Enigmatic Code

Programming Enigma Puzzles

Category Archives: enigma

Enigma 427: Settle some new scores

From New Scientist #1577, 10th September 1987 [link]

Football teams A, B, C and D are having a competition against each other, under a new method which has recently become popular. Under this method, 10 points are awarded for a win, 5 points for a draw, and 1 point for each goal scored. The situation when all but one of the matches had been played was as follows:

A, 9 points; B. 2 points; C, 24 points; D, 34 points.

Each side scored at least one goal in every match, but not more than seven goals were scored in any match.

Find the score in each game.



Enigma 426: Time and again

From New Scientist #1576, 3rd September 1987 [link]

Time and again, you’ve been asked to sort out letters-for-digits puzzles, where digits are consistently replaced by letters, different letters being used for different digits. Today, that recurring theme is used in a truly recurring way. The fraction on the left (which is in its simplest form) represents the recurring decimal on the right. Should you want an extra optional clue, I can also tell you that the last two digits of the numerator of the fraction are equal.

Enigma 426

What is AGAIN?


Enigma 1084: 1-2-3 triangles

From New Scientist #2240, 27th May 2000

The diagram shows a large triangle divided into 100 small triangles. There are 66 points that are corners of the small triangles.

You are to write 1 or 2 or 3 against each of the 66 corner points. The only restrictions are:

(a) the corners of the large triangle must be labelled 1, 2 and 3 in some order;
(b) each number on a side of the large triangle must be the same as the number at one end of that side.

Q1: Is it possible for you to write the numbers on so that there are precisely 10 small triangles with corners labelled 1, 2 and 3?

Q2: As Q1 but with 32 small triangles.

Q3: As Q1 but with 61 small triangles.

Q4: As Q1 but with 89 small triangles.


Enigma 1085: Cut and run

From New Scientist #2241, 3rd June 2000

Place your finger in the starting box in the grid and follow each instruction. You will find that you visit each square before finishing in the appropriate box.

Now cut up the board into six rectangles each consisting of two adjacent squares. Then put them back together to form a new three-by-four grid with the starting square one place lower than it was before. Do all this in such a way that, once again, if you follow the instructions you visit each square.

In your new grid, what instruction is in the top left-hand corner, and what is the instruction in the bottom right-hand corner?


Enigma 425: Them thar’ Hills

From New Scientist #1575, 27th August 1987 [link]

“I am old,” said Mr Methuselah, “but not as old as the Hills. Did you know that if you add up the ages of all the Hills exceptin’ Mr Hill you gets his age? And did you know that if you multiply the ages of all the Hills except Mr Hill you get a number which contains ones only, and as many ones as there are Hills, not counting Mr Hill? Every Hill has a different age less than 100, and every Hill’s age in years is odd, exceptin’, of course, Mr Hill.”

I didn’t know this. How could I? I had only just arrived in Rome, Georgia, the knew nothing of the locality. But once he had told me it sure set me to wondering:

How old is Mr Hill? How old is Mrs Hill? And how old are the Hillocks?


Enigma 1086: Stacking trays

From New Scientist #2242, 10th June 2000

Harry owns fewer than 500 snooker balls; he also owns four differently sized rectangular trays in which he can stack the balls, each tray being large enough to accommodate at least four balls along each side.

Harry starts the stacking by filling the base of the tray with as many balls arranged in rows and columns as its size allows; he then stacks further balls in layers, the balls in each layer covering the cavities between the balls in the layer below. If he starts with a square tray (though he does not necessarily have one that is square) the number of balls in each successive layer remains square until the top layer consists of just one ball. If he starts with a tray that is not square the top layers consists of a single row of two or more balls. So on top of a layer of 15 balls arranged 5 × 3 would rest a layer of eight (= 4 × 2) and then a top layer of three (= 3 × 1) balls.

If Harry stacks any one of his four trays as described above, he uses all his snooker balls to complete the task.

Everything stated above about Harry is also true for Tom. But Tom owns more snooker balls than Harry.

How many more?


Enigma 424: A round of fractions

From New Scientist #1574, 20th August 1987 [link]

Anne and Barbara have just played a round of golf consisting of 18 holes. Anne decided to keep an unusual record of the game. After each hole she formed the fraction consisting of her total score to that point divided by Barbara’s. Naturally she reduced each fraction to its lowest terms. For example, if, after seven holes, Anne had had a total score of 33 strokes Barbara one of 30 strokes, then Anne would have recorded the fraction 11/10. The fractions Anne obtained are as follows, in increasing order, not necessarily in the order they occurred in the game:

9/10, 25/27, 14/15, 17/18, 18/19, 19/20, 29/30, 31/32, 1, 68/67, 36/35, 28/27, 12/11, 7/6, 9/7, 7/5, 3/2, 5/3.

On the course that Anne and Barbara played on, each hole was par 4, that is, at each hole a player is expected to take four strokes. In their round, each girl, at each hole, scored par or a birdie or a bogey. A birdie is a score one less than par and a bogey is a score one more than par.

Which holes did Anne win, that is, take fewer strokes at, which did Barbara win, and which were shared?


Enigma 1087: Egyptian triangles

From New Scientist #2243, 17th June 2000

George has made a number of spinners for his children to select numbers when playing board games. Each has a circular disc divided into a number of equal sectors. He has made several discs of various sizes and with various numbers of sectors.

George’s son has discovered that three of the discs will fit together so that the marked radii form a triangle (shaded in the diagram) which includes just one sector on each disc.

Further research revealed that several other sets of three discs of suitable sizes and numbers of sectors can be used for form triangles of various shapes in this way, always including just one sector on each disc. George has found one set in which two of the discs have different prime numbers of sectors.

How many sectors are marked on the third disc of this trio?


Enigma 423: Four teams, more letters

From New Scientist #1573, 13th August 1987 [link]

In the following football table digits (from 0 to 9) have been replaced by letters. The same letter stands for the same digit wherever it appears and different letters stand for different digits. The four teams are eventually going to play each other once.

Enigma 423

(Two points are given for a win and 1 point to each side in a drawn match).

Find the score in each match.


Enigma 1088: That’s torn it

From New Scientist #2244, 24th June 2000

I had ten cards with a different digit on each and I tore one of them up. I used two of the remaining cards to form a two-figure prime, three others to form a three-figure prime, and the last four to form a four-figure prime.

If I told you the total of those three primes it would still be impossible for you to work out which digit I had torn up. In fact, the middle two digits of the total both equal the digit I tore up. That should enable you with a little brainpower to say what the total of the three primes is.

What is that total?


Enigma 422: Keeping fit by halves

From New Scientist #1572, 6th August 1987 [link]

In our local sports club everyone plays at least one of badminton, squash and tennis. Of those who don’t play badminton, half play squash. Of those who don’t play squash, half play badminton. Of those who play badminton and squash, half play tennis.

I play badminton only: there are two more players who play tennis only than there are who play badminton only.

If a player plays just two of the three games, then his or her spouse also plays just two of the three games.

The membership consists entirely of married couples and each of the three games is played by at least one member of each married couple.

How many people are there in the club, and how many of those play all three games?


Enigma 1089: Catch the buses

From New Scientist #2245, 1st July 2000

The island of Buss is divided into 3 counties, A-shire, B-shire and C-shire, each containing a number of towns. Each town in A-shire has just one road and that goes to a town in B-shire. Each town in B-shire, irrespective of how many roads it has to towns in A-shire, has just one road that goes to a town in C-shire. There are 3 bus companies, Red, Yellow and Green. Here are the buses for today:

• From each A-shire town one Red bus runs to a B-shire town.
• From each B-shire town one Yellow bus runs to a C-shire town.
• From each A-shire town one Green bus runs to a C-shire town.

Naturally the destinations are determined by the roads. A bus company is “economical” if no town is the destination of more than one of its buses today. A bus company is “covering” if every town in the county its buses today finish in is the destination of at least one of those buses. Which of the following statements are bound to be true:

(1) If the Green Co is economical then the Red Co is economical.
(2) If the Green Co is not economical then the Red Co is not economical.
(3) If the Green Co is covering then the Yellow Co is covering.
(4) If the Green Co is not covering then the Red Co is not covering.
(5) If the Green Co is economical and the Red Co is covering then the Yellow Co is economical.
(6) If the Green Co is covering and the Yellow Co is economical then the Red Co is covering.

This is another puzzle from an old paper copy of New Scientist that I recently found. It currently doesn’t appear in the on-line archives.


Enigma 421: Simple arithmetic

From New Scientist #1571, 30th July 1987 [link]

“Substitution sums,” said my friend and colleague Dr Addam, “are not at all difficult to concoct. Take, for example, this one.” And he wrote on a handy sheet of paper:

“The idea, as ever,” he continued, “is to substitute a different digit for each letter wherever it appears.”

I pondered for a minute. “I see what you mean,” I said. “There seem to be several solutions.”

“True enough,” said Addam; “so I’ll make it more difficult. Can you find me a solution in which the number THREE is divisible by 3, FOUR by 4, and …”

“SEVEN by 7?” I interrupted.

“No, that can’t be done. I was going to say FOURTEEN is divisible by 14.”

After a few minutes, I had the required solution. What was it?”


Enigma 1090: Week links

From New Scientist #2246, 8th July 2000 [link]

Today we are trying to crack a code. To each of the days:


there is assigned a different whole number from 1 to 10.

We know that, for any pair of days, their numbers have a factor larger than 1 in common if and only if their first three letters have a letter or two in common. So, for example, TUE and WED both have an E and so their numbers will have a factor larger than 1 in common, whereas the numbers for WED and THU will have no factor larger than 1 in common.

We also know that their numbers will satisfy:


What (in order MON to SUN) are their seven numbers?


Enigma 420: A princely sum

From New Scientist #1570, 23rd July 1987 [link]

The King of Zoz was asked by his son for 88,200 ducats to cover the latter’s gambling debts incurred at college.

“That is indeed a princely sum!” said the King, calling for his ornate coffer, the one containing single ducats only.

“How much is there in the old coffer of yours?” asked the Prince as two servants struggled with it it.

The father smiled indulgently but knowingly and replied: “If you were to multiply the number of ducats it now contains by the number of ducats would remain in it were I to remove 88,200 ducats, you would arrive at a perfect square.”

“There must be many numbers of ducats which fit those constraints,” said the son wistfully.

“Aye. And if you have the wit to deduce the number of different numbers of ducats there could be in the coffer consistent with my statement I shall know that you have learnt something and I shall pay your debt. If not, you can cover it yourself.”

Assuming there will be at least one ducat remaining in the coffer if the King removed 88,200 ducats, how many different possible numbers of ducats are there?


Enigma 1091: One’s best years

From New Scientist #2247, 15th July 2000 [link]

Mr Meaner has now retired from teaching. As a tough arithmetic exercise each year he used to ask his class to take the number of the year and find some multiple of it which consists simply of a string of ones followed by a string of zeros. For example in 1995 one girl in the class found that the 19-digit number 1111111111111111110 was a multiple of 1995!

Mr Meaner had been asking this question every year since he started training as a teacher. On the first occasion it was a reasonably straightforward exercise and most of the class found a multiple using as few digits as possible.

It is lucky that he did not ask the question a year earlier, for that year would have required over two hundred times as many digits as that first occasion did.

In what year did Mr Meaner first ask the question?


Enigma 419: Painting by numbers

From New Scientist #1569, 16th July 1987 [link]


1. You will need four copies of:

Enigma 419 - 1

labelled A, B, C, D.

2. Take A. The artist Pussicato signs the top row and you sign the bottom row; your signature must contain 9 letters.

3. Fill in B by using A as follows. Take each A square in turn, find the position of its letter in the alphabet and from that number subtract the appropriate multiple of 5 to leave a number from 1 to 5. Put that number in the corresponding B square.

4. Fill in C by using B as follows. Each square touches three or five other squares, including touching along a side or at a corner. Take each B square in turn and add up the numbers in the squares it touches. From the total subtract the appropriate multiple of 5 to leave a number from 1 to 5. Put that number in the corresponding C square.

5. Paint D by using C as follows.Number the five colours, Red, Blue, Green, Yellow, White, 1 to 5 in any order you like. Take each C square in turn and find the colour you have given its number. Paint the corresponding D square with that colour.


Enigma 419 - 2

A painter whose name involves only the first five letters of the alphabet produced:

Enigma 419 - 3

What was the painters name?


Enigma 1092: A prime age

From New Scientist #2248, 22nd July 2000 [link]

Marge, April, May, June, Julia and Augusta have all celebrated their birthday today. They are all teenagers and with the exception of the one pair of twins their ages are all different.

Today, only Marge and April have ages which are prime numbers, but the sum of the ages of all the girls is also a prime number. On their birthday last year, only May and June had ages which were prime numbers, but again the sum of the ages of all the girls was a prime number. On their birthday two years before that, only May and Julia had ages which were prime numbers, but even then, the sum of the ages of all the girls was again a prime number.

How old is Augusta?


Enigma 418: Let us divide

From New Scientist #1568, 9th July 1987 [link]

In the following division sum each letter stands for a different digit:

Re-write the sum with the letters replaced by digits.


Enigma 1093: Primed to spend

From New Scientist #2249, 29th July 2000 [link]

Bill’s credit card has the usual four four-digit numbers, which are in ascending order of size. All are prime numbers and the sum of the digits of each is the same.

The digits in the first number are all different and the third number is the first number reversed. The digits in the second number are all different and the fourth number is the second number reversed. The last digits of the four numbers are all different.

He has a hopeless memory for figures, but he can always work out his four-digit PIN from his card, because he can remember that it is equal to the difference between the first and third numbers (or the difference between the second and fourth numbers, which is the same) and happens to be a perfect square.

What is the fourth number?