Enigmatic Code

Programming Enigma Puzzles

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?


Puzzle 69: Division: letters for digits

From New Scientist #1120, 14th September 1978 [link]

For some reason Uncle Bungle does not like divisors. This has been left out in the latest division sum which he has produced with letters substituted for digits. Here it is:

Find the divisor and all the digits of the sum.


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.


Puzzle 70: Football five teams: new method

From New Scientist #1121, 21st September 1978 [link]

The new method of rewarding goals scored in football matches goes from strength to strength. In this method 10 points are given for a win, 5 points for a draw and 1 point for each goal scored. Once can get some idea of the success of the method from the fact that in the latest competition between 5 teams, when some of the matches had been played, each team had scored at least 1 goal in every match. They are eventually going to play each other once.

The points were as follows:

A   11
B    8
C   12
D    5
E   43

Not more than 9 goals were scored in any match.

What was the score in each match?


Enigma 1107: Factory work

From New Scientist #2263, 4th November 2000

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?


Tantalizer 477: Precognition

From New Scientist #1028, 25th November 1976 [link]

I overheard Professor Foresight discussing the results of a small precognition test the other day. It emerged that he had tossed a penny five times, inviting the thirteen members of his class to write down what was coming before each throw. Six students had done better than the rest, all scoring the same number, although no two had produced identical lists of guesses. Nor had any two of the remaining students produced identical lists.

It also emerged that the penny had not come up Heads all five times. Nor was the actual series Head, Tail, Tail, Tail, Head. Nor was it Tail, Tail, Head, Tail, Tail. At this point the discussion broke up and I was left wondering just what the actual series was. Given that each of these series just mentioned was the guess of one of the unsuccessful seven, can you oblige?


Enigma 1108: Every vote counts

From New Scientist #2264, 11th November 2000

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?


Tantalizer 478: Surprises

From New Scientist #1029, 2nd December 1976 [link]

King Ethelweed needed a new champion. So he commanded his three doughtiest knights to appear before him on the first Monday of the new year and bade them fight one another. They fought all day long until the eventide, when the king called a respite and awarded x ducats to the winner, y ducats to the second knight and z to the third. xy and z are positive descending whole numbers.

To the valiant knights’ dismay, the same happened on the next and each following day, until King Ethelweed at length declared himself satisfied. One each day the same prizes of xy and z were awarded, the being no ties on any day.

Thus it befell that Sir Kay gained the most ducats and became the king’s champion, even though he fared worse on the second day than on the first. Sir Lionel took home twenty ducats in all and Sir Morgan, despite winning top prize on the third day, amassed a mere nine.

Which was the final day and who won how many ducats on it?


Enigma 1109: All square in games

From New Scientist #2265, 18th November 2000

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?


Puzzle 71: All wrong, all wrong

From New Scientist #1122, 28th September 1978 [link]

A couple of one’s, a couple of two’s and a six;
All wrong, all wrong!

If only I thought that the puzzle was one I could fix,
I’d sing a song.

But as I feel sure that it’s rather too much for me,
My voice is muted.

Uncle Bungle’s my name and I fear that you must agree,
I’m rather stupid.

So please, I implore,
Continue the fight,
With tooth and with claw,
With main and with might,
To make wrong sums right.


The figures given are all incorrect. Write out the whole division sum.


Enigma 1110: Dots and lines

From New Scientist #2266, 25th November 2000

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.


Enigma 400: Potential difficulties

From New Scientist #1550, 5th March 1987 [link]

I asked Electrophorus what he was working on.

“You know that joining unlike terminals of a pair of batteries produces a voltage across the two free terminals equal to the sum of the voltages of the separate batteries. And connecting unlike terminals produces a voltage equal to the difference of the voltages of the separate batteries?”

“Yes”, I replied. “With a battery of 2 volts and one of 5 volts one obtains 3 volts (sources opposing) or 7 volts (sources reinforcing).”

“Well, before lunch I had three batteries, none of which had zero voltage, and a voltmeter with a holder that would accommodate only two batteries. So I measured the voltages across the free terminals of all possible pairwise combination of these three batteries both in the case where the voltages reinforced and where they opposed. I wrote on my blackboard the resulting six positive numbers in order of increasing magnitude.”

“When I returned from lunch eager to calculate the ratings of the three batteries, I found the three batteries gone and my blackboard wiped clean. I remember that the second smallest reading occurred twice. It was either 13 or 17 volts, I forget which. I had noticed, rather inconsequentially perhaps, that reversing the digits of this double reading produced another reading which occurred in my measurements.”

What were the ratings of the three batteries?


Tantalizer 479: Cat and five tales

From New Scientist #1030, 9th December 1976 [link]

Someone let the cat out. Who was it? That is rather hard to decide. Delia says it was one of the twins, meaning Bert or Claud. Alice says it was Bert; and Bert (shame on him!) says it was Claud. Meanwhile Claud says it was Delia; and Emma says it was not Claud.

So it is all a bit of a puzzle and you will be expecting to be told how many of them are right in what they say. But that would make it all much too easy, as you could then deduce who the culprit was. So you will just have to manage with what information you have.

Who let the cat out?


Enigma 1111: Base-age

From New Scientist #2267, 2nd December 2000

Fill in the following cross-figure. No answer begins with a zero. The same base is used for all the entries, but it is not necessarily 10.

1. A palindromic prime.
4. The square of the base being used.
5. A square.

1. Three times my son’s age.
2. A prime.
3. A palindromic square.

How old is my son?