Enigmatic Code

Programming Enigma Puzzles

Tag Archives: by: Keith Austin

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]

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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]

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 419: Painting by numbers

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

Instructions

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.

Example

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?

[enigma419]

Enigma 415: Buses galore

From New Scientist #1565, 18th June 1987 [link]

The Service Bus Company runs buses on the route shown by the map:

Enigma 415

Each bus starts its journey at the Terminus, T, and goes towards A. At each crossroads it goes straight across. The bus eventually enters T from B and leaves for C. It finally arrives at T from D, to complete its journey. The time to go from one crossroads to the next is three minutes and so it takes one hour to complete the journey.

Buses are timetabled to leave T on the hour and at various multiples of three minutes after the hour. When a bus completes a journey it immediately begins its next journey and so the timetable repeats each hour.

This means each bus reaches a crossroads at times 00, 03, 06, 09, …, 54, 57. The buses are timetabled so that no two buses ever reach the same crossroads at the same time. Also the buses are timetabled so that the maximum number of buses are running on the route.

How many buses are running on the route?

[enigma415]

Enigma 1097: Chessboard triangles

From New Scientist #2253, 26th August 2000 [link]

Take a square sheet of paper of side 1 kilometre and divide it into small squares of side 1 centimetre. Colour the small squares so as to give a chessboard pattern of black and white squares.

When we refer to a triangle, we mean a triangle OAB, where O is the bottom left corner of the square of paper, A is on the bottom edge of the paper and B is on the left hand edge of the paper.

Whenever we draw a triangle then we can measure how much of its area is black and how much is white. The score of our triangle is the difference between the black and white areas, in square centimetres. For example if OA = 3 cm and OB = 2 cm then we find the score of the triangle is 1/6 cm².

Question 1. What is the score of the triangle with OA = 87,654 cm and OB = 45,678 cm?

Question 2. What is the score of the triangle with OA = 97,531 cm and OB = 13,579 cm?

Question 3. Is it possible to draw a triangle on the paper with a score greater than 16,666 cm²?

[enigma1097]

Enigma 411: The third woman

From New Scientist #1561, 21st May 1987 [link]

The Ruritanian Secret Service has nine women agents in Britain: Anne, Barbara, Cath, Diana, Elizabeth, Felicity, Gemma, Helen and Irene. Any two of the women may or may not be in contact with each other.

To preserve security contacts are limited by the following rule: for any two of the women there is a unique third woman who is in contact with both of the women. The British Secret Service has so far discovered following pairs of women that are in contact: Anne and Cath, Anne and Diana, Cath and Barbara, Barbara and Gemma, Elizabeth and Felicity.

Which of the women are in contact with Helen? Who is the woman in contact with both Anne and Irene?

[enigma411]

Enigma 1101: Disappearing numbers

From New Scientist #2257, 23rd September 2000 [link]

This game starts when I give a row of numbers; some numbers in italic [red] and some in bold [green]. Your task is to make a series of changes to the row, with the aim of reducing it to a single number or to nothing at all. In each change you make, you select two numbers that are adjacent in the row and are of different font [colour], that is to say one is italic [red] and the other is bold [green]. If the numbers are equal, you delete them both from the row; otherwise you replace them by their difference in the font [colour] of the larger number.

For example, suppose I gave you the row:

3, 4, 3, 2, 5, 2.

One possibility is for you to go:

[I have indicated the pair of numbers that are selected at each stage by placing them in braces, the combined value (if any) is given on the line below in square brackets].

3, 4, 3, {2, 5}, 2
3, {4, 3}, [3], 2
{3, [1]}, 3, 2
[2], {3, 2}
2, [1]

You have come to a halt and failed in your task.

On the other hand you could go:

{3, 4}, 3, 2, 5, 2
[1], {3, 2}, 5, 2
{1, [1]}, 5, 2
{5, 2}
[3]

And you have succeeded in your task.

For which of the following can you succeed in your task?

Row A: 9, 4, 1, 4, 1, 7, 1, 3, 5, 4, 2, 6, 1, 4, 8, 3, 2.

Row B: 2, 3, 5, 9, 6, 3, 1, 4, 2, 3, 1.

Row C: 1, 2, 3, 4, 5, 6, …, 997, 998, 999, 1000, 3, 5, 7, 9, 11, …, 993, 995, 997, 999.

Row D: 3, 2, 1, 4, 5, 4, 3, 2, 4, 3, 7, 4, 1, 5, 1, 4, 2, 4, 3, 1, 2, 7, 9, 3, 7, 5, 3, 8, 6, 5, 8, 4, 1, 5, 2, 3, 1, 4, 10, 6, 3, 5, 7, 4, 1, 4.

I have coloured the numbers in italics red, and those in bold green in an attempt to ensure the different styles of numbers can be differentiated.

When the problem was originally published there was a problem with the typesetting and the following correction was published with Enigma 1104:

Due to a typographical error, three of the numbers in Enigma 1101 “Disappearing Numbers” appear in the wrong font. In each case, the following should have been printed in heavy bold type:

the second number 3 in the initial example;
the first number 5 in row B;
and the first number 4 in the second line of row D.

I have made the corrections to the puzzle text above.

[enigma1101]

Enigma 1105: Road ants

From New Scientist #2261, 21st October 2000 [link]

Take a large sheet of paper and a black pen and draw a rectangle ABCD with AB = 10 metres and BC = 2 metres. Now draw lines to divide your rectangle into small squares, each of side 1 centimetre. Place your diagram so that A is due north of D and B is east of A. In each small square draw the diagonal that goes from northwest to southeast. Let P and Q be the mid-points of AD and BC, respectively. Then there is a black line PQ; remove it and replace it by a red line.

Amber is a small ant who can walk along the black lines in your diagram. North of PQ she covers a centimetre in 1 minute, but south of PQ she can cover a centimetre in 30 seconds. She is to walk from C to A and she chooses the quickest route.

1. How long does Amber take on her journey? Give the time, to the nearest second, in hours, minutes and seconds.

Ben is another ant who walks along the black lines. North of PQ he goes at the same speed as Amber, but not south of PQ. The fastest time for Ben to get from C to A is 24 hours.

2. South of PQ, how long does Ben take to cover a centimetre? Give the time, to the nearest second, in minutes and seconds.

[enigma1105]

Enigma 406: The ritual

From New Scientist #1556, 16th April 1987 [link]

I entered the jungle clearing and found, at the centre, six stones numbered 1, 2, 3, 4, 5, 6. My native guide demonstrated the ancient ritual which was enacted on that site.

First I had to arrange the stones in any order I wished, so I put them as 421365. I then noted the number on the first stone — all counting is from the left — is was 4, and so I picked up the fourth stone and placed it at the right hand end. I obtained 421653. Next, I looked at the second stone, a 2, and moved the second stone to the end, to give 416532. I repeated the procedure for the third stone to give 416532, and then for the fourth, fifth and sixth stones, to give successively, 416523. 465231, 652314. The final arrangement was called the result of the ritual.

Just then the high priestess, Sarannah, entered. She arranged the stones in a certain order, carried out the ritual and obtained the result 314625. My guide explained that the arrangement Sarannah started with was the result of applying the ritual to a very special arrangement, which she could not tell me.

What was the arrangement that Sarannah started with?

[enigma406]

Enigma 402: A DIY puzzle

From New Scientist #1552, 19th March 1987 [link]

In the puzzle below, select some of the guest lists and discard the remainder. The lists you keep should make a puzzle which has exactly one solution and involves no more lists than is necessary.

Who did it?

by …

There has been a series of robberies at house parties recently. Each was clearly a one-man job — the same man each time — and each was an inside job. The possible suspects are Alan, Bryan, Chris, David, Eric, Fred, George, Harry, Ian, Jack, Ken and Len. The male guest list at the parties were as follows:

1. All but David, George and Len.
2. Bryan, Chris, Eric, Harry, Ian and Ken.
3. All but Bryan and Ken.
4. Chris and Ian.
5. All but Alan, Fred, Jack and Len.
6. Bryan, Chris, Ian and Ken.
7. All but Eric and Harry.
8. All but David and George.
9. All but Chris and Ian.
10. All but Alan and Fred.

Who carried out the robberies?

Which lists should you use in your puzzle?

What is the answer to your puzzle?

[enigma402]

Enigma 1110: Dots and lines

From New Scientist #2266, 25th November 2000 [link]

Matthew and Ben are playing a game. The board is a 1-kilometre square divided into 1-centimetre squares. The centre of each small square is marked by a red dot.

Matthew begins the game by choosing a number. Ben then selects that number of red dots. Finally Matthew chooses two of Ben’s selected dots and draws a straight line from one to the other. Matthew wins if his line passes through a red dot other than those at its ends; otherwise Ben wins.

What is the smallest number that Matthew can choose to be certain of winning?

In the magazine this puzzle was incorrectly labelled Enigma 1104.

[enigma1110]

Enigma 398: Down on the farm

From New Scientist #1548, 19th February 1987 [link]

Farmer O. R. Midear has crossed a turnip with a mangel to get a tungel, and he now has many fields of tungels. As a tungel expert, he looks at a tungel to see if it is red or not, if it is smooth or not, and if it is firm or not. He knows that if a tungel is red then it is smooth.

Regulations have just been introduced which put fields of tungels into classes A, B, C, D, E; a field may be in more than one class.

Class A: All fields containing no red tungel.
Class B: All fields containing no smooth tungel.
Class C: All fields in which every red tungel is firm.
Class D: All fields in which every smooth tungel is firm.
Class E: All fields containing a red tungel which is not firm.

To test Farmer Midear’s understanding of the regulations he was asked to say which of the following statements are true.

1. If a field is in A then it is in B.
2. If a field is in B then it is in A.
3. If a field is in C then it is in B.
4. If a field is in B then it is in C.
5. If a field is in C then it is in D.
6. If a field is in D then it is in C.
7. If a field is in D then it is not in E.
8. If a field is not in E then it is in C.
9. In every field there is a tungel such that if it is not red then the field is in A.

What should Farmer Midear’s answer be?

[enigma398]

Enigma 1114: Christmas changes

From New Scientist #2270, 23rd December 2000 [link]

Christmas is said to change things and so this enigma is an old puzzle with some changes.

Problem: You have to assign a digit to each of the 10 letters in the sum here:

When you have decided on an assignment of digits to the 10 letters, then your assignment is a solution of the problem if it satisfies at least one of the following conditions:

• At least one digit is assigned to more than one letter.
• When the digits are put into the addition sum it is not correct.
• The digit assigned to “U” is smaller than the digit assigned to “H”.

You need to find an assignment of digits to the 10 letters which is NOT a solution of the problem.

What is the value of BLEAT in your assignment?

[enigma1114]

Enigma 393: Decode the sum

From New Scientist #1543, 15th January 1987 [link]

In the following addition sum, different letters stand for different digits and the same letter stands for the same digit throughout.

Decode the sum.

[enigma393]

Enigma 1121: Families

From New Scientist #2277, 10th February 2001 [link]

There are six families, each consisting of mother, father and child. The mothers are Amber, Barbara, Christine, Dorothy, Ellen and Frances; the fathers are George, Harry, Inderjit, James, Kenneth and Lewis; the children are Matthew, Naomi, Oliver, Peter, Quentin and Rachel. The other day, everyone kept a diary of who they met in the 24-hour period; of course, everyone met the other two members of their family, but they also met other people. These are the diary records, given by initials:

A met G, J, L, M, N, P;
B met H, J, K, O, P, R;
C met I, K, L, M, O, Q;
D met G, I, L, P, Q, R;
E met H, I, K, M, N, R;
F met G, H, J, N, O, Q.

Also:

G met M, N, P;
H met O, P, R;
I met M, O, Q;
J met N, O, Q;
K met M, N, R;
L met P, Q, R.

If I told you who the wife of Inderjit is, then you could not work out who the father of Oliver is.

Question 1: Who is the wife of Inderjit?

If I told you who the mothers of Oliver and Quentin are, then you could work out who the mother of Peter is.

Question 2: Who are the mothers of Oliver, Quentin and Peter?

[enigma1121]

Enigma 390: Which statements are false?

From New Scientist #1539, 18th December 1986 [link]

Each of the following six statements is true or false or we cannot say whether it is true or false.

(1) Either 2 or 3 is the first true statement in the list of six.
(2) We can say both 4 and 5 are true.
(3) 6 is false and/or 4 is true.
(4) 1 is true and/or 6 is true.
(5) 3 is false and/or 1 is true.
(6) Both 2 and 5 are true.

Which of the six statements are false?

[enigma390]

Enigma 386: Triangle of stones

From New Scientist #1535, 20th November 1986 [link]

I emerged from the impenetrable jungle into a clearing, at the centre of which were 21 stones laid on the ground to form a triangle.

Enigma 386

Just then, three native girls approached the stones; each wore a coloured sarong, one red, one blue, one white. They painted the six stones in the bottom row, white, white, blue, blue, red, red, in that order from left to right, and placed a coconut on the single stone in the top row. My guide explained that this was a traditional game. The girls would go and turn in the order red, white, blue, red, white. At each turn the girl would move the coconut to a stone which touched the coconut’s present stone and which was on a lower row. The game ended when the coconut reached the bottom row, and the colour of its final stone indicate the winner.

My guide knew the girls and said that if red could not win she would try to help white to win, similarly white would help blue and blue would help red: all the girls knew this.

After the game they exchanged the colours on two of the stones and played again. Blue won the second game.

Who won the first game and what was the row of colours for the second game?

[enigma386]

Enigma 1127: Lights out

From New Scientist #2283, 24th March 2001 [link]

There are 13 lights, A, B, C, …, L, M, in the dormitory and each one has its own switch. To save matron having to operate all 13 switches, 34 pairs of lights are connected. They are:

AB, AC, AD, AJ, BC, BE, BI, CE, CF, CG, CH, CI, CK, CM, DE, DF, DG, DI, DJ, DK, DL, DM, EF, EG, EH, EI, EJ, EL, FJ, FM, GH, GM, KL, LM.

Whenever a switch is operated it changes its own light and each of the lights connected to it from on to off, or from off to on. When matron enters the dormitory at 9.00 pm all the lights are on. As an example, if she operates switch A then lights A, B, C, D and J go off. If she then operates switch J then lights A, D and J come back on and lights E and F go off.

Question: Which switches should matron operate so that all the lights are off when she has finished.

See also Enigma 1137.

[enigma1127]

Enigma 381: Island airlines

From New Scientist #1530, 16th October 1986 [link]

Come with us now to those 10 Pacific islands with the quaint native names A, B, C, D, E, F, G, H, I, J. These are served by Lamour and Sarong airlines. Each morning one plane from each airline leaves each island bound for another of the islands. No two planes from the same airline are going to the same island and the two planes leaving any island go to different destinations.

The planes all carry out the return journeys overnight and everything is repeated the next day and so on.

Each island has a beautiful queen. One morning, each queen left her island on the Lamour plane, staying overnight on the island she reached, and left the next morning on the Sarong plane. At the end of their two-day journey the queens of A, B, C, D, E, F, G, H, I, J found themselves on the islands of C, F, B, H, J, D, E, I, G, A, respectively.

Some time later, the queens made similar journeys, again starting from their home islands, but this time taking the Sarong planes on the first day and the Lamour planes on the second day. This time the Queens of A, B, C, D, E, F, G, H, I, J ended up on the islands of J, A, B, I, F, H, E, C, G, D respectively.

As the sun slowly sits in the west we say a fond farewell to a £10 book token, which will be sent to the first person to tell as the details of the 10 Lamour routes.

[enigma381]