any length. An example of an

for the above sum is:answer537 + 4726 = 5263.

We are concerned with letter sums that have exactly one answer. For each such letter sum we produce its A-number by taking its answer and discarding the + and =. For example if the above letter sum had only the answer given above then its A-number would be:

53747265263.

What are the four smallest A-numbers we can get by this process?

[enigma1174]

25 March 2016

Posted by on **From New Scientist #1486, 12th December 1985** [link]

“I’m not broke, but you’ve got more money than I have,” Bubbles complained.

“True,” said Hippocrene, blushing. “But less than twice what you’ve got. Now listen. If you square the money I’ve got, and add that to what you’ve got, you get the square of what Johnny’s got. See?”

“No,” said Bubbles, toying nervously with her beads. “I don’t see how you can square money.”

“That’s easy. Just express it as pounds. I mean the square of £1.20 is 1.44. The square of 15½p is 0.024025 — that is, 0.155². And so on. OK?”

“Yes,” said Bubbles, with a cheerful wink. “Go on.”

“And if you square yours and add that to mine, you get the square of Johnny’s too. Isn’t that amazing?”

“Moderately so.”

When this conversation took place, the halfpenny was still in use.

How much had Hippocrene and Bubbles respectively?

[enigma338]

21 March 2016

Posted by on **From New Scientist #2331, 23rd February 2002** [link]

In the following statement digits have been consistently replaced by capital letters, different letters being used for different digits:

UN,NEUFandCENTare all perfect squares.Find the numerical value of the square root of (

UN×NEUF×CENT).

For some reason this puzzle is labelled as **Enigma 1112** in the magazine.

This puzzle is somewhat similar to **Enigma 1291** (also set by Richard England).

[enigma1175]

18 March 2016

Posted by on **From New Scientist #1485, 5th December 1985** [link]

I was visiting my Uncle Ever-Clever, the inventor. “What are all these little cubes and boxes?” I asked. “Ah,” he said, “that’s a mathematical game I’ve been working on involving cubelets, each of whose faces is either black or white. I realised that I’d have my work cut out making each cubelet individually, so I hit upon the idea of taking a large cube of black wood and an equal cube of white wood, then painting the black one white and the white one black.”

“How would that help?” I asked woodenly.

“Well, when you saw them up you obtain cubelets having various combinations of white and black faces. Every distinguishable combination of black faces and white faces manufacturable by these means occurs exactly once in a complete set of my Chock-a-Block cubes.

“And do all-white and all-black each count as combinations?” I asked.

“Of course, you blockhead!” was the affectionate replay as he closed the Brewster window (he suffered from sunspots).

“Well, I transformed the two painted cubes without wastage of wood into equal cubelets in such a way that, when they were sorted into complete sets, the amount of wood left over was the minimum possible. These boxes each contain a full set of my cubelets; those over there between the Wimshurst bicycle and the Luminous Moondial are the ones left over… But you must be famished, dear boy; let me ring for a pot of Logwood Tea and some Dwarfstar Cake.”

How many boxes were there, how many cubelets did they each contain and how many cubelets were left over?

[enigma337]

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