Enigmatic Code

Programming Enigma Puzzles

Puzzle 20: “I do not wish to be, I’d like to add”

From New Scientist #1071, 29th September 1977 [link]

I do not wish to be,
I’d like to add.
Being is a matter of degree,
I know this well,
But my decision just to be no more
Is not a fad.
It comes from an experience of life second to none.
It means that I have said, no love, no laughter,
Mechanisation’s what I want and see what follows after.
It is not mad
To want to add,
Nor is it bad.
It leads to certainty, and that is what I have a passion for,
“Why yes, perhaps it is, perhaps it is not”, no more, no more.
For now the great adventure, the hour when I cease to exist,
And just become a computer, no life but also no risk.
Perhaps you this my no-life has no future.
To show you’re wrong I offer you a test.
Letters for digits, add something to “COMPUTER”
And you will find certainty, peace and rest.

Write the sum out with numbers substituted for letters.


Enigma 504: Hooray for Hollywood

From New Scientist #1656, 18th March 1989 [link]

Twentieth Century Lion Studios has just held a week-long film festival celebrating 60 years of talking pictures. It first selected seven of the studio’s legendary stars. It then chose seven of the studio’s classic films so that each of the 21 possible pairs of stars appeared together in one of the films.

The stars were selected from Fred Astride, Humphrey Bigheart, Joan Crowbar, Bette Daybreak, Judy Garage, Clark Gatepost, Katherine Hipbone, Barbara Standup, James Student, Spencer Treacle, John Weighing. The films were chosen from the following list in which each film is given with its stars:

Cosiblanket, JC, JG, BS
Top Hit, BD, JG, KH
Stagecrouch, HB, CG, KH
A Star is Bone, HB, JC, JS
Mildred Purse, CG, ST, JW
High None, HB, JG, ST
King Koala, FA, JG, CG
Random Harpist, BD, KH, JS
Now Forager, JC, BD, CG
Mrs Minimum, CG, JS, JW
The Adventures of Robin Hoop, KH, BS, JW
The Maltese Foghorn, FA, JC, ST
Mr Deeds goes to Tune, BS, JS, ST
Meet me in St Lucy, BD, CG, BS
Gone with the Wine, FA, HB, BD
Singing in the Rind, FA, JC, KH
Mutiny on the Bunting, BD, ST, JW
The Best Years of our Lifts, FA, HB, BS
Double Identity, FA, JG, JS

Which seven films were selected?


Puzzle #07: Amveriric’s boat

From New Scientist #3233, 15th June 2019 [link]


The billionaire Mr Amveriric keeps a yacht in a private dock in the Mediterranean. It is tethered to the quay by a rope.

Last time his staff tied up the boat, they left too much slack in the rope, so the boat is now 1 metre away from the quay when the rope is taut. Hearing that a storm is on the way, Amveriric realises that the boat might get smashed against the wall by the buffeting wind, so he sends his henchman, Benolin Chestikov, to shorten the rope.

Seeing that the boat is 1 metre from the wall, Benolin decides he will pull the rope horizontally by 1 metre, and as he pulls the boat moves in horizontally.

Will the boat reach the wall or not? (And can you prove to yourself without resorting to trigonometry?)


Enigma 1006: Booklist

From New Scientist #2161, 21st November 1998

I have just been looking at the top six best-selling books listed in today’s paper. They are numbers from 1 (for the best selling book) to 6, and after each book its position in last week’s list is given.

It turns out that the same six books were in the list last week. For each of the books I multiplied the number of last week’s, and I got six different answers.

Now, if you asked my the following questions — then you would get the same answer in each case:

1. How many primes are there in that list of six products?
2. How many perfect squares are there in that list of six products?
3. How many perfect cubes are there in that list of six products?
4. How many odd numbers are there in that list of six products?
5. How many of the six books are in a higher place this week than last?

For the books 1-6 this week, what (in that order) were their positions last week?


Tantalizer 427: Pub crawl

From New Scientist #978, 4th December 1975 [link]

Peter Pickle has drawn up this handy map of the twenty pubs in his town. On crawling nights he starts with a pint at The Swan and then moves off along the lines stopping at each pub he passes. (He may visit the same pub more than once).

He follows a formula on stepping out of The Swan: P, Q, R, Q, P, Q, P, S, S, P, S, P, Q, R, Q. In the formula P, Q, R and S stand for north, east, south and west (not necessarily in that order). The final Q brings him to The Bull (the red dot on the map) for the first and only time.

Can you mark the Swan on the map?


Enigma 503: Early in the season

From New Scientist #1655, 11th March 1989 [link]

Five football teams are to play each other once. After some of the matches had been played — in fact no team had played more than two matches — some details of the situation looked like this:

Enigma 503

(Two points are given for a win and one point to each side for a draw).

Find the score in all the matches that had been played.


Puzzle #06: Darts challenge

From New Scientist #3233, 8th June 2019 [link]

Everyone knows that on a regular dartboard you can score 180 with three darts, by getting three treble twenties. However, there are scores below 180 that you can’t get with three darts.

What is the lowest score you can’t get with three darts? And for that matter, what is the lowest score that you can’t get with two darts? And with one dart? (You can of course score zero with a dart, simply by missing the board).


Enigma 1007: Triangular and Fibonacci cubes

From New Scientist #2162, 28th November 1998

Harry, Tom and I were trying to find a 3-digit triangular number, a 3-digit Fibonacci number and a 3-digit perfect cube that between them used 9 different digits. (Triangular numbers are those that fit the formula n(n+1)/2; in the Fibonacci sequence the first two terms are 1 and 1, and every succeeding term is the sum of the previous two terms).

We each found a valid solution. Mine had one number in common with Harry’s solution and a different number in common with Tom’s solution; otherwise, the numbers used in our solutions were all different.

List in ascending order the numbers used in my solution.


Puzzle 21: Cricket (4 teams)

From New Scientist #1072, 6th October 1977 [link]

A, B, C and D have all played each other once at cricket.

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

A, B, C and D got 10, 11, 17 and 14 points respectively, and you are told that only one match was won outright.

Find the result of each match.


Enigma 502: Fill this up!

From New Scientist #1654, 4th March 1989 [link]

Here is a partly filled-in Magic Square in which I have replaced digits with letters, different letters being used consistently for different letters.

In the complete Magic Square the sum of the three numbers in each row, the sum of the three numbers in each column and the sum of the three numbers in each long diagonal is the same, namely PUT.

Enigma 502

Please fill this up with numbers (put a number in every square).


Puzzle #05: Murphy’s law of socks

From New Scientist #3232, 1st June 2019 [link]

I am convinced that my washing machine eats socks. Every time I wash a load, another sock disappears. Last week I ran out of socks, so I bought myself three new pairs.

What is the chance that, after my first three washes, I will be left with three odd socks? Indeed, what is the chance that I will have even one pair intact?


Enigma 1008: Triangles and squares

From New Scientist #2163, 5th December 1998

I have a solid irregular tetrahedron. Its four triangular faces are coloured red, yellow, blue and green, with the green face having the largest perimeter.

The lengths of the six edges are all different and each equals a number of centimetres which is a perfect square. For example, one of the edges is 16 centimetres long.

What are the lengths of the three sides of the green face?


Tantalizer 428: Sisters of mercy

From New Scientist #979, 11th December 1975 [link]

Faith, Hope and Charity had “adopted” an old couple in their neighbourhood and a random one of the drops in each morning to jolly things along. Tom and Annie, the oldsters, take it in good part, especially since they started having a flutter on who the next ministering angel will be.

“Tell you what”, Tom proposed slyly one evening, “Let’s have an extra bet. Who do you bet it will be for the next two days?”
“Faith both days”, said Annie.
Tom replied, “And I bet it will be Hope, followed by Faith. £1?”
“Very well”, said Annie, “but what if we are both wrong?”
“Then the bet stands until such time as Faith arrives either for the second day running (and you win) or on the day after Hope (and I win).”
“Done”, said Annie.

What are Tom’s chances of winning?


Enigma 501: A reciprocal arrangement

From New Scientist #1653, 25th February 1989 [link]

“As you insist on disturbing my peace of mind with puzzles”, remarked Potter to Kugelbaum as they sat down to drinks at the Maths Club, “it is only fair that you submit to the same fate”. Kugelbaum agreed. “The Egyptians expressed fractions as sums of reciprocals”, continued Potter. “For example, they wrote 3/8 = 1/8 + 1/4. That and the notion of getting one over you inspires this puzzle:”

“The smallest integer, U, such that 1/U may be expressed as the sum of exactly two reciprocals in exactly and only two distinct ways is 2; for 1/2 = 1/4 + 1/4 and 1/2 = 1/3 + 1/6 and there are no other ways of doing it. The second smallest is 3, since 1/3 may be expressed as the sum of exactly two reciprocals in exactly and only two distinct ways: 1/3 = 1/6 + 1/6 and 1/3 = 1/4 + 1/12. Again there are no other ways. But the third smallest value of U is not 4, since 1/4 may be expressed as the sum of two reciprocals in three distinct ways.”

“Yes”, said Kugelbaum in reply, “and namely 1/8 + 1/8; 1/6 + 1/12; and, of course, 1/5 + 1/20. What is your question?”

Potter drew himself up: “What is the eighth smallest value of U, such that 1/U is expressible as the sum of exactly and only two reciprocals in exactly and only eight distinct ways?”

Kugelbaum’s eyes glazed over and the cogs began to whir. In fact, Potter didn’t even know if the question had an answer and so when Kugelbaum gave the answer, he had to take it on trust. Given that Kugelbaum is never wrong, what was his answer?


Puzzle #04: Which door?

From New Scientist #3231, 25th May 2019 [link]

You may have heard of the US game show “Let’s Make a Deal”, in which the star prize is hidden behind one of three doors and the contestant has to pick the lucky door.

Now there is a new game show, “Let’s Make a Bigger Deal”, hosted by Jayne Brody. There are five doors, A, B, C, D and E, and contestant Nico is allowed to choose three. If the prize is behind one of them, he wins. Nico picks doors A, B and D.

As always happens on the show, to build drama, Brody opens three doors (two of them Nico’s) that she knows don’t have the prize behind them: A, D and E. Two remain closed: Nico’s (B) and C. Brody says: “Nico, do you want to stick with B, or switch to C? You can phone a friend if you want”.

Nico likes this idea and rings his friend Leah: “Hi Leah, there are two doors left. Should I choose door B or door C?”.

Which should Leah suggest? And should Nico follow Leah’s advice?


Enigma 1009: Squared square

From New Scientist #2164, 12th December 1998

The diagram shows the simplest solution to the classical problem of dissecting a square into a number of smaller squares all with sides which are integers, no two the same. Unfortunately, the dimensions (several of which are prime numbers) have been deleted.

By studying the diagram with care can you determine the side of the outer square?


Puzzle 22: Crimes assorted

From New Scientist #1073, 13th October 1977 [link]

When Professor Knowall and I came to the Island of Imperfection we had hoped for a long quiet holiday. But it was not to be. Our arrival seemed to have triggered off a whole series of misdemeanours and our hope of sitting in the sun perhaps under a palm tree, and have a few gorgeous lady Wotta Woppas fanning our fevered brows, did not look like becoming more than a hope.

There are three tribes on the island — the Pukkas who always tell the truth, to Wotta Woppas who never tell the truth, and the Shilla-Shallas who make statements which are alternately true and false or false and true.

The police on the island, who are all good Pukkas and true, had told us that on the day before there had been three separate actions against the laws of the island; one of them was a case of dangerous drinking, one of provoked assault, and one of bobbery (there has been a lot of this on the island recently, it means robbing a bobby).

Since we knew that the police were all Pukkas there seemed to me to be a lot to be said for letting them do the investigations and then using our minds to interpret and come to conclusions on whatever they found. But the Professor would not have this.

“Duty before palm trees and gorgeous Wotta Woppas, my dear Sergeant Simple”, he said, “A detective must detect, and the sun seems to have sharpened up my wits and my enthusiasm”.

I don’t quote know how he did it, but very shortly the Professor had in front of us three inhabitants of the island who had been persuaded for perhaps forced to make statements. These statements will of course conform to their tribal customs, and fortunately we knew that there was one man from each tribe. It seemed pretty clear from the investigations that we had made, that there was no reason why the crimes could not all have been committed by the same person.

The three inhabitants of the Island, whom I shall call A, B and C, spoke as follows:

(1) C was the dangerous drinker;
(2) I am not the most truthful of the three of us;
(3) When asked whether C was a Pukka, B said “Yes”.

(1) A was the man who was guilty of provoked assault;
(2) A is a Wotta-Woppa;
(3) C was guilty of bobbery.

(1) One of us was guilty of more than one offence;
(2) B was not guilty of provoked assault;
(3) A was guilty of dangerous drinking.

It will not surprise my readers to know that the Professor did not take long to arrive at his conclusion.

What can you say about who was guilty of the three offences?


Enigma 500: Child’s play

From New Scientist #1652, 18th February 1989 [link]

The children at the village school have a number game they play. A child begins by writing a list of numbers across the page, with just one condition, that no number in the list may be bigger than the number of numbers in the list. The rest of the game involves writing a second list of numbers underneath the first; this is done in the following way. Look at the first number — that is, the left-hand one, as we always count from the left. Say it is 6, then find the sixth number in the list — counting from the left — and write that number in the first place in the second row — so it will go below the 6. Repeat for the second number in the list, and so on. In the following example, the top row was written down, and then playing the game gave the bottom row:

6,  2,  2,  7,  1,  4, 10,  8,  4,  2,  1
4,  2,  2, 10,  6,  7,  2,  8,  7,  2,  6

The girls in the school use the game to decide which boys are their sweethearts. For example, Ann chose the list of numbers:

2,  3,  1,  5,  6,  4

For a boy to become Ann’s sweetheart he has to write down a list of numbers, play the game, and end with Ann’s list on the bottom row.

Bea chose the list:

2,  3,  2,  1,  2

and Cath the list:

3,  4,  5,  6,  7,  1,  2,  5,  7,  3,  6,  9

Find all the lists, if any, which enable a boy to become the sweetheart of Ann, of Bea, and of Cath.

Enigma 1736 is also called “Child’s play”.


Puzzle #03: Cube shadow

From New Scientist #3230, 18th May 2019 [link]

At midday at her home in Ecuador, Natalia holds a solid cube 1 metre above the ground and it casts a shadow. She rotates the cube a bit and finds that the smallest shadow she can create is a square. What is the shape of the largest shadow she can produce with the cube at noon and how much bigger is it than the square shadow?


Enigma 1010: Christmas list

From New Scientist #2165, 19th December 1998

As usual, this Christmas I shall be given a diary. So I shall have to transfer information from my current diary into it. This includes the dates of 11 birthdays which I have to remember, no two of which are in the same month.

If you write each of these different dates as a number (so that, for example, New Year’s Day would become 11, and Christmas Day would become 2512) and write the 11 numbers in increasing order, then they have a surprising property. The difference between the first and the second is the same as the difference between the second and the third, which is the same as the difference between the third and the fourth, and so on through the list. Also no two adjacent numbers arise from birthdays in consecutive months. More of the 11 birthdays will fall on Saturdays in 1999 than did in 1998.

What are the dates of the Saturday birthdays in 1999?


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