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Programming Enigma Puzzles

23 May 2018

Posted by on **From New Scientist #1003, 3rd June 1976** [link]

Four snails set off down the garden path just as dawn broke. Fe and Fi kept pace with each other, a modest but steady shuffle which had taken them a mere eight yards by the time Fo and Fum had reached the rhododendron. Fo was so puffed that he stopped for an hour’s rest and even Fum, who carried straight on, was reduced to the pace of Fe and Fi.

Fo started again just as Fe and Fi came level with him and surged away at his previous pace. Fe promptly accelerated and kept level with him but Fi continued as before. Fe was this one yard ahead of Fi at the end of the path but half an hour behind Fum.

How long is the path?

[tantalizer452]

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9 May 2018

Posted by on **From New Scientist #1004, 10th June 1976** [link]

No child was left out of the school play. Each was an angel, a bunny or a demon.

“Were you a bunny, dear?”, Granny asked Tom.

“No”, said Tom firmly.

“He was!”, said Dick.

“He wasn’t!”, said Harry.

“How many of you were bunnies?”, Granny asked.

“Just one”, said Harry.

“Not none”, said Dick.

“More than one”, said Tom.Any angel had made two true statements, any bunny one and any demon none.

Who was what?

[tantalizer453]

25 April 2018

Posted by on **From New Scientist #1005, 17th June 1976** [link]

From a complete pack I take two cards chosen from the kings, queens and jacks. I mark one “A” and the other “B”. Your task is to identify A and B, given that the ranking of suits from the top is Spades, Hearts, Diamonds, Clubs and that “if” does not mean the same as “only if”:

1. If A is black, B is a jack.

2. A is a queen, only if B is a diamond.

3. A is a heart, only if B is black.

4. A is a king, if B is a spade.

5. A is a club, if B is a king.

6. A is a heart, if B is a heart.

7. If the higher is a queen, the lower is a heart.

8. The lower is a jack, only if the higher is a heart.

9. If the lower is red, B is a spade.

10. The higher is red, only if A is not a king.

[tantalizer454]

12 April 2018

Posted by on **From New Scientist #1006, 24th June 1976** [link]

Ballistico is a wild Guatemalan game for two, played by tossing a dead hen over a barn. The Projecteador (or thrower) stands on one side, the Manudor (or catcher) on the other. If the hen lands within ten metres of the manudor’s position and he fails to catch it, the projecteador scores a ping. Otherwise the manudor scores a pong. Each ping or pong counts as one point.

The players change role after each throw. In other words the projecteador for one throw is the manudor for the next. In the last game I watched 5 pings were scored and the game was won by 7 points to 6.

Was the original projecteador the winner of the loser?

[tantalizer455]

11 April 2018

Posted by on **From New Scientist #1007, 1st July 1976** [link]

Feeling mortal, Lord Woburn summoned his daughters, Alice and Beatrice, to hear about his will. “I have decided to leave you my hippos”, he announced. “There are either 9 or 16 of them but you do not know which. Each of you will inherit at least one and I shell tell each of you privately how many the other will be getting”.

He was as good as his word. “How many shall I be getting?” Alice asked Beatrice nervously afterwards. Beatrice refused to say but asked, “How many shall I be getting?”. Alice refused to say and again asked, “How many shall I be getting?”. You should know that each lady is a perfect logician, who never asks a question she knows or can deduce the answer to.

I think this proves that a square deal on the hippopotonews is equal to the sum of the squaws on the other two sides. At any rate how many hippos was each to receive?

The puzzle as presented above is flawed, in that the situation described would not arise. An apology was published with **Tantalizer 460**.

[tantalizer456]

28 March 2018

Posted by on **From New Scientist #1008, 8th July 1976** [link]

Now that Uncle Arthur can’t get about, he watches the world through a sunny little window in the sitting room. It is a diamond-shaped, mullioned window made up of 49 small diamond-shaped panes, separated by lead bars. Just below the window there is a bee hive and sleepy bees are often to be seen ambling up the glass. Uncle Arthur has noticed that they always move in a series of straight lines, passing through the middle of each pane, crossing from pane to pane at the mid point of a bar and moving in an upward direction.

He points out that there are many possible routes a bee can take from bottom to top and would like to know exactly how many.

[tantalizer457]

14 March 2018

Posted by on **From New Scientist #1009, 15th July 1976** [link]

If you must have your operation at St. Vitus’ Hospital, choose your surgeon with care. There are four in residence and no two of them are equally safe. Here are six bits of information to cheer you up while you wait:

1. Cutaway is the most lethal.

2. Anyone safer than Borethrough is safer than Axehead.

3. Divot is not the safest.

4. Anyone safer than Divot is no less lethal than Cutaway.

5. Borethrough is not the safest.

6. Anyone safer than Axehead is safer than Divot.Do I hear you complain that the six statements cannot all be true? Quite right — I put one false one in for diplomatic reasons. And now can you rank the butchers starting with the safest?

[tantalizer458]

7 March 2018

Posted by on **From New Scientist #1010, 22nd July 1976** [link]

Our horticultural club had a little competition on Monday, with three events. For Vegetables you could enter either 1 cabbage or 2 turnips or 3 leeks or 4 potatoes; for Flowers either 2 hollyhocks or 4 lupins or 6 roses or 8 gladioli; for Fruit either 3 pears or 4 apples or 5 quinces or 6 strawberries. There were 5 competitors each of whom entered for two events.

Arthur Acorn displayed 12 items in all, Bill Barley 11, Crissie Canteloupe 9, Dahlia Dennis 6 and Edward Earthy 5. The prize for the entry judged best not only in its event but also in the whole show went to Crissie. She was in fact the only person to show that kind of item. You could deduce what it was, if I told you exactly what the other four competitors entered.

So what was it?

[tantalizer459]

21 February 2018

Posted by on **From New Scientist #1011, 29th July 1976** [link]

The Patagonian navy used to have 16 class I gun boats with 5 guns apiece, some class II gun boats with 4 guns apiece and some class III gun boats with 3 guns apiece. That was until the admiral was ordered to crush a revolution which had broken out on seven scattered islands simultaneously.

After some thought and more gin he split his fleet into seven flotillas, each of six ships and each with a different number of guns. There was at least one gun boat of each class in each flotilla. This savage armada sailed truculently into the mist one grey dawn and was never seen again.

How many boats of class II were lost?

[tantalizer460]

20 February 2018

Posted by on **From New Scientist #1012, 5th August 1976** [link]

Here are some cryptic clues, each indicating a colour of the rainbow:

1. Buzzing bottle.

2. What I do when I stub my toe.

3. Spring innocent.

4. Cold and depressed.

5. Iris for massed voices.

6. Almost violent.

7. Danger to health.

8. Mixed teenagers with no teas.

9. Stockings in gowns.

10. Butterfly in charge of the fleet.

11. Hammer with wings.A certain number of them refer to the colour of Aunt Bea’s new bonnet. If I told you how many, you could work out what colour that is. So I shan’t. But you can. So what is it?

[tantalizer461]

31 January 2018

Posted by on **From New Scientist #1013, 12th August 1976** [link]

A fragment of a prophecy lately unearthed says that the Oxford vs. Cambridge boat race of 1980 will take the usual form (8 oarsmen in each boat and so on) but will end in a tie. So an occupant of each boat will be picked and random and these two will decide the event by a mile race run on foot. This is unlikely to be a cliff-hanger, however, as the chances are 2:1 that exactly one of them will have a wooden leg. Since this is an unmistakeable handicap, Oxford is therefore likely to be the winner, despite having at least one wooden leg in the boat.

An impossible prophecy? Not at all, if you avoid a small catch. Assuming that no one in either boat has two wooden legs, can you work out how many wooden legs there will have to be in the Cambridge boat?

[tantalizer462]

17 January 2018

Posted by on **From New Scientist #1014, 19th August 1976** [link]

The notice in the magistrates retiring room at Bulchester court reads baldly, “Monday: Smith, Brown, Robinson”. These are the surnames of next Monday’s bench, which will, as always, include at least one man and one married woman. All male magistrates at Bulchester happen to be married. These facts are known to all magistrates.

The court being a large and new amalgamation, Smith, Brown and Robinson know nothing about each other. But Smith, on being told the sex of Brown, could deduce the sex of Robinson and the marital status of both. And Robinson, being told only that Smith could do this, could deduce the sex and marital status of Smith and Brown.

What can you deduce about the trio?

[tantalizer463]

3 January 2018

Posted by on **From New Scientist #1015, 26th August 1976** [link]

The Pentathlon at the West Wessex Olympics is a Monday-to-Friday affair with a different event each day. Entrants specify which day they would prefer for which event — a silly idea, as they never agree.

This time, for instance, there were five entrants. Each handed in a list of events in his preferred order. No day was picked for any event by more than two entrants. Swimming was the only event which no one wished to tackle on the Monday. For the Tuesday there was just one request for horse-riding, just one for fencing and just one for swimming. For the Wednesday there were two bids for cross-country running and two for pistol-shooting. For the Thursday two entrants proposed cross-country and just one wanted horse-riding. The Friday was more sought after for swimming than for fencing.

Still, the organisers did manage to find an order which gave each entrant exactly two events on the day he had wanted them.

In what order were the events held?

I don’t think there is a solution to this puzzle as it is presented. Instead I would change the condition for Thursday to:

For Thursday two entrants proposed cross-country and just one wanted

fencing.

This allows you to arrive at the published answer.

[tantalizer464]

20 December 2017

Posted by on **From New Scientist #1016, 2nd September 1976** [link]

Paul Pennyfeather, you will recall from Evelyn Waugh’s novel, was sent down from Oxford and went to teach in Dr Fagan’s horrid school at Llanaba Abbey. There he found that a class of the beastliest boys could be kept quiet till break by offering a prize of half a crown for the longest essay, irrespective of all possible merit. Now read on:

“Sir”, remarked Clutterbuck after break, “I claim the prize”.

“But you”, Paul protested feebly, “have written only one-third as many words as Ponsonby, one-fifth as many as Briggs and one-eighth as many as Tangent”.

“Nonetheless, Sir, Dr Fagan would certainly wish me to have the prize”.

And so it proved. You might also like to know that the oldest of these four boys wrote 2222 more words than the second oldest and used more full stops in his essay than any of them who wrote less words than the youngest.

Where was Clutterbuck in the order of age, and how many words did he write?

[tantalizer465]

6 December 2017

Posted by on **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

notally 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 doesnotally 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

notx thennoty”. But they could not answer the original question. Assuming all the hypotheticals to be true, can you?

[tantalizer466]

5 December 2017

Posted by on **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]

22 November 2017

Posted by on **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]

8 November 2017

Posted by on **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]

25 October 2017

Posted by on **From New Scientist #1021, 7th October 1976** [link]

“My formula for life is Wit multiplied by Will”, Uncle Ernest announced for the umpteenth time. “I daresay you don’t know what that gives you, young Tommy”.

“Oh yes I do, Uncle”, Tommy replied cheekily.

“Success”, snapped Uncle Ernest.

“Thirst”, retorted Tommy.

Tommy, of course, had made a cryptarithmetic problem of it:

WIT × WILL = THIRST.

Each different letter stands for a different digit.

What is the value of THIRST?

[tantalizer470]

11 October 2017

Posted by on **From New Scientist #1022, 14th October 1976** [link]

Hop, Skip and Jump live in different houses in Tantalus St., which is numbered from 1 to 100. Here is what they have to say about the matter:

Hop: “My number is divisible by 7. Skip is much too fat. Jump’s number is twice mine.”

Skip: “Hop lives at 28. My number is one third of Jump’s. Jump and I are

notboth even.”Jump: “Hop lives at 91. Skip lives at 81. My number is divisible by 4.”

One of them has thus made three true statements, another three false and the remaining fellow has alternated, uttering either true, false, true or false, true, false.

Who lives where? And is Skip much too fat?

A correction to this puzzle was published with **Tantalizer 473**. The problem statement above has been revised accordingly.

[tantalizer471]

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