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Programming Enigma Puzzles

25 January 2021

Posted by on **From New Scientist #1759, 9th March 1991** [link]

The Sunnycourt Tennis Club has 22 members, and each year it organises a Saturday afternoon tournament consisting of a number of singles matches. The matches are selected so that everyone in the club has at least one game, but the total number of games is always less than 22.

Last year, after the list of games had been settled, the secretary looked to see if there 8 games which involved 16 different players. As there were 8 courts available she had hoped to take a photograph with them all in use. Unfortunately, the list did not include 8 such games.

During this last winter, the club decided that this year’s tournament should involve one game fewer than that year. The secretary then did some calculations and found that however the games were selected, she was certain to be able to find 8 games in the list which involved 16 different players.

How many games were there in the tournament

year?last

[enigma605]

22 January 2021

Posted by on **From New Scientist #1758, 2nd March 1991** [link]

Below is an addition sum with letters substituted for digits. The same letter stands for the same digit wherever it appears, and different letters stand for different digits.

Write out the sum with numbers substituted for letters.

[enigma604]

18 January 2021

Posted by on **From New Scientist #1757, 23rd February 1991** [link]

Start with ** and write down its square. Then write down the next square and the next until the final one you write down is the square of **, each of the squares having three digits. Now count up all the occurrences of the digits in that list of squares; here’s an inventory:

* noughts;

* ones;

* twos;

* threes;

** fours;

* fives;

* sixes;

* sevens;

* eights;

* nines.Unfortunately the word processor seems unable to print digits, and replaces them with asterisks. But I can tell you that if the correct digits were here instead of the asterisks and you then took a grand count of the occurrences of the actual numerical digits in this Enigma (from “Start” to “list?” at the end) then, quite apart from the fours and fives and so on, there would be * noughts, * ones, * twos and * threes (where these last four digits are all different).

What was the two-figure number whose square started your list, and what was the two-figure number whose square ended your list?

[enigma603]

15 January 2021

Posted by on **From New Scientist #1756, 16th February 1991** [link]

As usual, letters have replaced digits in this addition sum, with different letters consistently being used for different digits:

We can also tell you that the damsel is in her prime, that is,

SUEis a prime.Find the value of the

SUM.

[enigma602]

11 January 2021

Posted by on **From New Scientist #1755, 9th February 1991** [link]

There are a number of ways of moving about a grid. The diagram below shows one way of moving from the origin O to the point A.

The coordinates of O are (0, 0) and those of A are (4, 5). Like most normal grids, movement is permitted only along grid lines — diagonal moves are banned. But unlike normal grids, this one has its own particular rule governing movement — it is permitted only in positive directions, that is, to the right or upwards.

Find:

(a)the number of routes from O to A in the diagram above; and

(b)the number of routes, on an expanded grid, from O to a point B whose coordinates are (20, 6).

[enigma601]

8 January 2021

Posted by on **From New Scientist #1754, 2nd February 1991** [link]

The 10 towns in Colouritania are joined by various roads. The map shows the 10 towns as small circles; I had drawn those roads I can remember. There are some other roads, each joining a pair of towns. There are no road junctions outside the towns, but one road can cross over another by means of a bridge.

Each day, for the past 1022 days, at 0600 hours, the 10 towns have each been allocated a colour, blue or yellow, with at least one town of each colour. The pattern of blue and yellow among the 10 towns has been different on each day.

On each of these 1022 days, at 0605 hours, the following descriptions have been applied: a

roadis said to begreenif it joins a blue town and a yellow town; atownis said to begreenif it has more green roads leaving it than non-green roads, or the same number of each.Which of the following statements are true and which are false?

(a) We can say for certain there was a day when all the roads were green.

(b) We can say for certain there was not a day when all roads were green.

(c) We can say for certain there was a day when no road was green.

(d) We can say for certain there was a day when all the towns were green.

[enigma600]

4 January 2021

Posted by on **From New Scientist #1753, 26th January 1991** [link]

In a competition between three local football teams, the sides have played each other once. The points that they get for their matches are according to the new and, it is hoped, better system which the Lancashire football association has devised in the belief that it will lead to more goals. In fact, each side scored at least one goal in each match.

In this new system, 10 points are awarded for a win, 5 points for a draw, and 1 point for each goal scored.

I was originally told by a normally reliable source — the association’s secretary — that Anchester United had scored 11 points, that Boldon Wanderers had scored 14 points, and that Clackburn had scored 16 points. Unfortunately, however, the secretary later confided in me that due to his excessive consumption of local ale, each of the figure he had produced was incorrect.

Worse, all the secretary could remember was that these totals were in fact either 1 more or 1 less than the correct figure.

What was the score in each match?

[enigma599]

1 January 2021

Posted by on **From New Scientist #1752, 19th January 1991** [link]

In Susan Denham’s Enigma last week she related details of an unusual frame of snooker (and she reminded readers of the basic rules). Reading that teaser reminded me of a frame which I once saw between the two great players Dean and Verger.

Dean had a break; Verger followed this with a break of 1 point fewer; Dean followed this with a break of 1 point fewer; Verger followed this with a break of 1 point fewer; and so on, up to and including the final clearance.

In any one break no colour (apart from the red) was potted more than once. Dean never potted the blue. Each player potted the black twice as many times as his opponent potted the yellow. The brown was potted more times than the green.

How many times (in total) did the winner pot each of the colours? (Red, Yellow, Green, Brown, Blue, Pink, Black).

[enigma598]

28 December 2020

Posted by on **From New Scientist #1751, 12th January 1991** [link]

In snooker there are 15 red balls worth one point each. If a player pots a red it stays in the hole and he (or she) is allowed to try to pot one of the colours yellow, green, brown, blue, pink or black (worth 2-7 points in that order). If a colour is potted it is brought out again and the player can try for another red, and so on. This continues until all the reds have gone. Then the remaining six colours are potted in ascending order.

The total points achieved in one such run is called a “break”. For example:

red + pink + red + black + red

would be a break of 16.

Having completed the break, the player sits down and lets the other player try for a red and continue the break, and so on. At the end the winning player is the one with the higher grand total of points. A player may choose not to pot the final black if the result is already determined without it. No other rules concern today’s story.

Stephens and Hendry play a frame of snooker. Stephens starts with a break of 3, Hendry follows with a break of 4, Stephens with a break of 5, and so on, and this pattern continues to the end. As usual in quality snooker, the black was potted more times that the yellow, and the pink was potted more times that the blue.

How many times did Stephens pot the brown? And how many times did Hendry pot the brown?

[enigma597]

25 December 2020

Posted by on **From New Scientist #1750, 5th January 1991** [link]

While I was pulling a celebratory cracker, a fascinating toy fell out. It was a puzzle which first required the construction of four cubes from cruciform “flat-outs”, all consisting of six squares each of which portrayed a knave from one of the playing card suits, denoted by S, H, D and C in the first figure.

Then the challenge was to place the cubes in a vertical column (with flat faces) such that all suits appeared on all four vertical sides.

In the end I managed it, and the second figure shows my column (from top-to-bottom being in the order shown above) with a couple of the suits filled.

Fill in the other seven visible faces in the diagram above.

*Happy Christmas* from **Enigmatic Code**!

[enigma596]

21 December 2020

Posted by on **From New Scientist #1748, 22nd December 1990** [link]

Christmas is celebrated on Argos-3 just as in the rest of the known Universe, except that on the remote world (“Argos-3, Twinned with Earth”, says the brochure) it falls on 25 Dodecembrius.

Their houses are uniform in style, shape and size and are all to be found on one side only of the street. The house number is thus also the ordinal number of the house counted from the beginning of the street. The first house is 1, the second 2 and so on up to the last house in the street, whose number in some cases can be as high as 10¹².

With such long street, it was hardly amazing that Santa felt a little put out to discover that the last parcel of his delivery in the small hours of the 25th was cryptically addressed:

Many house number in Christmas Street have the property that if you remove their first digit (that is, the left-hand end digit) you are left with a 25th of the number you started with. Well, deliver this parcel to the 25th such house in the street.

With a sigh, Santa started from house Number 1 and zoomed along the street in his quantum sled until he reached the house in question. To his amusement it was his own house. He went in and placed the parcel under his Christmas Cactus.

At what number Christmas Street does Santa live?

This completes the archive of *Enigma* puzzles from 1990. There is now a complete archive of puzzles from November 1974 to December 1990, and also from September 1997 to December 2013, comprising of 1441 *Enigma* puzzles, as well as 127 from the *Tantalizer* series and 90 from the original *Puzzle* series. There are 351 *Enigma* puzzles remaining to post, so 80% of *Enigma* puzzles are now available on the site.

[enigma595c] [enigma595]

18 December 2020

Posted by on **From New Scientist #1748, 22nd December 1990** [link]

The Reverend Brown watched as the Sunday School children sorted the Christmas food parcels for the local people in need. The children had drawn a diagram on the floor, so:

The children showed Rev Brown how they used the diagram to sort the parcels into order according to their weight. Ten children took a parcel and stood at the top of one of the 10 vertical lines. They then walked down their lines, keeping abreast, until they reached the first horizontal line, BI.

The two children in lines B and I compared their parcels and, as I was lighter, they exchanged places. If B had been lighter or if the weights had been equal they would have stayed where they were.

All 10 children they carried on until they reached the next horizontal line, AC, when the process was repeated. They then carried on until they reached the bottom of the vertical lines, at which point the parcels were in order, with the lightest at A and the heaviest at J.

The example the children showed Rev Brown was as follows:

Rev Brown asked if the diagram would sort correctly every possible row of 10 parcels, whatever they weighed. The children did not know, but they said they had tested every possible row of 10 parcels each weighing 1 or 2 pounds and they all had been sorted correctly.

Which of the following rows can we say

for certainwill be sorted correctly by the diagram?(a) 3 2 2 1 2 1 2 1 1 1

(b) 3 2 1 3 2 1 3 2 1 1

(c) 5 8 7 2 1 6 4 3 9 2

[enigma595b] [enigma595]

14 December 2020

Posted by on **From New Scientist #1748, 22nd December 1990** [link]

My Christmas present to my nephew is a marvellous new electronic toy. When you press a button, 36 letters appear evenly spaced around a circle. A spot starts at the centre of the circle, moves straight to one of the letters, straight to another letter, and so on — thus spelling out a word. Sometimes it pauses at a letter to indicate a double-letter. At the end of the word, the spot goes straight back to the centre pauses briefly there before starting on the next word.

You can choose one of many themes, and so I tried it out by choosing “Christmas”.

The spot spelt out the following messages:

UNTO US A CHILD IS BORNand:

DING DONG MERRILY ON HIGHand then it spelt out just one of the following words:

PARTRIDGE

DOVES

HENS

BIRDS

RINGS

GEESE

SWANS

MAIDS

LADIES

PIPERS

DRUMMERS

LORDSEach time the spot moved during this entire display, it moved a distance equal to either the radius or the diameter of the circle.

Which of those twelve words did it spell out?

[enigma595a] [enigma595]

11 December 2020

Posted by on **From New Scientist #1747, 15th December 1990** [link]

Dorothy came to a part of the Land of Oz where there were 16 towns, no three of them in a line, as shown on the map:

When Dorothy saw there were no roads she felt sad that the people from different towns did not meet one another. So she decided to build some yellow-brick roads.

When the wicked Witch of the West heard of Dorothy’s plan she confronted the girl. “There are four conditions that your roads must satisfy”, she commanded.

“(a) Each road must be a straight line from one town to another;

(b) no two roads must cross;

(c) a town cannot have more than one road;

(d) before you start I shall colour 8 towns Red and 8 towns Green — each road must join a Red town and a Green town.”The Witch then had a long think about which colour to give each town. She knew that Dorothy would build as many roads as possible for the given colours. The Witch wanted to choose the colours so that the fewest number of roads could be built. Eventually [the Witch] worked out her best choice of colours, and painted the towns.

How many roads could Dorothy then make?

In the original puzzle text “Dorothy” appears where I have placed “the Witch” in square brackets.

[enigma594]

7 December 2020

Posted by on **From New Scientist #1746, 8th December 1990** [link]

It was, I’m afraid. typical of my Uncle that he should have torn up the sheet of paper which gave particulars of the numbers of matches played, won, lost, drawn and so on, of four local football teams that were eventually going to play each other once.

Not only had he torn it up, but he had also thrown away more than half of it. This was all that was left:

(2 points are given for a win and 1 for a draw).

As if to emphasise his muddled ways, he had also got one of the remaining figures wrong, but “fortunately” it was only 1 out (that is, 1 more or less than the correct figure).

Each side played at least 1 game, and not more than 7 goals were scored in any match.

Calling the teams A, B, C and D, in that order, find the score in each match.

[enigma593]

4 December 2020

Posted by on **From New Scientist #1745, 1st December 1990** [link]

Even in Utopia, it seems necessary to search for the best possible leader. My Utopian friend has just sent me a letter outlining how they do it:

“The Dextrous Party elected a new leader last week. Their procedure may interest you. They use the usual exhaustive ballot system so that any Dextrous MP who can get five of his fellows to agree to vote for him may stand; whoever comes last must drop out, leaving the others to fight it out in a second ballot and so on until there are only two left in a final ballot.”

“Over the years, however, an additional refinement has been introduced. Now each candidate must state before the first ballot which one of the other candidates he or she is prepared to serve under. If the preferred candidate polls more than him, he drops out as well as the one coming last, and instructs his followers to vote for his preferred candidate. (If his preferred candidate has had to drop out as well, he still drops out if the preferred candidate polled more, but in this case his supporters may vote as they see fit).”

“In this way, the contest is usually decided after one or two ballots rather than three or four. This time, Boodle chose Poodle, while Noodle was the choice of both Poodle and Doodle. Two of the four candidates went through to a second ballot (nobody tied) with Poodle polling the same number of votes each time.”

What was the order of the candidates on the first ballot? And who won in the end?

[enigma592]

30 November 2020

Posted by on **From New Scientist #1744, 24th November 1990** [link]

Your task this week is to find a particular whole number about which I’ll give you some clues. Unfortunately, some of them aren’t true:

- At least one of the last two statements in this list is true.
- This is either the first true or the first false statement in the list.
- There exist three consecutive false statements.
- The difference between the numbers of the last true statement and the first true statement is a factor of the unknown number.
- The sum of the numbers of the true statements is the unknown number.
- This is not the last true statement.
- Each true statement’s number is a factor of the unknown number.
- The unknown number equals the percentage of these statements which are true.
- The number of different factors which the unknown number has (excluding 1 and itself) is more than the sum of the numbers of the true statements.
- There are no three consecutive true statements.
What is the number?

[enigma591]

27 November 2020

Posted by on **From New Scientist #1743, 17th November 1990** [link]

Pussicatto was recalling his early days in Germany in the 1920s when he was a student at the Bowows. His professor had given everyone in his class a canvas divided into 12 numbered squares, so:

The students were to paint each square either red or green. The professor played the class a record and said that only one way of colouring the canvas captured the spirit of the music.

The students began to paint their canvases and the professor walked round looking at their work. After a while he stopped the class and declared sadly that everyone had already painted at least one square the wrong colour even though they were only a little way into the task.

The partial paintings at that points of the 26 students in the class were as follows:

- 1R, 2R, 5R, 6R
- 1G, 4R, 11G, 12G
- 2R, 5G, 8G, 9G
- 6R, 10G, 11R
- 3G, 4R, 5R, 6R, 7G
- 1G, 9G, 10R
- 3G, 5R, 9G, 12R
- 8R, 11R, 12R
- 5G, 8G, 9R
- 4G, 7R, 8G, 10G
- 1G, 11R, 12G
- 8R, 12G
- 1G, 3R, 4R, 7R
- 3R, 5R, 7G
- 2G, 5R, 6G
- 3G, 9R, 10R, 12R
- 1G, 3G, 11G
- 3G, 4G, 5R, 7G
- 1R, 2G, 7R
- 8R, 10G, 11G, 12R
- 2G, 5G, 9G
- 1R, 2R, 6G, 7R
- 4G, 7R, 10R
- 3G, 4R, 6G, 7G, 8G
- 6R, 11G
- 8R, 10R, 11G, 12R
For example, the 26th student had painted square 8 Red, square 10 Red, square 11 Green and square 12 Red.

What is the correct painting?

[enigma590]

23 November 2020

Posted by on **From New Scientist #1742, 10th November 1990** [link]

In this division sum, each letter stands for a different digit. Rewrite the sum with the letters replaced by digits.

[enigma589]

20 November 2020

Posted by on **From New Scientist #1741, 3rd November 1990** [link]

I have one of those popular 15-tile puzzles in which the 15 pieces move around in a 4×4 frame by sliding a piece into the gap, another piece into the gap thus created, and so on. In this way, lots of other (but not all) arrangements can be achieved. When new the arrangement was:

I have now moved the pieces into an arrangement with the gap and the lighter/darker pieces as in the pattern shown below:

Also:

(i) all the even digits are again in the top half;

(ii) the four-figure number in the top row is 12 times another number with all

itsdigits even;(iii) the four-figure number in the second row is a perfect square;

(iv) a three-figure number formed by some arrangement of the three digits in the bottom row divides exactly a certain number of times into the four-figure number in the third row (and, as a bonus clue, that “certain number” is a perfect square).

Fill in the numbers in [the grid given above].

[enigma588]

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