Enigmatic Code

Programming Enigma Puzzles

Enigma 401: Uncle bungles the answer

From New Scientist #1551, 12th March 1987 [link]

It is true, of course, that there are rather a lot of letters in this puzzle, but despite that I though that for once Uncle Bungle was going to write it out correctly. In fact there was no mistake until the answer but in that, I’m afraid, one of the letters was incorrect.

This is another addition sum with letters substituted for digits. Each letter stands for the same digit whenever it appears, and different letters stand for different digits. Or at least they should, and they do, but for the mistake in the last line across.

Which letter is wrong?

Write out the correct addition sum.

As it stands the puzzle has no solution. New Scientist published the following correction with Enigma 404:

Correction to Enigma 401, “Uncle bungles the answer”. Unfortunately, as a result of a printer’s error, New Scientist managed to bungle the question. We will publish the correct question, in full, in our issue of 9 April, as Enigma 405. In the meantime, our apologies to those who were thwarted by the mistake.

[enigma401]

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 321: Going to pieces

From New Scientist #1469, 15th August 1985 [link]

I have a puzzle consisting of eight jigsaw type pieces. Each sturdy piece has been cut from a piece of card 2 inches by 3 inches, by removing one or more of the six 1 inch by 1 inch squares into which it can be divided. For example, one of the pieces is as shown:

The card is the same on both sides and the pieces can be used either way up. For three-quarters of the pieces (like the one shown) this is no advantage, but the other pieces are different when turned over. No two pieces of the puzzle are the same.

The pieces can all be put together to form a large rectangle and, although this can be done by several different arrangements of the pieces, you can only get a large rectangle of one particular size.

What is the size of the large rectangular jigsaw which we can make?

[enigma321]

Enigma 267: Just one at a time

From New Scientist #1414, 26th July 1984 [link]

I have in mind a five-figure number. It satisfies just one of the statements in each of the triples below.

The sum of its digits is not a multiple of 6.
It is divisible by a number whose units digit is 3.
Its middle digit is odd.

The sum of its digits is odd.
It has a factor which is not palindromic.
It is not divisible by 1001.

It has two or more different prime factors.
It is not a perfect square.
It is not divisible by 5.

What is the number?

Due to industrial action New Scientist was not published for 5 weeks between 19th June 1984 and 19th July 1984.

This brings the total number of Enigma puzzles available on the site to 804, just over 45% of all Enigma puzzles published.

[enigma267]

Enigma 251: Could deece be love?

From New Scientist #1398, 23rd February 1984 [link]

“… She loves me … she loves me not … she loves me …”

The voice, Alice decided, was coming from behind the privet hedge. Peering over she saw a squat figure rolling a cube. He stopped and appeared to concentrate deeply as he counted on his fingers. “She loves me not!” he declared suddenly, and he picked up the cube to throw again. Alice coughed politely and the figure looked up. It was the red knight.

“Bless me!” he exclaimed.

“Oh, I do hope I didn’t disturb you, it’s just I did wonder …”

“What I was doing? Well I’m playing a game with this deece, of course!”

“Deece?” enquired Alice.

“Gracious, surely you know what a deece is. Like dice only it has two ‘sixes’ and no ‘one’.”

“But that’s cheating.”

“Not at all, it makes things much more exciting. Now where was I? Oh yes, I keep rolling the deece until the product of all the throws contains the number of the last throw. So if my score was 120 and I threw a three then the matter would be resolved.”

“The matter?”

“Courtship, my girl. Whether she loves me or not.”

Alice felt it wrong to ask who “she” was.

“Why, one thousand five hundred and … confound it, you’ve made me forget the rest. Let’s see, this last throw was … yes … and the one before it was a three …”

Alice sensing some arithmetic in the air, decided that it was time to leave.

What, meanwhile, were the throws that the knight had made so far, in the correct order?

[enigma251]

Enigma 240: The missing *

From New Scientist #1386, 1st December 1983 [link]

In an * to economise * the wording * my Enigmas * am this * experimenting with * new system * writing. In * system I * every third * with a *. It should * be possible * work out * is being * while saving * 33 per cent of * words in * process. This * puzzle is * five men, * names are (* conveniently) Arnold, *, Cedric, Derek * Eric. All * play for * local football *. One of * is goalkeeper * the other * play at * forward, outside *, outside right * centre half. * a quite * coincidence, the * also live * a row * five terraced * next to * other. Cedric * next door * the goalkeeper. * centre forward’s * are Derek * the outside *. The centre * lives at * end of * row, next * a player * has scored * more goal * Basil (who * not live * to the {*} right). Eric * at outside *. And finally, * lives fewer * away from * than Derek * from the *.

What are * positions of * five men? (* will appreciate * in cases * the word * in doubt, * most appropriate * be used, * known!)

Note: There appears to be a misprint in the original puzzle statement. I have inserted the * in braces into this puzzle so that every third word is replaced by a *. In this form the puzzle has a unique solution, without it I could not find a satisfactory solution.

[enigma240]

Enigma 218: Relatively speaking

From New Scientist #1364, 30th June 1983 [link]

The professor during his lecture on relativity asked: “If I am in a spacecraft travelling at half the speed of light and pass another craft travelling in the opposite direction at a quarter of the speed of light, what is our relative velocity?”

“Three quarters of the speed of light,” replied one student.

“You weren’t paying attention at my last lecture,” said the professor. “We proved that, according to the special theory of relativity, when two velocities are to be added then the result is not their sum but this,” he broke off to write $(v_1 + v_2) / (1 + v_1 v_2 / c^2)$ on the board then continued, “where c is the velocity of light — 300 000 kilometres per second.

“Is it possible for two equal rational velocities to be added so that the result is an integral number of 1000 kilometres (we shall say megametres) per second?”

“No professor,” answered a bright student. “But if the speed of light is decreased by an integral number of megametres per second then it is possible.”

“But you can’t reduce the speed of light! — It is constant,” protested the professor.

“But we can imagine it to be less,” persisted the student.

The professor then suggested the amount his student had taken as the velocity of light.

“I took it as more than that, professor.”

“In that case I calculate what you took as the speed of light and all the possible sums of the equal velocities.”

Can you?

Note: I am waiting for a phone line to be connected at my new house, so I only have sporadic access to the internet at the moment.

[enigma218]

Enigma 194: The plot thickens

From New Scientist #1340, 13th January 1983 [link]

“So what did you think of that?” smiled George, pulling on his coat as the titles drifted up the screen.

“I don’t know,” said Edith, still sitting with a dazed expression. “These modern films do get so involved.”

“How do you mean?”

“Well, for example, I still don’t know why they had to strangle that world famous neurosurgeon while he was in the shower.”

“That was because of the conversation he had with Boris and the heroine’s solicitor when they were flying in that damaged jet.”

“Boris was the blond chap who later watched in horror as that psychopathic psychiatrist fellow took the fatal plunge from the multi-story inferno …”

“No, no, that was Charlie.”

“So who was the bald one who was running away from that UFO thingy?”

“Oh, he was the one who was savaged by the giant shark after falling out of the boat.”

“And that was Derek, wasn’t it?”

“Wrong again, love. Derek was the man from the CIA who was peering through the eyes of that Rembrandt when somebody stabbed the chap who was mistaken for the solicitor.”

“That means,” said Edith thinking hard, “that the one who lived happily ever after must have been Alvin, the drug-smuggling baseball player.”

“My love, you’re getting very confused. Remember, Alvin snuffed it shortly after the visit from Charlie’s widow.” There was a brief silence. “So … how did Eric fit into all this?”

Since Eric was one of only five men being described in the conversation, I’m sure you can answer that, and also work out the sequence in which the characters met their fates!

[enigma194]

Enigma 192: Merry Christmas

From New Scientist #1337, 23rd December 1982 [link]

Mr Pickwick and his friends, Mr Snodgrass, Mr Tupman and Mr Winkle spent last Christmas together. “No children this year, alas,” observed Mr Pickwick on Christmas morning, “I am very fond of children.” But just then there was a knock on the front door. Opening it, Mr Pickwick beheld more than half a dozen children, who thereupon sang God Rest Ye Merry Gentlemen. “Bless my soul!” he beamed and, fetching a tin of striped humbugs from the mantelpiece, shared them out equally and exactly among the children. The tin had once held a gross of humbugs but Mr Pickwick had already eaten some. Yet there were still enough (more than a hundred) to ensure that each child would receive more than half a dozen. In fact, if you knew how many Mr Pickwick had eaten himself, you could work out exactly how many each child got.

With the carollers gone, it was time for presents. As usual each person gave, and each received, one scarf, one pair of gloves and one bottle of port. Each gave a present to each. Mr Pickwick gave gloves to the person who gave Mr Snodgrass a scarf; and Mr Winkle gave port to the person who gave Mr Tupman gloves.

The dinner was a true feast — a sizzling goose, which weighed 8lb plus half its own weight, pursued by a pudding decked with holly and enriched with as many silver coins as you could place bishops on a chess board without any attacking any square occupied by another. Afterwards came cigars. “I would have you know”, remarked Mr Pickwick, puffing contentedly, “that if cigar-smokers always told the truth and others never did, then Mr Snodgrass would say that Mr Winkle would deny being a cigar smoker. Furthermore Mr Tupman would say that Mr Winkle would deny that Mr Snodgrass smokes cigars”. After these and other pleasant exchanges the quartet retired a trifle unsteadily to bed.

Thus were Mr Pickwick’s Yuletide jollifications exceedingly merry. He wishes similar Christmastide celebrations for revellers everywhere.

A few questions remain.

(1) How many humbugs had Mr Pickwick eaten himself?
(2) Who gave a bottle of port to whom? (four answers)
(3) What did the goose weigh?
(4) How many coins were hidden in the pudding?
(5) Was Mr Winkle a cigar-smoker?
(6) What are the four words of a, b, c, d letters in the sentences in italics [or bold] such that:
a³ – a = 20b and b² = 2bc and 2b² = c²d.

This puzzle completes the archive of Enigmas from 1982.

There are now 650 puzzles on the site, with a full archive from the start of Enigma in 1979 to the end of 1982 (192 puzzles) and also puzzles from January 2005 up to the end of Enigma in December 2013 (457 puzzles) — which is just over 36.5% of all Enigma puzzles.

[enigma192]

Enigma 1324: Rhombic squares

From New Scientist #2483, 22nd January 2005

Imagine you have two identical isosceles right-angled triangles. Lay one down so that one short side is horizontal and one is vertical. Lay the second one down so that a short side of the second is against a short side of the first, the two hypotenuses being parallel, so forming a rhombus, one example being as shown.

You are asked to place a digit at each of nine points, the four corners of the rhombus, the midpoint of each of the four sides, and the midpoint of the shorter diagonal such that eight numbers read horizontally and vertically from the top are all different perfect squares.

Alison and Bertha have each found a different solution. One of Alison’s two-digit squares is 81.

Which two did Bertha find?

[enigma1324]

Enigma 183: The heart has its reasons

From New Scientist #1328, 21st October 1982 [link]

Romeo and Juliet sometimes play a little game of deduction. They extract all the hearts from a pack of cards, discard the ace, shuffle the other 12 and take one each. They look at their own but do not show the other.

Anyone who can prove that his/her own card is the higher must say so at once and scores five points. Anyone who can name the other’s card must do so at once and scores 10 points. The maximum possible score is thus 15. As soon as one player performs either or both of these feats, the other has one further chance to score and the game stops. Neither may lie or suppress information and each knows the other to be a perfect logician. (Any actual proving is done after the game).

Being sporting, they often say a bit more than the mere grunt which the rules require for a player who is in no position to score. For instance, Romeo opened a recent game by remarking:

“I do not know which of us has the higher card.”

“Nor do I,” said Juliet.

“Nor do I,” said Romeo.

“Nor do I,” said Juliet.

“Nor do I,” said Romeo.

At this point Juliet scored and Romeo went on to win the game.

Which card did Romeo hold?

I think this puzzle is flawed, in that there is not a unique answer for the card that Romeo holds.

[enigma183]

Enigma 1336: Rectangles

From New Scientist #2495, 16th April 2005

George has a rectangular piece of paper, 3 inches by 4 inches, marked with a 1-inch grid. He is wondering in how many ways he can mark a rectangle (which may be a square) on it, following the grid lines.

He has identified eight [*] different possible sizes and shapes, and various different places in which each can be marked, ranging from 1 to 17 different positions per shape. The total is 60.

He now has a larger rectangular piece of paper of integer dimensions (more than 100 square inches) and he has tackled the same problem. Instead of 60, he has calculated a much larger number which is the product of four consecutive primes.

What are the dimensions of this piece of paper?

[*] I think there is a mistake in this puzzle, in that it should read “He has identified nine different possible sizes and shapes…”. It all seems to make sense if you make that change.

[enigma1336]

Enigma 161: Two times table

From New Scientist #1306, 20th May 1982 [link]

Lunchtime at Bramfield School is currently being taken up with a five-a-side soccer competition, in which five teams are playing in a league and will eventually play each other just once. Young James, who has shown himself to be more capable of tackling sums than footballers, is keeping a record of the results from the touchline, and has the table below as his latest check:

It is surprising, even with James’s keen eye for numbers, that he has spotted that the sum of all the numbers in this table is twice what it was earlier in the competition (when the league had no clear leader[*]).

What had been the results at that stage?

[*] This phrase is intended to imply two or more teams had the same number of points at the top.

I don’t think there is a unique solution to this puzzle (as the problem statement implies).

[enigma161]

Enigma 147: Think of a number

From New Scientist #1292, 11th February 1982 [link]

“Think of a two-digit number.”

“Right.”

“Call it AB for the moment. Now think of a one-digit number.”

“Right.”

“Call it C. Does ABC make a two-digit number?”

“Yes, it does.”

“Call that one DE. Now write down the five-digit number ABCDE and divide it by 581. What do you get?”

“Well, AB was 45. C was 6 and so DE was 51. So my answer is 78.573149 or thereabouts.”

“Bad luck! If it had come out as a whole number, I would have paid for the drinks.”

What whole number would have done the trick?

This puzzle does not have a unique solution.

Enigma 52 and Enigma 262 are also called “Think of a number”.

[enigma147]

Enigma 140: Muddy fields

From New Scientist #1284, 17th December 1981 [link]

I am at X, mounted on my velocipede, and thirsty: so I want to reach the river, which runs straight along BYC, as soon as I can. I have to keep within three fields on the plan. In each of them I can maintain a steady speed, at 30 mph within XBYZ (including the boundary), and at rates which I know (but you don’t yet) in AXZD and ZYCD (including their boundaries).

If I tell you that there are precisely three different quickest routes from X to the river, that should enable you to work out:

(a) my speed in AXZD,
(b) my speed in ZYCD.

Note: In New Scientist #1294 (with Enigma 149) an apology was published stating that this puzzle does not actually have a solution.

[enigma140]

Enigma 133: Are you positive?

From New Scientist #1277, 29th October 1981 [link]

On a 3-by-3 array like the one shown, I have coloured some of the little squares black and left the rest white. The four 2-by-2 corners of my pattern look like:

These are not necessarily the right way up and some, unfortunately, may be the negatives of what they should be. Sorry! But if I told you the colour of the middle square in the top row of my pattern and I told you the number of black squares in my pattern, then you’d be able to work out my pattern (the right way up too!).

Find my pattern, or the negative of it, whichever takes less shading.

Note: As stated I don’t think this problem has any solutions, but it is possible to add reasonable extra conditions to the problem in order to give a unique solution.

[enigma133]

Enigma 99: Harem scarem

From New Scientist #1243, 5th March 1981 [link]

The six wives of Sheikh Inbed the Terrible are in something of a tizz. They have just heard that they are to be reshuffled in their present ranking. At least two will rise, at least two fall; and no-one will have her same place after the sheikh-up.

Each has issued a prediction. “Bathsheba and I will both be demoted,” predicts Armenia. “Euterpe will be placed above both Delilah and Fatima,” Bathsheba opines. Casbah, the present No. 3 wife, expects that Delilah will outrank Fatima. Delilah predicts a rise of exactly one place for herself. Euterpe’s guess is that she herself will go down and that Casbah will outrank Armenia. “I shall fall and Casbah will rise,” predicts Fatima.

Sheikh Inbed would have you know that those about to rise are right and those about to fall wrong.

Can you list the ladies in order of rank (No. 1 first) both before and after?

[enigma99]

Enigma 98: Oh brother!

From New Scientist #1242, 26th February 1981 [link]

Of the three brothers Alpha, Beta and Gamma one always lies, one always tells the truth, and the third one is unreliable (sometimes telling the truth and sometimes not). Yesterday the thinnest said that his name was Alpha, and on hearing that Beta said that the younger of his two brothers would definitely agree to the correctness of that statement.

The youngest said that the eldest was the fattest, at which point Alpha accused him of being a consistent liar. The fattest and the elder of his two brothers chuckled on hearing Alpha say this, but they agreed that Gamma was the youngest.

Describe Gamma (e.g. unreliable, medium build, youngest).

Note: As stated I don’t think this puzzle has a unique description for Gamma (although it does for Beta).

[enigma98]

Enigma 97: Cross keys

From New Scientist #1241, 19th February 1981 [link]

Knowing guests at the Bramfield Astoria Hotel always book rooms with odd numbers since while these are all situated on the eastern side of the building overlooking the beach, the even numbered rooms overlook the abattoir (beyond the cemetery). With this in mind, Derek Smith decided to book nine “consecutive” odd-numbered rooms at the Astoria for the annual holiday of himself and eight friends.

On arrival the friends chose a room each, deposited their luggage and then went out on the town. When they returned, however, they were all somewhat inebriated and having collected their batch of keys they could not remember whose was which. However, Alf, Joe and Fred knew that they were each next door to two “prime-numbered” rooms, while Bill knew that his number was four larger than Joe’s. Charlie knew he was twice as far from Fred as he was from Harry, while Eddie was next door but one to Graham.

The keys were distributed to fit these facts, but alas nobody ended up in the room with his luggage (and Alf was landed with a larger room number while Graham went to a room with a two digit number).

In which room did Derek end up?

Note: As stated there appear to be multiple different solutions to this problem.

[enigma97]

Enigma 53: Calculating birthdays

From New Scientist #1196, 28th February 1980 [link]

“When where you born?” Jamie asked Camilla.

“I’ll tell you what,” she said, “I’ll write the date on your calculator as a five-figure number, made up of the date, month and last two digits of the year — for example, if it was St Valentine’s Day, 1963, I would put 14263. There we are. You see that I was born in the 1950s. How many days are there in my birthday’s month.”

“Thirty” replied Jamie confidently.

“That’s right. Now, do you see that if I were to divide the number formed by the last two digits by an odd, prime factor of the whole number the result would be a different factor of the whole number? Now I’ll multiply the whole number by that odd prime factor.”

“Why, that makes the six-figure date of my fourth birthday, and its year is 20 more than your birth-year!” exclaimed Jamie.

What was Jamie’s date of birth?

Note: As posed this puzzle seems to have no solutions. But by relaxing the condition on Camilla’s birth month having 30 days I was able to find a unique solution (which is the same as the published solution).

[enigma53]