Enigmatic Code

Programming Enigma Puzzles

Category Archives: enigma

Enigma 1101: Disappearing numbers

From New Scientist #2257, 23rd September 2000

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}

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.


Enigma 410: Most right

From New Scientist #1560, 14th May 1987 [link]

The addition sums which Uncle Bungle has been making up recently, with letters substituted for digits, have been getting longer and more complicated. And no one will be surprised to hear that in the latest one everything is not as it should be. In fact one of the letters is wrong.

Here it is:

What can you say about the letter which is wrong? What should it be? Find the correct sum.


Enigma 1102: The Apathy Party

From New Scientist #2258, 30th September 2000

George called a meeting to inaugurate the National Apathy Party, open to anyone who has never voted in a General Election. He hopes to be the next Prime Minister. The turnout was phenomenal, but he managed to seat them all in Wembley Stadium (capacity 80,000). George proposed that the President and the Committee should be chosen by chance, rather than by ballot. The delegates had been allocated sequential membership numbers of arrival — George, of course, being No. 1. He proposed that one number be chosen at random by the computer — that member would be the President. All members whose numbers divide exactly into the President’s number would be on the Committee. Apathy reigned — this totally undemocratic procedure was agreed.

The computer produced an odd membership number for the President and the number of committee members, including George and the President, was an odd square greater than 10.

What was the President’s membership number?


Enigma 409: Hands and feet

From New Scientist #1559, 7th May 1987 [link]

There are six footpaths through our extensive local woods, one linking each pair of four large oaks. I decided to go on a long walk starting at one of the oaks, keeping to the footpaths, ending back where I started, covering each of the footpaths exactly twice, and never turning around part-way along a path.

Whenever I was at an oak my watch showed an exact number of minutes, and in the previous 30 seconds up to and including arriving at the oak or in the 30 seconds after leaving the oak the hour and minute hands of the watch were coincident.

I set out sometime after 6am and I was back home before midnight on the same day. I walked at a steady pace from start to finish.

What time was I at the oak at the start of my round walk, and what time did I get back there at the end of the day?


Enigma 1103: Brush strokes

From New Scientist #2259, 7th October 2000

Our sign painter has an odd way of calculating his charges. For each continuous brush-stroke (which can be any shape but must not go over the same ground twice) he charges £1. He paints capital letters in a simple style and does not use two strokes where one would do. So, for example his U, E, G and H would cost £1, £2, £2 and £3 respectively.

My house number is a three-figure prime and I have asked the sign painter to spell out the three different digits (so that, for example, 103 would be ONE NOUGHT THREE and would cost £24). For my house number the cost in pounds equals the sum of the three digits and is also a prime.

What is my [house] number?


Enigma 408: Royal numbers

From New Scientist #1558, 30th April 1987 [link]

In the following subtraction sum, each letter stands for a different digit. Replace the letters with digits.


Enigma 1104: Odd and even squares

From New Scientist #2260, 14th October 2000

In the following statement digits have been consistently replaced by capital letters, different letters being used for different digits:

ONE and NINE are odd perfect squares, FOUR is an even perfect square.

Find the numerical value of the square root of (NINE × FOUR × ONE).


Enigma 407: Bug-in-a-box

From New Scientist #1557, 23rd April 1987 [link]

A collector was ordering a thin rectangular box to house insects captured in the field.

“Each of its edges must measure a whole number of inches”, he told the carpenter.

“Well, that leaves a lot of room for manoeuvre”, the other remarked.

“That’s exactly the point”, the collector said. “You see, they cling to the edges of the box for security, poor little mites, and they detest traversing the same edge twice … But though they are confined to the edges they can still dream of unconquerable space. In fact I don’t know which is more important, the volume or the length of the edges. You’d better make the volume in cubic inches equal to the sum of all the 12 edges of the box”.

“Well, that narrows is down a little”, said the carpenter.

“Oh dear, I hope not”, said the collector. “Please make it as big as you can consistent with these specifications”. So saying he rushed out, almost snagging his net on the door handle.

The carpenter produced the box just as he was bidden. Assuming bugs do refuse to traverse any part of an edge previously trodden by them, what is the furthest a captive bug can walk before it becomes bored?


Enigma 1105: Road ants

From New Scientist #2261, 21st October 2000

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.


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?


Enigma 1106: Not a square unused

From New Scientist #2262, 28th October 2000

Harry, Tom and I each found a set consisting of a 4-digit perfect square, a 3-digit perfect square and a 2-digit perfect square that between them used nine different digits; but none of us could add a 1-digit square with the unused digit because 9, 4, 1 and 0 all appeared in each of our three sets. No two of us found exactly the same set; none of the squares in my set appeared in either Harry’s set or Tom’s set. There is one further set that none of us found whose unused digit is again not itself a perfect square.

List in ascending order the three squares in this set that none of us found.


Enigma 405: Uncle bungles the answer

From New Scientist #1555, 9th April 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.

Enigma 405

Which letter is wrong?

Write out the correct addition sum.

Note: This is a corrected version of Enigma 401.


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.

Enigma 401

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.


Enigma 1107: Factory work

From New Scientist #2263, 4th November 2000 [link]

When it comes to factor problems it is often quicker to use a bit of cunning logic than to resort to a computer or even a calculator, and here is one such puzzle.

Write down a four-figure number ending in 1 and then write down the next eight consecutive numbers, and then write down the nine numbers obtained by reversing the first nine. For example:

3721    1273
3722    2273
3723    3273
3724    4273
3725    5273
3726    6273
3727    7273
3728    8273
3729    9273

You could then count how many of all those numbers have a factor greater than 1 but less than 14: in this example there are actually eleven of them.

Your task now is to find a four-figure number ending in 1 so that, when you carry out this process, fewer than half the numbers have a factor greater than 1 but less than 14.

What is that four-figure number?


Enigma 404: Regular timepiece

From New Scientist #1554, 2nd April 1987 [link]

Enigma 404

My daughter has a regular hexagonal clock without numerals, as illustrated. I tried to fool her recently by rotating it and standing it on a different edge, but she recognised that the hands did not look quite right.

On the other hand, my son, has a clock on a regular polygon, again without numerals, which I can stand on any different edge and make the clock show the wrong time with its hands in apparently legitimate positions.

How many edges does this regular polygon have?


Enigma 1108: Every vote counts

From New Scientist #2264, 11th November 2000 [link]

George was nominated for president of the Golf Club. There was only one other candidate, and the president was elected by a simple ballot of the 350 members, not all of whom in fact voted.

The ballot papers were taken from the ballot box one at a time and placed in two piles — one for each candidate — with tellers keeping a count on each pile.

George won (what did you expect?), and furthermore his vote was ahead of his opponent’s throughout the counting procedure.

“That must be a one-in-a-million chance,” said the demoralised loser.

“No,” said George. “Now that we know the number of votes we each received, we can deduce that the chance of my leading throughout the count was exactly one in a hundred.”

How many members did not vote?


Enigma 403: Taking stock

From New Scientist #1553, 26th March 1987 [link]

“What animals have you in that barn there?” said the man from the ministry.

The farmer beamed. “Pigs, cows and ducks, sir.”

“How many are there, though?”

“Oh, quite a few, really, sir.”

“I need figures, man!” persevered the would-be census taker.

“If it’s figures you’ll be wanting, sir,” replied the farmer, “I can tell you that multiplying the number of horns by the number of legs by the number of wings gives 720.”

“Yes, but how many of each animal are there?” snapped the other exasperatedly.

“Telling you the number of cows alone wouldn’t enable you to deduce the number of ducks and pigs. Telling you the number of ducks alone wouldn’t enable you to deduce the number of pigs and cows. But telling you the number of pigs would enable you to deduce the number of cows and ducks right enough. So I reckon you can work out how many cows and ducks there be in yonder barn even if I don’t tell you the number of pigs in it.”

With that he ducked into a hen-house.

How many cows, ducks and pigs were there in the barn?


Enigma 1109: All square in games

From New Scientist #2265, 18th November 2000 [link]

At tennis a set is won by the first player to win 6 games, except that if the score goes to 5 games all, the set is won either by 7 games to 5 or by 7 games to 6. The first person to win three sets wins the match.

Sometimes at the end of a match each player has won exactly the same number of games. This happened when André beat Boris in a match in which no two sets contained the same number of games.

You will have to work out whether André won by 3 sets to 1 or by 3 sets to 2, but if I told you how many games in total each player won you would be able to deduce with certainty the score in each of the sets that André won and in the set or each of the sets that Boris won.

What was the score in each of the sets that André won? (Give each set’s score in the form x-y, André’s score always comes first).


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?


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.