Enigmatic Code

Programming Enigma Puzzles

Enigma 1169: Snookered

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?


Enigma 342: A full set of dominoes

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.

Enigma 342


Enigma 1170: Folderol

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:

Enigma 1170

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?


Enigma 341: Spot the score

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:

Enigma 341

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

Find the score in each match.


Enigma 1171: Irregular hexagon

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?


Enigma 340: Unmagic cross-figure

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:

Enigma 340


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.


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 unmagic 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.

Find the 3×3 answer.


Enigma 1172: Plant a tree

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 different distances. 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?


Enigma 339b: Happy Christmas

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]

Enigma 1173: Six months

From New Scientist #2329, 9th February 2002

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.


Enigma 1173

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?


Enigma 339a: Christmas dice

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 this 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 to much more than that. And with this pair 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.”

So, 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]

Enigma 1174: Small sums

From New Scientist #2330, 16th February 2002

An example of a letter sum is:


That is to say word + word = word, where each word is a string of letters of any length. An example of an answer for the above sum is:

537 + 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:


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


Enigma 338: Square money

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?


Enigma 1175: French squares

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:

UNNEUF and CENT are 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).


Enigma 337: Chock-a-block

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?


Enigma 1176: Football’s ups and downs

From New Scientist #2332, 2nd March 2002

In our local football league each of the four teams plays each of the others once each season. Here is a table showing the results at the end of last season, with the teams in alphabetic order.

Enigma 1176

Unfortunately every entry is wrong, being one more or one less than it should be. No two games had the same score.

What were the scores in the six games?

Thanks to Hugh Casement for providing a transcript for this puzzle.


Enigma 336: Seven teams, two leagues

From New Scientist #1484, 28th November 1985 [link]

Three football teams (A, B and C) are to play each other once, and four other football teams (D, E, F, and G) are also to play each other once. After some — or perhaps all — of the matches had been played, two separate tables were drawn up giving some details of the games played, won, lost etc.

But in these two tables, just to make things more interesting, letters have been substituted for digits. In both tables the same letter stands for the same digit (from 0 to 9) whenever it appears, and different letters stand for different digits.

The tables looked like this:

Enigma 336

(Two points are given for a win, and one point to each side in a drawn match).

Find the score in each match.


Tantalizer 500: Three kings

From New Scientist #1051, 12th May 1977 [link]

Tommy’s homework included the question:

“Pippin, Quorum and Rudolph were all Tantalusian monarchs. In what order did they reign?”

Tommy consulted Uncle George, who remarked:

“Well, Pippin was the first Tantalusian monarch to shoot an elephant. No monarch since Quorum has shot a hippopotamus. And, if Quorum is the most recent of the three, then Pippin was the earliest. That gives you enough to go on, assuming you know who else shot what.”

“But, Uncle, I don’t.”

“No matter — you can work that out too, if you bear in mind that everything I’ve just told you is relevant.”

Who reigned in what order and shot what?


Enigma 1177: Photocopy

From New Scientist #2333, 9th March 2002 [link]

George drew a rectangle on a piece of paper and marked a dot 2 centimetres to the right and 1 centimetre up from the bottom left hand corner, as measured by perpendicular grid lines. (Assume the dot has zero size).

He then made a photocopy, reduced in scale so that the length of the diagonal of the copied rectangle equalled the short side of the original. He rotated the copy and placed it so that three of its corners lay on the sides of the original as shown.

Enigma 1177

George was then surprised to find that the copy of the dot coincided exactly with the original.

What were the dimensions of the original rectangle?


Enigma 335: Square up

From New Scientist #1483, 21st November 1985 [link]

My calculating nephew has been foxing me again. He entered a three-figure number on the calculator, showed it to me, and then squared the number but tried to trick me by showing me the square upside down.

I could soon see that there was something wrong because although I could see a number it had no digits in common with the number that had been squared — whereas I know that their first digits should have been the same. Further investigation would have shown that the original number and its inverted square had no factors larger than 1 in common.

Now, without a computer, and without hours of work, you should be able to tell me what I want to know. What was the original three-figure number? And what number did my nephew show me trying to kid me that it was the square?


Enigma 1178: Last two digits

From New Scientist #2334, 16th March 2002 [link]

The teacher challenged the children to find a particular 5-digit number consisting of 5 different digits such that when they multiplied it by 2 they created another 5-digit number consisting of the other 5 digits.

By giving the children the last two digits of the number that they had to find the teacher ensured that there was only one possible answer; but I shall be less kind and merely tell you that the last digit of the number they had to find was 1.

What was that number ending in 1 that they had to find?



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