Enigmatic Code

Programming Enigma Puzzles

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°).

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.

[enigma364]

Enigma 1148: Four-way fairway tie

From New Scientist #2304, 18th August 2001

After they had each played four rounds in the golf tournament Bernhard, Colin, Darren and Ernie all ended up with the same total score even though the scores for the 16 individual rounds that they played were all different. Each player’s score for each round was in the 60s or 70s (that is to say between 60 and 79 inclusive). Bernhard’s score in each of his four rounds was a prime number, Colin’s score in each of his four rounds was a semi-prime (the product of two prime numbers), Darren’s and Ernie’s scores in each of their four rounds were numbers that are neither prime nor semi-prime. Darren’s best (lowest) round was better than Ernie’s best round, and Darren’s worst round was better than Ernie’s worst round.

List in ascending order Ernie’s scores for the four rounds.

[enigma1148]

Enigma 363: Incomplete statistics

From New Scientist #1512, 12th June 1986 [link]

Six football teams — A, B, C, D, E and F — are to play each other once. After some of the matches have been played a table giving some details of the matches played, won, lost, etc., looked like this:

Enigma 363

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

Find the score in each match.

This is the 1000th puzzle posted to the site in the enigma category. However, since there are a couple of duplicate puzzles (Enigma 83 duplicates Enigma 9, and Enigma 1770 duplicates Enigma 1757) I’m holding off claiming we’ve got to 1000 Enigma Puzzles until next week (and it also means I don’t have to celebrate the 1000th puzzle by posting a football problem).

[enigma363]

Enigma 1149: Egyptian fraction

From New Scientist #2305, 25th August 2001

The Ancient Egyptians could not comprehend fractions with numerators greater than 1, such as 8/11. If required to divide 8 sacks of wheat evenly among 11 bakers, each could first be given 1/2 a sack, then from the residue each could be given 1/5 of a sack, but not as much as 1/4. With modern arithmetic we can use this logic to calculate that 8/11 = 1/2 + 1/5 + 1/37 + 1/4070.

There are several ways of representing 8/11 as the sum of reciprocals without including such unfriendly fractions as 1/4070. George’s friend Henry has discovered that 8/11 = 1/2 + 1/6 + 1/22 + 1/66, which he likes because the reciprocal denominators are all even divisors of his house number, 66. George has matched this achievement by expressing 8/11 as the sum of the reciprocals of a set of different odd divisors of his odd house number, which is less than 200.

What is George’s house number?

[enigma1149]

Enigma 362: On the face of it

From New Scientist #1511, 5th June 1986 [link]

I have a cube and on each face there is a different digit (written in modern digital style, so that, for example, a 2 would look the same either way up). My viewpoint from the front stays the same in all that follows; it is the cube which moves. By a “top twist” I mean that the front face in view moves to the top, by a “right twist” that the front face in view moves to the right, etc., each time bringing a new face into view.

I start by looking at the face in view. Bottom twist, look again, bottom twist, look again, bottom twist, look again. In all I have read off, quite correctly placed, a four-figure perfect square.

Left twist and ready to start again. Look at the face in view, left twist, look again, top twist, look again, top twist, right twist, look again. I have now read off, quite correctly placed, another four-figure perfect square.

Now for a fresh start. I look at a face, left twist, look again, left twist, look again, left twist, look again. I have read off, quite correctly placed, a four-figure number which is exactly twice a perfect square.

Top twist, top twist and ready to start again. Look at the face in view, left twist, look again, left twist, look again, left twist, look again. This time I have read off, quite correctly placed, a four-figure number which is not a perfect square.

What is that last number?

[enigma362]

Enigma 1150: Cubic meter

From New Scientist #2306, 1st September 2001

My car has a five-figure display for the milometer, showing the total miles travelled since the car was made, and a three-figure display for an odometer (which can be set to zero at the beginning of a trip) to measure the distance of a journey. When I got the car both meters were showing perfect cubes with no zeros. I also noticed that both notched up the next mile simultaneously.

I never reset the odometer, it just keeps going round and round, going back to zero when it reaches one thousand. Once when it was at zero the milometer was again showing a perfect cube. (In fact it was the second occasion since I bought the car that the odometer had been at zero, but you don’t need to know that). I have run the car for several years since then (although the milometer has never gone back to zero yet).

As I parked this morning I noticed that again both meters were displaying perfect cubes with no zeros.

What were the two meters displaying this morning?

[enigma1150]

Enigma 361: How many furlongs to the mile?

From New Scientist #1510, 29th May 1986 [link]

“How can one inch, one foot, one yard, one furlong and one mile be only 249 inches?” asked Albert, who was looking at a calculation in the Tarizania University library.

“Simple,” replied the librarian. “That sum was done before we adopted the English system of metrication.”

“How many furlongs to the mile were there?”

“Perhaps you would like to work out that prime number if I tell you that there were more furlongs to the mile than yards to the furlong, more yards to the furlong than feet to the yard and more feet to the yard than inches to the foot,” said the librarian.

What is the answer to Albert’s question?

[enigma361]

Enigma 1151: Workers’ bonus

From New Scientist #2307, 8th September 2001

Each of the workers of the Manufacturing Company joined the company on 30 June in some year before 1996. The company’s profits for each year ending 30 June are always the same, £P, where P is a whole number between 2000 and 2500. On the 30 June each year, each worker, W, receives a bonus of P × (A/B) pounds, rounded, if necessary, to the nearest pound, with 50p going up; where A = the number of years W has worked for the company, and B = the sum of the numbers of years worked for the company by all the workers. Worker Frances found that in each of the years 1996, 1997 and 1998 her figure P × (A/B) was a whole number and she received bonuses of £315, £336 and £350, respectively.

Question 1. What is P?

Frances joined the company at least one year later than every other worker. Suppose that all the workers live for 1000 years and over that 1000 years the company’s annual profits remain at £P and no worker joins or leaves the company.

Question 2. What is the first year when all the workers will receive the same bonus?

[enigma1151]

Enigma 360: For the time boing

From New Scientist #1509, 22nd May 1986 [link]

“This is my favourite clock,” said Mr Fescu, the Curator of the House of Clocks. “It has a curious mechanism which prevents it from stopping or being started except exactly on the hour. It only chimes on the hour and normally emits a number of bongs equal to the hour it is striking. But if it stops and is restarted at a later hour less than 12 hours on it emits all the chimes it missed between the stopping and starting times, not counting those of the hour at which it is restarted.”

“You mean if it stops at 9:00 and is restarted at 8:00 it emits 70 bongs?”

“Precisely,” said my guide, clutching and enormous key. “And what is more interesting is that certain numbers of bongs are special, in that they occur between one unique pair of stopping and starting times only. If you heard 70 bongs it could only signify that the clock stopped at 9:00 and was restarted at 8:00. Given the time at which the clock has stopped this time, there is only one hour at which I could rewind it to yield a magic number of bongs, and Lo!” he said, hopping from foot to foot and gesticulating at the church clock just visible in the distance, “that hour is arrived.”

So saying he moved the hands to the right time, wound the mechanism up and kicked it, whereupon it emitted an uneven number of bongs and a prime number at that.

What was the time; when had the clock stopped; how many bongs did the clock emit?

[enigma360]

Enigma 1152: Tet on the Nile

From New Scientist #2308, 15th September 2001

Pharaoah Tetrakamun was bored with the rectangular [based] pyramids at Gizah so he commanded his architect to design him a tetrahedral monument. The six edges (all different) were to measure cubes of successive integers, in cubits. The face with the largest perimeter was to form the base. Eventually the architect found the lowest possible set of six successive cubes.

In ascending order, what were the lengths of the three sides of the base?

[enigma1152]

Enigma 359: Neat odd quad

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

Enigma 359

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.

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 1153: Luconacci numbers

From New Scientist #2309, 22nd September 2001

In the Fibonacci sequence the first two terms are 1 and 1, and each subsequent term is the sum of the previous two terms; so the sequence starts 1, 1, 2, 3, 5. Less well known is the Lucas sequence, whose first two terms are 1 and 3, and each subsequent term is the sum of the previous two terms; so the sequence starts 1, 3, 4, 7, 11. In the Tribonacci sequence (so named by Mark Feinberg) the first three terms are 1, 1 and 2, as in the Fibonacci sequence, and each subsequent term is the sum of the previous three terms; so it starts 1, 1, 2, 4, 7.

Harry, Tom and I were looking to find a 2-digit Fibonacci number, a 2-digit Lucas number and a 2-digit Tribonacci number that used six different digits. We each found a different solution; our three Fibonacci numbers were all different from each other; our three Lucas numbers were all different from each other; and our three Tribonacci numbers were all different from each other. None of the numbers in my solution appeared in either Harry’s or Tom’s solution.

List in ascending order the numbers in my solution.

[enigma1153]

Enigma 358: Add up the scores

From New Scientist #1507, 8th May 1986 [link]

In the following football table and addition sum letters have been substituted for digits (from 0 to 9). The same letter stands for the same digit wherever it occurs and different letters stand the different digits.

The four teams are eventually going to play each other once – or perhaps they have already done so. The score in each match is different.

Enigma 358 - Table

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

Enigma 358 - SUM

Find the scores in the football matches and write the addition sum out with numbers substituted for letters.

[enigma358]

Enigma 1154: Funny money

From New Scientist #2310, 29th September 2001 [link]

After the Apathy Party swept into power in the General Election, George, its founder, realised one of the hazards of government: impractical proposals are liable to become law because no one has properly assessed the consequences. The latest is so bizarre that even the Apathy Party must reject it?

The Chancellor of the Exchequer thinks it would be fun to abolish the current coinage and mint just one denomination of coin and one of note. He has proposed an absurd pair of values, each a whole number of pence. George realises that although a large sum of money, such as £999.99, can be paid exactly in several different ways, there are precisely 10,000 amounts that can each be paid exactly using only one combination of the proposed notes and coins. Worse, there is a smaller, but still substantial, number of amounts that cannot be paid exactly using any combination of one or both of the denominations.

How many different amounts cannot be paid exactly?

See also Enigma 1194.

Thanks to Hugh Casement for providing a transcript of this puzzle.

[enigma1154]

Enigma 357: Present and correct

From New Scientist #1506, 1st May 1986 [link]

There were six boys in the tutorial group and they had five different teachers. Each teacher knew some of the boys by Christian name and the rest by surname. Here is how four of the teachers referred to the group:

1. Andy, Bob, Charlie, Thompson, Underwood, Vardy;

2. Andy, Bob, Smith, Thompson, Underwood, Vardy;

3. Charlie, Dave, Eddie, Smith, Thompson, Wilkinson;

4. Andy, Charlie, Dave, Smith, Thompson, Underwood.

The fifth teacher was still learning their names but already knew at least two, namely the Dave in the group and the Vardy in the group.

Name the six boys (Christian name and surname each).

[enigma357]

Enigma 1155: Ending up with numbers

From New Scientist #2311, 6th October 2001 [link]

To test my nephew’s arithmetic I get him to write down a three-figure number and then to write down the next twenty consecutive numbers as well. Then he has to “reverse” each of the numbers (so that 237 would become 732, and 540 would become 45, and so on). So, for example, if his starting number was 185 then the twenty-one numbers he would get would be: 581, 681, 781, 881, 981, 91, 191, 291, 391, 491, 591, 691, 791, 891, 991, 2, 102, 202, 302, 402 and 502.

I then ask him to cross out all those numbers which are divisible by 2, then from what’s left to cross out all those numbers divisible by 3, then from what’s left to cross out all those numbers divisible by 5, then from what’s left to cross out all those numbers divisible by 7, and finally from what’s left to cross out all those numbers divisible by 11. So, for example, if he started with 185 (as above) then he would end up with just the six numbers 881, 191, 391, 491, 691 and 991.

On one occasion recently he chose his three-figure starting number, carried out the above process and when he had finished he was left with 14 numbers. By some neat logic you can work out what the starting number was.

What was it?

[enigma1155]

Enigma 356: Counting house

From New Scientist #1505, 24th April 1986 [link]

“My fortune,” said the king to his vizier, “consists of a number of identical gold pieces. I remember their number by a set of peculiar numerical circumstances. I do not speak their number openly, as there are spies and eavesdroppers at court. Instead I shall tell you indirectly.”

“My fortune cannot be divided equally among my sons without splitting gold pieces. Nor could it be so divisible, unless the number of my sons were to run into thousands. My three youngest sons live in the palace, and of the other 15 some have perished in the late wars against the heathen.”

“Now the number of pieces of gold in my fortune has the following curious property: if one multiplies it by the number of my sons living, one obtained a 10-digit number in which all the digits from 0 to 9 appear, with one exception. If any number of digits be cut off the right of this number, the remaining digits form a number which is divisible without remainder by the number of digits cut off. You have a reputation of being a calculator, and so will know the number I mean, and thus the number of gold pieces in my counting house.”

The vizier bowed and reached for his pen case and parchment.

How many sons and how many gold pieces did the king possess?

[enigma356]

Enigma 1156: The tax process

From New Scientist #2312, 13th October 2001 [link]

The people on the island of Fairshare have their own tax process. If A is the average income for the people on the island then only people earning more than A pay tax. If a person earns P, which is more than A, then that person pays tax (P−A)²/P.

When all the tax has been collected, then it is shared between the people earning less than A, in proportion to the amounts their incomes fall short of A.

There are 10 people on the island, C, D, E, F, G, H, I, J, K and L. Their final incomes after the tax process were:

C = F£117,
D = F£112,
E = F£103-58,
F = F£90,
G = F£60-47,
H = F£53-52,
I = F£51-15,
J = F£46-57,
K = F£44-20,
L = F£41-99 (that is to say 41 Fairshare pounds and 99 pence, where there are 116 pence to the Fairshare pound).

What were the original incomes of E, H and K?

See also Enigma 1253.

[enigma1156]

Enigma 355: Diffy-dist

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

Enigma 355

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 1157: Never the same result

From New Scientist #2313, 20th October 2001 [link]

Albion, Borough, City, Rangers and United have played another tournament. This time each team played each of the other teams twice. The two matches that any two teams played against each other had different results, even in the case of City, which did not draw any of its matches.

Two points were awarded for a win and one point for a draw. After each team had played each of the other teams once, United was in the lead, one point ahead of Rangers, which was one point ahead of City, which was one point ahead of Borough, which was one point ahead of Albion.

But by the end of the tournament the positions were completely the opposite: Albion finished one point ahead of Borough, which was one point ahead of City, which was one point ahead of Rangers, which was one point ahead of United.

If you knew the results of the first match between Borough and United you could deduce with certainty the results of all the other matches played.

Give the result of the first match Borough played against each of the other four teams, naming the opponents in each match.

[enigma1157]