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Programming Enigma Puzzles

22 October 2014

Posted by on **From New Scientist #2442, 10th April 2004**

We are playing ordinary noughts and crosses but with two extra rules:

(i) A player may not go in the centre square if going in the centre square puts that player in a position which is better for them than every other position they could have gone to instead.

(ii) If after eight turns no one has won and the centre square is empty then the game ends and is declared a draw.

For the purpose of rule (i), position P is better for a player than position Q if either (a) the player can force a win from P but not from Q, or (b) the player can force a draw or better from P but not from Q.

If, in a game, O goes first and each player plays as well as possible, is the result a win for O, a win for X or a draw?

**Note:** I am still waiting for a phone line to be connected at my new house, so I only have sporadic access to the internet at the moment. The latest estimate is that I’ll have a connection by the end of October 2014.

20 October 2014

Posted by on **From New Scientist #1377, 29th September 1983** [link]

I was constantly being kept in at school to do extra work on decimals. So many unhappy memories are revived now that my son brings home his decimal homework: but thank goodness for calculators!

Unfortunately, the homework calls for exact answers, but my calculator displays only six digits after the decimal point. But that was no disadvantage for last night’s homework. For the teacher had listed all the integers greater than 1 whose reciprocals were recurring decimals such that the first six digits occurring after the decimal point occurred again in that order as the 7th-12th digits and the 13th-18th etc. So, for example, the first two numbers in the list were 3 and 7 because ⅓=0.3333333333333… and ⅐=0.142857142857… My son’s homework was to work out the decimals of all the reciprocals. So, for once, my limited calculator display was sufficient to check his answer.

How many numbers had the teacher listed?

**Note:** I am still waiting for a phone line to be connected at my new house, so I only have sporadic access to the internet at the moment. The latest estimate is that I’ll have a connection by the end of October 2014.

18 October 2014

Posted by on **From New Scientist #2443, 17th April 2004**

Triangular numbers are integers that fit the formula n(n+1)/2, such as 1, 3, 5, 10, 15. Your task is to put a digit in each of the eight outer squares of a 3×3 grid, so that the numbers that you read across the two outer rows and down the two outer columns are four different 3-digit triangular numbers.

The numbers must also be such that with a digit of your choice in the central square you could read another 3-digit triangular number across the central row and with a different digit of your choice in the central square you could read another 3-digit triangular number down the central column. No number may start with a zero.

What are the triangular numbers that you could read across the central row and down the central column?

**Note:** I am still waiting for a phone line to be connected at my new house, so I only have sporadic access to the internet at the moment. The latest estimate is that I’ll have a connection by the end of October 2014.

16 October 2014

Posted by on **From New Scientist #1376, 22nd September 1983** [link]

Below is an addition sum with letters substituted for digits. The same letter stands for the same digit wherever it appears, and different letters stand for different digits.

Write the sum out with numbers substituted for letters.

**Note:** I am still waiting for a phone line to be connected at my new house, so I only have sporadic access to the internet at the moment. The latest estimate is that I’ll have a connection by the end of October 2014.

13 October 2014

Posted by on **From New Scientist #2444, 24th April 2004**

Some players entered a “round robin” tennis tournament, where each player plays each of the others once, with each match resulting in a win for one of the players. At the end of the tournament I noted how many matches each of the players had won. The men were pretty pathetic, winning only one match each. Even so, Alan beat Barbara in their match. On the other hand, Christine beat David: had that result been reversed all the women would have won the same number of matches.

How many men and how many women entered the tournament?

**Note:** I am still waiting for a phone line to be connected at my new house, so I only have sporadic access to the internet at the moment. The latest estimate is that I’ll have a connection by the end of October 2014.

12 October 2014

Posted by on **From New Scientist #1375, 15th September 1983** [link]

I’ve been trying to draw the smallest possible circles with a precise number

nof lattice-points on their circumferences. Here are my results forn=4 andn=8. Each has it’s centre at a point (½, ½), and the radii are ½√2 (about 0.707) and ½√10 (about 1.581).I find the problem trickier when

nis odd. What is the radius of the smallest circle with exactly five lattice-points on its circumference, please? And where will its centre be?

The original article seems to have miscalculated ½√10.

**Note:** I am still waiting for a phone line to be connected at my new house, so I only have sporadic access to the internet at the moment. The latest estimate is that I’ll have a connection by the end of October 2014.

11 October 2014

Posted by on **From New Scientist #2445, 1st May 2004**

George’s blue wheelie bin (used for recyclable paper) was emptied by the council on 1 April 2004. It is emptied every fourth Thursday, and we presume that this schedule will continue indefinitely, in spite of official holidays. The next collection will be on 29 April giving, unusually, two collections in one calendar month. Looking to the future, when are the first years in which the following events will occur?

(1) One calendar month in the year has no collection day.

(2) There are two collections in one calendar month in the year.

(3) No month in the year has two collection days.

(4) Two months in the year each have two collection days.

(5) February has two collection days.

**Note:** I am still waiting for a phone line to be connected at my new house, so I only have sporadic access to the internet at the moment. The latest estimate is that I’ll have a connection by the end of October 2014.

9 October 2014

Posted by on **From New Scientist #1374, 8th September 1983** [link]

At the car factory where I work the latest model is pouring off the assembly line. Their chassis numbers started at 1 and they have been numbered consecutively since.

In the quality control department we have to check some of the cars. The first one we checked was one of the early ones off the line. Then we checked all cars whose chassis numbers were multiples of the first number which we had checked.

Unfortunately this system let too many mistakes go undetected and so the manager asked us to check an extra car last week. He also introduced new rules by which we would subsequently have to check all cars whose chassis numbers were the sum of two numbers (possibly the same) which we had already checked.

Car number 77777 has just gone through unchecked and we are now checking number 77778. Our union is demanding extra staff in this department because we now realise that under the new rules we are going to have to check

everycar from now on.How many of the first 77777 cars were checked?

**Note:** I am still waiting for a phone line to be connected at my new house, so I only have sporadic access to the internet at the moment. The latest estimate is that I’ll have a connection by the end of October 2014.

8 October 2014

Posted by on **From New Scientist #2446, 8th May 2004**

Consider the two sentences: “The number of a’s in this sentence is one” and “The number of e’s in this sentence is seven”. The first of those sentences is true and the second is false. We consider all the possible sentences of the form of those two sentences with any of the 26 letters of the alphabet in place of a and e and any word for a number between 1 and 20, inclusive, at the end of the sentence. So we are looking at 26 × 20 = 520 sentences in all.

How many of these sentences are true?

**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. The latest estimate is that I’ll have a connection by the end of October 2014.

6 October 2014

Posted by on **From New Scientist #1373, 1st September 1983** [link]

Another letters for numbers problem. It may leave you at sea, for it is a tease to keep you on your toes. You see, each letter stands for a

differentdigit each time it appears.What is a

TEASE?

**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. The latest estimate is that I’ll have a connection by the end of October 2014.

4 October 2014

Posted by on **From New Scientist #2447, 15th May 2004**

I have constructed a cyclical chain of 3-digit numbers, where each number starts with the digit (which is never 0) that was the last digit of the previous number in the chain, and the numbers in the chain are alternately perfect squares and triangular numbers.

The chain consist of as many numbers as is possible, consistent with the requirements outlined above and the stipulation that no number may appear in it more than once.

How many numbers are there in the chain?

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

2 October 2014

Posted by on **From New Scientist #1372, 25th August 1983** [link]

In the following fully-worked division, all the digits have been replaced by blanks except for one. And before you start counting your blessings in having at least one digit to go on, I must tell you that the 5 is, in fact, wrong.

What is the five-digit quotient?

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

29 September 2014

Posted by on **From New Scientist #2448, 22nd May 2004**

I laid out a row of identical saucers and gave my nephew some marbles, all the same. I then set him the following problem:

“If I ask you to place all the marbles in the saucers, with at least one marble in each, in how many ways can this be done?”

After some thought he came to the correct conclusion that there were between a thousand and ten thousand such arrangements, and then he gave up. So I then asked him the following simpler question:

“How many of these arrangements will be symmetrical, in the sense that if I reversed the order of the saucers the pattern remains the same?”

This time he gave the correct answer of 20.

What was the correct answer to the first question?

**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.

26 September 2014

Posted by on **From New Scientist #1371, 18th August 1983** [link]

The picture shows a cuboctahedron, A semi-regular solid with six square faces and eight triangular ones. It has 12 vertices, lettered from A to L.

Can you place the numbers 1 to 14, one on each face, so that the numbers round each vertex add up to the same “magic sum”?

That magic sum should be made as small as possible, please.

Please let me know:

(A) What the magic sum is.

(B) What numbers are on the faces having an edge in common with the 7-face.

**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. The current estimate is that the line will be connected at the end of September 2014.

25 September 2014

Posted by on **From New Scientist #2449, 29th May 2004**

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

UN and NEUF are odd perfect squares, ZERO is an even perfect square.

Please send in the numerical value of the square root of (UN × NEUF × ZERO).

**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. The current estimate is that the line will be connected at the end of September 2014.

24 September 2014

Posted by on **From New Scientist #1370, 11th August 1983** [link]

Here is a magic square, that is each row, each column each of the long diagonals adds up to the same total. But the numbers in the boxes have been replaced on the usual letters-for-digits principles: each letter stands for the same digit throughout, and different letters stand for different digits.

What about the sum of the rows etc? Can it be

anynumber? It sounds like it, because the sum of each row is NE!What is ENIGMA?

**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. The current estimate is that the line will be connected at the end of September 2014.

19 September 2014

Posted by on **From New Scientist #2450, 5th June 2004** [link]

Susan Denham’s recent Enigma “Natural Numbers” prompted me to look for a French version. Once again I have assigned a number to each letter of the alphabet and the numbers, which are not all different, include negative numbers, zero, positive numbers and fractions.

I can tell you that:

U + N = 1

D + E + U + X = 2

T + R + O + I + S = 3

Q + U + A + T + R + E = 4

C + I + N + Q = 5

S + I + X = 6

S + E + P + T = 7

H + U + I + T = 8

N + E + U + F = 9

D + I + X = 10

O + N + Z + E = 11

D + O + U + Z + E = 12

T + R + I + E + Z + E = 13

Q + U + A + T + O + R + Z + E = 14

Q + U + I + N + Z + E = 15Please send in S + A + I + N + T + T + R + O + P + E + Z.

**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. The current estimate is that the line will be connected at the end of September 2014.

18 September 2014

Posted by on **From New Scientist #1369, 4th August 1983** [link]

In the following addition sum all the digits are wrong. But the same wrong digit stands for the same correct digit whenever it appears, and the same correct digit is always represented by the same wrong digit.

2 4 5 3 4 1 9 5 4 2 4 4 7 1 1 --------- 9 1 0 8 6 =========Find the correct addition sum.

There are now 712 puzzles on the site, which is exactly 40% of the total number of Enigma puzzles (1780). Although because some Enigmas share numbers (particularly at Christmas) this probably slightly underestimates that total number of Enigmas that I have yet to publish.

**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. The current estimate is that the line will be connected at the end of September 2014.

16 September 2014

Posted by on **From New Scientist #2451, 12th June 2004** [link]

“Multiplying by 9/5 and adding 32,” I explained to my clever nephew George, “is useless in practice. What you need is some memorable equivalents, like 10 °C being 50 Fahrenheit. Here’s one I’ve invented: 16 °C = 61 °F. See, to get from one to the other you just reverse the two digits.”

“Actually 16 °C = 60.8 °F,” I said.

“So 61 is near enough,” I said.

“Near enough is not exactly right.”

“But you cannot do it exactly,” I objected sourly.

“You can’t, because you insist on boring old base 10. But I bet I can, using other bases,” George retorted. Off he went to investigate, and was soon back. “-90 °C = -130 °F,” he said, “and to base 21 this says -46 °C = -64 °F. I have other examples, including two between the freezing and boiling points of water.”

What were the two examples that George found? Give your answers in the form

x°C =y°F wherexandyare written in base 10 (andxlies between 0 and 100).

**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. The current estimate is that the line will be connected at the end of September 2014.

15 September 2014

Posted by on **From New Scientist #1368, 28th July 1983** [link]

Admirals Drake, Jellicoe, Nelson and Raleigh were lately piped aboard HMS Beefeater for purposes of splicing the mainbrace. Each arrived immaculate in his well-cut uniform, did his share of the work to the full and departed with dignity barely askew. But (sh!) Drake presently realised that he had somehow come to be wearing a pair of trousers too small for him and a similarly tight feeling presently crept over Jellico. Nelson and Raleigh too found themselves clad in the wrong breeches.

Each promptly dispatched the trousers in his possession to one of the others. Each received one pair, not sent by the admiral to whom he had sent one and, alas, not his own. Nelson received a pair too tight for him.

Well, shiver my timbers, they then did the same again. Each sent off the pair in question. Each received a pair, not from the Admiral he had fired one at, not one which he had had already and still not his own. Drake received a pair too large for him, his own being now lodged with either Jellicoe or Nelson.

Who now had whose breeches?

**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. The current estimate is that the line will be connected at the end of September 2014.

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