Enigma 364: Wrong-angled triangle

From New Scientist #1513, 19th June 1986 [link]

“We Yorkshireman,” said my friend Triptolemus, “like a puzzle as a cure for insomnia, instead of counting sheep. Have you got a nice simple question, without a mass of figures to remember?”

So I said, “If a wrong-angled triangle has whole-number sides and an area equal to its perimeter, how long are its sides?”

He slept on the the problem and gave me the answer next morning.

Can you?

(A wrong-angled triangle is of course the opposite of a right-angled triangle. Instead of two of its angles adding up to 90°, it has two angles differing by 90°).

News

There are now 1000 Enigma puzzles on the site, with a full archive of puzzles from Enigma 1 (February 1979) up to this puzzle, Enigma 364 (June 1986) and also all puzzles from Enigma 1148 (August 2001) up to the final puzzle Enigma 1780 (December 2013). Altogether that is about 56% of all the Enigma puzzles ever published.

I have been able to get hold of most of the remaining puzzles up to the end of 1989 and from 2000 onwards, so I’m missing sources for most of the puzzles originally published in from 1990 to 1999. Any help in sourcing these is appreciated.

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Enigma 359: Neat odd quad

From New Scientist #1508, 15th May 1986 [link]

I call ABCD an odd cyclic quadrilateral, or “odd quad” for short. It has four corners A, B, C, D, and four straight sides AB, BC, CD, DA, so it’s a quadrilateral. The corners lie on a circle, so it’s cyclic. And it’s odd because — well, what is its area? I have decided to define that as what the sides cut off from the outside world, that is, the sum of the shaded areas.

A neat odd quad has the lengths of its four sides all different positive whole numbers and its area is a whole number too.

Can you find a neat odd quad with an area less than 30? What are the lengths of:

(a) The sides AB, CD, which don’t cross?
(b) The crossing sides BC, AD?

[enigma359]

Enigma 355: Diffy-dist

From New Scientist #1504, 17th April 1986 [link]

Diffy-dist is a game for two, played on a regular 6 × 6 square array of points, numbered as in the picture. Each player marks a point at each turn, the only restrictions being that the distance between any pair of marked points must be different from the distance between any other pair. Suppose, for instance, points 1 and 5 and 23 have been marked. Now 9 cannot be marked, because 9 – 1 and 9 – 5 are equidistant; nor 35, because 35 – 5 and 1 – 23 are equidistant.

The result of a recent game was unusual. We ended with six points marked. Of their numbers, one was prime, two were odd, three were consecutive, and four were multiples of 3.

What were the six numbers?

[enigma355]

Enigma 351: Four flies

From New Scientist #1500, 20th March 1986 [link] [link]

Four unfriendly flies are sitting watching the cricket. Each is on the boundary of the ground, which is perfectly circular. Their names, in order round the edge, are At, Bet, Cot and Dut.

The length of each of the straight lines At-Bet, Bet-Cot, Cot-Dut and Dut-At is an exact whole number of flymins (this is, the flies’ unit of distance). Two of those lines actually have the same length. Furthermore, the total of these four lengths is precisely the same number as the area (in square flymins) of the quadrilateral At-Bet-Cot-Dut.

If I tell you that that area does not include the pitch (which is at the centre of the ground), can you tell me the four lengths?

[enigma351]

Enigma 347: Trellis

From New Scientist #1496, 20th February 1986 [link]

I went out on Tuesday to buy a piece of trellis to cover a gap which I measured 30 inches wide × 48 inches high. Allowing half an inch spare on all sides for fixing, I needed at least 31 inches × 49 inches.

I found at the shop that they sold the stuff in various sizes. Size 5, for instance, which is shown in the picture, has five rows of five studs (on which the whole thing swivels); size 6 has six rows of six studs and so on.

In every size the studs were 5 inches apart along the slats, which were 1 inch wide, with each end rounded off in a semicircle of ½-inch radius centred on the last stud.

When I told the manager my measurements, he said,” I’m afraid size 11 will be just too small. If you pull it out to 31 inches wide, the height will be just short of 49 inches. So you’d better take size 12. We charge 5p per stud, so that will be £7.20… Yes, certainly we’ll change it if the size should be wrong.”

When I got home, I was horrified to find that I had mis-measured the height of the gap. It was really 84 inches, not 48 inches. So I needed 31 inches × 85 inches. Back I went to the shop, where, having done my homework meanwhile, I was able this time to ask myself for the size most economical of trellis that would meet my requirements exactly.

What was the price of this trellis?

[enigma347]

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.

[enigma342]

Enigma 339a: Christmas dice

From New Scientist #1487, 19th 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 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?

[enigma338]

Enigma 334: Biggabagg’s Bettamix

From New Scientist #1482, 14th November 1985 [link]

My two elder sisters love Mr Ivor Fitzclarence Biggabagg’s seductive sweet-counter. His “Bettamix”, a random mixture of liquorice and blackcurrant gums, is splendid value, they reckon: each buys a quantity every week. When they get home, they allow me to pick one gum for myself. Isn’t that nice of them?

I hate liquorice and love blackcurrant, but unfortunately they won’t let me just look and choose a blackcurrant gum. They do however let me first inspect the contents of each bag. Then I can either choose one bag at random and then pick a gum at random from it or mix up the contents of both bags and then pick a gum at random.

You will be glad to hear that I have now worked out a simple rule for deciding which of the two methods gives me the better chance of a blackcurrant gum.

I was surprised last week to find that it didn’t matter which method I chose. Anne’s bag contained 6 liquorice and 14 blackcurrant gums. Pam’s held 18 liquorice and quite a few blackcurrant.

Before I had a chance to choose, Anne removed some of the blackcurrant gums from Pam’s bag and asked me, in this new situation, to decide which method. And again I was surprised to find that it didn’t matter which method I chose.

How many blackcurrants did Anne remove?

Note: A correction was issued to this puzzle in New Scientist #1485 (along with the solution), the puzzle statement above has been modified accordingly.

[enigma334]

Enigma 329: Clear short circuit

From New Scientist #1477, 10th October 1985 [link]

An n-circuit is a closed path of n different points and n different legs. Every leg runs along a grid-line and every point is a junction of grid-lines. Legs do not overlap, but they may cross.

A clear circuit is one that you cannot make a circuit with just some of the points. Thus the 5-circuit A is not clear. Points 1, 2 and 5 would make a 3-circuit: so would points 3, 4 and 5. But B is clear.

The length of a circuit is just the sum of the lengths of the legs. Thus A has length 7, and B has length 11.

Can you find a clear 12-circuit with a length of 21 (or less)?

A similar problem to Enigma 325.

[enigma329]

Enigma 325: Clear thin circuit

From New Scientist #1473, 12th September 1985 [link]

 

 

An n-circuit is drawn on a triangular grid made up of equilateral triangles. It is a closed path of n different points and n different legs. Every leg runs along a grid-line and every point is a junction of grid-lines. Legs do not overlap, but they may cross.

A clear circuit is one that you cannot make a circuit with just some of the points. Thus the 5-circuit A is not clear. Points 1, 2 and 5 would make a 3-circuit: so would points 3, 4 and 5. But B is clear.

The fatness of a circuit is the area it encloses. Thus A has fatness 5 (measured by the number of little triangles enclosed), and B has fatness 7.

Can you find a clear 9-circuit with a fatness of 9 (or less)?

Happy Christmas from Enigmatic Code!

[enigma325]

Enigma 320: Triangular farm

From New Scientist #1468, 8th August 1985 [link]

Edgar’s farm is in the shape of an equilateral triangle (ABC in the diagram). It is divided into seven fields by three straight hedges, AQ, BR and CP. The fields such as BQT are 8 acres, and those such as ASUR are 22 acres. All I want to know is: what is the exact acreage of the middle field, STU?

[enigma320]

Enigma 316: The min-factor game

From New Scientist #1464, 11th July 1985 [link]

This is a game between you and the Devil. It starts with the natural numbers from 1 to N written in a row. You and the Devil play alternately, you first. The rules are simple:

(a) You take any number you choose (subject to rule (d) below) from those remaining in the row and delete it from the row.
(b) He then deletes from the numbers remaining in the row all those which are factors of the number you just took.
(c) Go to (a).
(d) You can never take a number which has no factor remaining in the row; that is, your take must permit the Devil in his turn to delete at least one number.

The game stops when you can legally take no more numbers, and you want the sum S of all the numbers you have taken to be as small as possible.

The picture records a game with N=7 and S=6. You did very well. Now try with N=30.

How small can you make S?

[enigma316]

Enigma 312: Six-stamp sheet

From New Scientist #1460, 13th June 1985 [link]

They are now printing stamp-books with stamps of different values on the same sheet. Let us take this a stage further, and design a sheet of 3 × 2 stamps, so that you can make up a postage of any whole number of pence from 1 up to N by tearing out a connected set of one or more stamps.

“Connected” means edge-connected, not just corner-touching, so that, for instance, the sheet illustrated achieves only N=5, since neither 4+2 nor 5+1 is a connected set, and so 6 is impossible.

Make a sketch showing how to make N as big as possible, and state what N is.

[enigma312]

Enigma 307: Seven a side

From New Scientist #1455, 9th May 1985 [link]

The team of seven has remained unchanged since 1981. The Chancellor described the annual team photographs to me as follows:

1981: Fred, Graham, Hermann, Jack; and men from Trinity, Unity and Varsity.
1982: Ken, Fred, Graham; and men from Unity, Varsity, Westminster and Sanctity.
1983: Graham, Hermann, Jack, Ken; and men from Varsity, Sanctity and Trinity.
1984: Jack, Ken, Fred; and men from Westminster, Unity, Trinity and Sanctity.

“Can you match up the names and colleges?”

“Surely”, I said, “there is some duplication, isn’t there?”

“Yes. There are three men of the same name. But there are not three men from one college”.

I found I still couldn’t do a complete matching, but I could say with certainty that ______ college had two men in the team.

Which college, and which two names?

[enigma307]

Enigma 303: Some dominoes

From New Scientist #1451, 11th April 1985 [link] [link]

Most of a full set of dominoes has been arranged in a 7×7 block with a hole in the middle. Please mark in the boundaries between the dominoes. There is only one answer.

[enigma303]

Enigma 299: Edge-equal tetra

From New Scientist #1447, 14th March 1985 [link]

The first picture shows a bird’s eye view of a tetrahedron with a number (a positive whole number) on each face, vertex and edge. The repeated 1 is on the bottom face which is out of sight.

The numbering is “edge-equal”, because every edge-number is the sum of the numbers on the two adjacent faces and also the sum of the numbers on the two adjacent vertices. 10 = 4 + 6 = 2 + 8, and 7 = 2 + 5 = 6 + 1, and so on.

That numbering has all the 8 face-numbers and edge-numbers different, which is nice, but if you count in the 6 edge-numbers too there is a duplication of 5, 7 and 8.

The second picture is for you to fill in an edge-equal numbering with all 14 numbers different. I had already put in the largest number, 21. The smallest number is to go on the bottom face, please.

[enigma299]

Enigma 295: The max-multiple game

From New Scientist #1443, 14th February 1985 [link]

This is a game between you and the Angel. It starts with the natural numbers from 1 to N, written in a row. You and the Angel play alternately, you first. The rules are:

(a) You take any number you choose (subject to D below) from those remaining in the row, and delete it from the row.
(b) The Angel deletes from the numbers remaining in the row all these which are multiples of the number you just took.
(c) Go to (a).
(d) You can never take a number which has no multiple remaining in the row; that is, your take must permit the Angel in his turn to delete at least one number.

The games stops when you can legally take no more numbers, and you want the sum S of all the numbers you have take to be as large as possible.

The picture records a game with N=9 and S=8. You could have done better. Now try with N=35. How large can you make S?

Also, today is (Spoiler Alert!) Cheryl’s Birthday!

[enigma295]

Enigma 290: Dice with a difference

From New Scientist #1438, 10th January 1985 [link]

Throwing two dice will give you a number from 2 to 12. Of course, some numbers are more likely that others. The probability of 2 for instance is 1/36; of 3 is 2/36; of 4 is 3/36; …; of 7 is 6/36; …; of 8 is 5/36; …; of 12 is 1/36.

That is true of two ordinary 6-sided dice, each bearing the letters of ENIGMA (which stand for the numbers one to six).

It is also true of this special pair of dice I have made — one with 9 sides bearing the letters IMAGINING, the other with 4 sides bearing the letters of GAGS. (S is a positive integer).

I’m not going to tell you how I constructed 9-sided and 4-sided dice. But I did, and they are fair dice. Can you interpret the MEANINGS of these fascinating facts?

[enigma290]

Enigma 288a: Multiplets

From New Scientist #1435, 20th December 1984 [link]

A multiplet is a given set of words, and the game is to connect them with a network of as few link-words as possible. In the network, any two adjacent words are related in one of three ways:

(a) one letter is changed: e.g. MINUS → MINES
(b) one letter is added/removed: e.g. PLUS → PLUMS
(c) anagram: e.g. PLUMS → LUMPS or PLUMS → SLUMP

The picture shows a network for the multiplet “MINUS, ZERO, PLUS”, with 11 link-words.

For your Christmas prize, the multiplet is the first 10 natural numbers, “ONE, TWO, …, TEN”. Can you connect them with 15 or fewer link-words? If so, how?

All link-words must be of three or more letters. They must be in the Concise Oxford Dictionary, or simple inflexions of words in it. Words there marked as abbreviations are not allowed; so “SEN” for instance is not permitted. Common words are preferable to obscure ones.

[enigma288a] [enigma288]