Enigmatic Code

Programming Enigma Puzzles

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.

[enigma427]

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Puzzle 59: Cricket

From New Scientist #1110, 6th July 1978 [link]

ABC and D are having a cricket competition with each other in which eventually there are all going to play each other once. Points are awarded as follows:

To the side that wins: 10
To the side that wins on the first innings in a drawn match: 6
To the side that loses on the first innings in a drawn match: 2
To each side for a tie: 5
To the side that loses: 0

The latest news I have about their points is as follows:

21; 10; 9; 6.

Find the result of each match.

[puzzle59]

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?

[enigma426]

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.

[enigma1084]

Tantalizer 466: Diplomacy

From New Scientist #1017, 9th September 1976 [link]

In the early years of this century, when the lamps were going out all over Europe, Lord Grey asked his advisers whether Bosnia would ally with Austria. They gave thought to the question but could only answer in hypotheticals.

If, they said, Bosnia allies with Austria, Montenegro will ally with Hungary. If Slovakia allies with Denmark, Finland will ally with Serbia. If Albania allies with Slovakia, Poland will ally with Transylvania. If Montenegro allies with Hungary, Latvia will ally with Poland. If Bosnia allies with Austria, then, unless Slovakia allies with Denmark, Finland will ally with Serbia. If Poland allies with Transylvania, Finland will not ally with Serbia. If Latvia allies with Poland then, if Rumania allies with Austria, Poland will ally with Transylvania. If Montenegro allies with Hungary and Albania does not ally with Slovakia, then Rumania will ally with Austria.

They added, of course, that “if x then y” implies neither “if y then x” nor “if not x then not y”. But they could not answer the original question. Assuming all the hypotheticals to be true, can you?

[tantalizer466]

Tantalizer 467: Nine men went to mow

From New Scientist #1018, 16th September 1976 [link]

Nine men went to mow, went to mow a meadow. So write 9 in one of the meadows. One man then went home and the rest went through a gate into the next meadow. Write 8 in that meadow. Another man went home. Write 7 in the next meadow. Then 6, then 5, then 4, then 3, then 2, then 1. Always use a gate. Do not enter the same meadow twice.

Now try adding the three digit number on the top line to that on the middle line. If you have mown the meadows in the right order, they add to the number on the bottom line. If not, go back to the start of the song and begin again. It can be done!

This puzzle was re-published 9 years later as Enigma 328.

[tantalizer467]

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?

[enigma1085]

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?

[enigma425]

Puzzle 60: Uncle Bungle gets the last line wrong

From New Scientist #1111, 13th July 1978 [link]

“If you know what you are adding you should be able to add it”, said Uncle Bungle when I complained to him that though the first two lines of his addition sum were legible, the last line across was not. I can of course see what he means by this but he seems to forget that in his addition sum letters have been substituted for digits. The same letter stands for the same digit whenever it appears and different letters stand for different digits.

In the last line across Uncle went haywire and as four of the letters were illegible I have replaced them by blanks. The sum reads like this:

Find the correct addition sum (from the remark that Uncle made I gathered that in the final sum all 10 digits appear).

[puzzle60]

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?

[enigma1086]

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?

[enigma424]

Tantalizer 468: Shell fire

From New Scientist #1019, 23rd September 1976 [link]

M. Champignon and M. Escargot are the chefs at the famous Café d’Amour and then have long worked happily together. But lately, alas, they have both fallen in love with the same waitress. They have decided there is only one way to settle the matter. Next Tuesday morning they will take a carton of six fresh eggs and hard boil two of them. The young lady will then replace these two in the carton and reposition them randomly.

This done, Champignon will pick an egg at random and try to break it on his head. If it proves hard boiled, the lady is his. If not, the turn will then pass to Escargot who will pick from the remaining five and try his luck. If he too is crossed in love, both chefs will join the Foreign Legion.

What are the odds that (a) Champignon, (b) Escargot, (c) neither will win the lady?

[tantalizer468]

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?

[enigma1087]

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.

[enigma423]

Puzzle 61: A division sum

From New Scientist #1112, 20th July 1978 [link]

Find the missing digits.

[puzzle61]

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?

[enigma1088]

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?

[enigma422]

Tantalizer 469: Revised version

From New Scientist #1020, 30th September 1976 [link]

Mr Chips was sodden with gloom after marking the R.I. test. Admittedly no one had scored nought, which was unusual. (In fact the scores were all different). But, alas, the other pupils had done better than the only four Christians in the class.

After some soul-searching he realised that it was his duty to adjust the marks. So he began by adding to each pupil’s score the number of marks gained in the test by all the other pupils. That was better but the resulting list left something to be desired. So he then subtracted from each pupil’s new score three times his or her original score. That was much more satisfactory. The scores were all positive and totalled 116 marks in all. The heathen Blenkinsop was rightly bottom and could therefore be made to write out the 119th psalm.

How many did Blenkinsop score before and after Mr Chips did his duty?

[tantalizer469]

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.

[enigma1089]

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?”

[enigma421]