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Programming Enigma Puzzles

17 November 2017

Posted by on **From New Scientist #1573, 13th August 1987** [link]

In the following football table digits (from 0 to 9) have been replaced by letters. The same letter stands for the same digit wherever it appears and different letters stand for different digits. The four teams are eventually going to play each other once.

(Two points are given for a win and 1 point to each side in a drawn match).

Find the score in each match.

[enigma423]

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15 November 2017

Posted by on
13 November 2017

Posted by on **From New Scientist #2244, 24th June 2000**

I had ten cards with a different digit on each and I tore one of them up. I used two of the remaining cards to form a two-figure prime, three others to form a three-figure prime, and the last four to form a four-figure prime.

If I told you the total of those three primes it would still be impossible for you to work out which digit I had torn up. In fact, the middle two digits of the total both equal the digit I tore up. That should enable you with a little brainpower to say what the total of the three primes is.

What is that total?

[enigma1088]

10 November 2017

Posted by on **From New Scientist #1572, 6th August 1987** [link]

In our local sports club everyone plays at least one of badminton, squash and tennis. Of those who don’t play badminton, half play squash. Of those who don’t play squash, half play badminton. Of those who play badminton and squash, half play tennis.

I play badminton only: there are two more players who play tennis only than there are who play badminton only.

If a player plays just two of the three games, then his or her spouse also plays just two of the three games.

The membership consists entirely of married couples and each of the three games is played by at least one member of each married couple.

How many people are there in the club, and how many of those play all three games?

[enigma422]

8 November 2017

Posted by on **From New Scientist #1020, 30th September 1976** [link]

Mr Chips was sodden with gloom after marking the R.I. test. Admittedly no one had scored nought, which was unusual. (In fact the scores were all different). But, alas, the other pupils had done better than the only four Christians in the class.

After some soul-searching he realised that it was his duty to adjust the marks. So he began by adding to each pupil’s score the number of marks gained in the test by all the other pupils. That was better but the resulting list left something to be desired. So he then subtracted from each pupil’s new score three times his or her original score. That was much more satisfactory. The scores were all positive and totalled 116 marks in all. The heathen Blenkinsop was rightly bottom and could therefore be made to write out the 119th psalm.

How many did Blenkinsop score before and after Mr Chips did his duty?

[tantalizer469]

6 November 2017

Posted by on **From New Scientist #2245, 1st July 2000**

The island of Buss is divided into 3 counties, A-shire, B-shire and C-shire, each containing a number of towns. Each town in A-shire has just one road and that goes to a town in B-shire. Each town in B-shire, irrespective of how many roads it has to towns in A-shire, has just one road that goes to a town in C-shire. There are 3 bus companies, Red, Yellow and Green. Here are the buses for today:

• From each A-shire town one Red bus runs to a B-shire town.

• From each B-shire town one Yellow bus runs to a C-shire town.

• From each A-shire town one Green bus runs to a C-shire town.Naturally the destinations are determined by the roads. A bus company is “economical” if no town is the destination of more than one of its buses today. A bus company is “covering” if every town in the county its buses today finish in is the destination of at least one of those buses. Which of the following statements are bound to be true:

(1) If the Green Co is economical then the Red Co is economical.

(2) If the Green Co is not economical then the Red Co is not economical.

(3) If the Green Co is covering then the Yellow Co is covering.

(4) If the Green Co is not covering then the Red Co is not covering.

(5) If the Green Co is economical and the Red Co is covering then the Yellow Co is economical.

(6) If the Green Co is covering and the Yellow Co is economical then the Red Co is covering.

This is another puzzle from an old paper copy of **New Scientist** that I recently found. It currently doesn’t appear in the on-line archives.

[enigma1089]

3 November 2017

Posted by on **From New Scientist #1571, 30th July 1987** [link]

“Substitution sums,” said my friend and colleague Dr Addam, “are not at all difficult to concoct. Take, for example, this one.” And he wrote on a handy sheet of paper:

“The idea, as ever,” he continued, “is to substitute a different digit for each letter wherever it appears.”

I pondered for a minute. “I see what you mean,” I said. “There seem to be several solutions.”

“True enough,” said Addam; “so I’ll make it more difficult. Can you find me a solution in which the number THREE is divisible by 3, FOUR by 4, and …”

“SEVEN by 7?” I interrupted.

“No, that can’t be done. I was going to say FOURTEEN is divisible by 14.”

After a few minutes, I had the required solution. What was it?”

[enigma421]

1 November 2017

Posted by on **From New Scientist #1113, 27th July 1978** [link]

Across:1a. The digits are all even.

4a. Odd, and even when reversed.

5a. 19 is a factor of this.

Down:1d. A perfect square.

2d. A perfect square when reversed.

3d. Each digit is 1 less than the one before.(There are no 0’s).

[puzzle62]

30 October 2017

Posted by on **From New Scientist #2246, 8th July 2000**

Today we are trying to crack a code. To each of the days:

MON, TUE, WED, THU, FRI, SAT, SUN

there is assigned a different whole number from 1 to 10.

We know that, for any pair of days, their numbers have a factor larger than 1 in common if and only if their first three letters have a letter or two in common. So, for example, TUE and WED both have an E and so their numbers will have a factor larger than 1 in common, whereas the numbers for WED and THU will have no factor larger than 1 in common.

We also know that their numbers will satisfy:

MON + TUE < WED + THU < FRI + SAT

What (in order MON to SUN) are their seven numbers?

[enigma1090]

27 October 2017

Posted by on **From New Scientist #1570, 23rd July 1987** [link]

The King of Zoz was asked by his son for 88,200 ducats to cover the latter’s gambling debts incurred at college.

“That is indeed a princely sum!” said the King, calling for his ornate coffer, the one containing single ducats only.

“How much is there in the old coffer of yours?” asked the Prince as two servants struggled with it it.

The father smiled indulgently but knowingly and replied: “If you were to multiply the number of ducats it now contains by the number of ducats would remain in it were I to remove 88,200 ducats, you would arrive at a perfect square.”

“There must be many numbers of ducats which fit those constraints,” said the son wistfully.

“Aye. And if you have the wit to deduce the number of different numbers of ducats there could be in the coffer consistent with my statement I shall know that you have learnt something and I shall pay your debt. If not, you can cover it yourself.”

Assuming there will be at least one ducat remaining in the coffer if the King removed 88,200 ducats, how many different possible numbers of ducats are there?

[enigma420]

25 October 2017

Posted by on **From New Scientist #1021, 7th October 1976** [link]

“My formula for life is Wit multiplied by Will”, Uncle Ernest announced for the umpteenth time. “I daresay you don’t know what that gives you, young Tommy”.

“Oh yes I do, Uncle”, Tommy replied cheekily.

“Success”, snapped Uncle Ernest.

“Thirst”, retorted Tommy.

Tommy, of course, had made a cryptarithmetic problem of it:

WIT × WILL = THIRST.

Each different letter stands for a different digit.

What is the value of THIRST?

[tantalizer470]

23 October 2017

Posted by on **From New Scientist #2247, 15th July 2000**

Mr Meaner has now retired from teaching. As a tough arithmetic exercise each year he used to ask his class to take the number of the year and find some multiple of it which consists simply of a string of ones followed by a string of zeros. For example in 1995 one girl in the class found that the 19-digit number 1111111111111111110 was a multiple of 1995!

Mr Meaner had been asking this question every year since he started training as a teacher. On the first occasion it was a reasonably straightforward exercise and most of the class found a multiple using as few digits as possible.

It is lucky that he did not ask the question a year earlier, for that year would have required over two hundred times as many digits as that first occasion did.

In what year did Mr Meaner first ask the question?

[enigma1091]

20 October 2017

Posted by on **From New Scientist #1569, 16th July 1987** [link]

Instructions1. You will need four copies of:

labelled A, B, C, D.

2. Take A. The artist Pussicato signs the top row and you sign the bottom row; your signature must contain 9 letters.

3. Fill in B by using A as follows. Take each A square in turn, find the position of its letter in the alphabet and from that number subtract the appropriate multiple of 5 to leave a number from 1 to 5. Put that number in the corresponding B square.

4. Fill in C by using B as follows. Each square touches three or five other squares, including touching along a side or at a corner. Take each B square in turn and add up the numbers in the squares it touches. From the total subtract the appropriate multiple of 5 to leave a number from 1 to 5. Put that number in the corresponding C square.

5. Paint D by using C as follows.Number the five colours, Red, Blue, Green, Yellow, White, 1 to 5 in any order you like. Take each C square in turn and find the colour you have given its number. Paint the corresponding D square with that colour.

ExampleA painter whose name involves only the first five letters of the alphabet produced:

What was the painters name?

[enigma419]

18 October 2017

Posted by on **From New Scientist #1114, 3rd August 1978** [link]

“Never tell them more than you need”, as Professor Knowall has so often said. “And pay them the compliment, my dear Sergeant Simple, of supposing that they are capable of putting one and one together to make two”.

As my readers will know, the professor, though he does not often have the time to turn his attention to anything other than crime, is very interested in football and likes making up and solving football puzzles. His remarks about putting one and one together to make two seemed rather silly to me at first, but I soon realised what he meant when he showed me the puzzle.

It was about four football teams, and gave some information concerning the number of matches played, won, lost and so on. But of the 24 pieces of information that one might have expected only 12 were given. One did indeed need to put one and one together to make two.

The information was as follows:

“That ought not to be too hard for you, my dear Sergeant”, he said, “but I must add also the information that not more than seven goals were scored in any match”.

I’m afraid it was too much for me, but I hope that the readers will be able to find out the score in each match.

(Each team is eventually going to play each other once).

[puzzle63]

16 October 2017

Posted by on **From New Scientist #2248, 22nd July 2000**

Marge, April, May, June, Julia and Augusta have all celebrated their birthday today. They are all teenagers and with the exception of the one pair of twins their ages are all different.

Today, only Marge and April have ages which are prime numbers, but the sum of the ages of all the girls is also a prime number. On their birthday last year, only May and June had ages which were prime numbers, but again the sum of the ages of all the girls was a prime number. On their birthday two years before that, only May and Julia had ages which were prime numbers, but even then, the sum of the ages of all the girls was again a prime number.

How old is Augusta?

[enigma1092]

13 October 2017

Posted by on **From New Scientist #1568, 9th July 1987** [link]

In the following division sum each letter stands for a different digit:

Re-write the sum with the letters replaced by digits.

[enigma418]

11 October 2017

Posted by on **From New Scientist #1022, 14th October 1976** [link]

Hop, Skip and Jump live in different houses in Tantalus St., which is numbered from 1 to 100. Here is what they have to say about the matter:

Hop: “My number is divisible by 7. Skip is much too fat. Jump’s number is twice mine.”

Skip: “Hop lives at 28. My number is one third of Jump’s. Jump and I are

notboth even.”Jump: “Hop lives at 91. Skip lives at 81. My number is divisible by 4.”

One of them has thus made three true statements, another three false and the remaining fellow has alternated, uttering either true, false, true or false, true, false.

Who lives where? And is Skip much too fat?

A correction to this puzzle was published with **Tantalizer 473**. The problem statement above has been revised accordingly.

[tantalizer471]

9 October 2017

Posted by on **From New Scientist #2249, 29th July 2000**

Bill’s credit card has the usual four four-digit numbers, which are in ascending order of size. All are prime numbers and the sum of the digits of each is the same.

The digits in the first number are all different and the third number is the first number reversed. The digits in the second number are all different and the fourth number is the second number reversed. The last digits of the four numbers are all different.

He has a hopeless memory for figures, but he can always work out his four-digit PIN from his card, because he can remember that it is equal to the difference between the first and third numbers (or the difference between the second and fourth numbers, which is the same) and happens to be a perfect square.

What is the fourth number?

[enigma1093]

6 October 2017

Posted by on **From New Scientist #1567, 2nd July 1987** [link]

We have a small snooker table at home, everything being in a reduced form of the real thing. The point system is the same: that is, 1 for a red, with each potted red enabling the player to try for one of six colours with points from 2 to 7. (For example, the blue is worth 5). At the end of the frame the six colours are potted in turn.

Last Saturday, I played with my daughter and the frame was completed in just one visit to the table by each of us. I opened, potted a red with my first shot, then potted a colour (which, of course, was brought out again) and then, whenever I successfully potted a colour after a red, it was always that same colour. Then I made a mistake (without any penalties) and my daughter took over. She, too, always followed a red by a particular colour, but a different one from mine. She cleared the table and we drew on points: we decided to replay the next day.

Surprisingly, all that I said about Saturday’s frame could be said about Sunday’s, but this time we drew with one more point each than on the previous day.

How many times, in total for the two frames, did I pot the blue?

How many balls does my small table have?

[enigma417]

4 October 2017

Posted by on **From New Scientist #1115, 10th August 1978** [link]

Each digit in the addition sum below is wrong. But the same wrong digit stands for the same correct digit wherever it appears, and the same correct digit is always represented by the same wrong digit.

Find the correct addition sum.

[puzzle64]

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