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Programming Enigma Puzzles

11 May 2018

Posted by on **From New Scientist #1598, 4th February 1988** [link]

Six boys from my class have joined together to form a secret society. The each have a different three-digit number, but each of the six numbers uses the same three digits in some different order.

The boys have noticed that, for any two of them, their numbers have a common factor larger than 1 precisely when their names have at least one letter in common. So, for example, Tom’s number and Sam’s number have a common factor larger than 1, whereas Bob’s and Tim’s numbers do not. Ken’s number is prime.

The sixth member of the society is one of Ian, Ben, Rod, Rob, Jak, Vic and Pat.

Who is the sixth member, and what is Bob’s number?

[enigma447]

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7 May 2018

Posted by on **From New Scientist #2218, 25th December 1999**

When I was at school I was given a Christmas puzzle to do. So, as far as I can remember it, I’ve reproduced it for you to try:

“Four different numbers larger than 6 have been placed in some of the circles of the Christmas star:

Put the numbers 1 to 6 in the remaining circles (one of them in each) so that the four numbers on each straight line add up to the same total.”

Now that I’ve tried this again I realise that I’ve made a mistake somewhere, because the puzzle as stated is impossible. In fact, it turns out that my only error is that one of the four numbers which I have placed on the star is incorrect.

Which one is incorrect, and what should it be?

Thanks to Hugh Casement for providing the sources for a large number of *Enigma* puzzles originally published between 1990 and 1999, including this one.

[enigma1063]

23 April 2018

Posted by on **From New Scientist #2221, 15th January 2000**

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

TEN is two away from a perfect cube

and:

there are TEN cubes not more than THOUSAND.

What is the numerical value of THEN?

[enigma1065]

13 April 2018

Posted by on **From New Scientist #1594, 7th January 1988** [link]

When bells are played in a particular sequence, a “change” is a different sequence obtained from the first by at least one pair of bells which were consecutive the first time reversing their order. Any number of pairs can do this, but no bell is involved in more than one move. So, for example, if four bells are played in the order ABCD, then the possible changes are: ABDC, ACBD, BACD and BADC.

Our local bell-ringing group is very keen. A number of them met last night, including one newcomer. They had a bell each and they rang them in a particular order (with the newcomer ringing first). Then, to test themselves, they decided to write down all the possible changes from that original sequence. They each had a piece of paper and in a few minutes each (including the newcomer) had written down some of the possible changes. They had each written the same number and, between them, they had included all the possible changes exactly once.

They then decided to choose one of these changes to play, but thought they had better choose one in which the newcomer’s bell still played first. So they deleted from their lists all those changes in which the newcomer’s bell had changed places: they all had to delete the same number of possibilities.

Including the newcomer, how many of them were there?

[enigma443]

9 April 2018

Posted by on **From New Scientist #2223, 29th January 2000**

A number of players entered for a knockout tennis tournament. Some of them played in the first round games, the rest being given “byes” into the second round so that thereafter there were normal rounds in which all remaining players took part, leading eventually to quarter-finals, semi-finals and the final.

Overall the tournament took a week, with the same number of games being played each day.

Actually a whole-number percentage of the entrants were knocked out in the first round.

What percentage?

[enigma1067]

6 April 2018

Posted by on **From New Scientist #1592, 24th December 1987** [link]

After our successful pantomime production in which I played the leading lady, I gave my little costarring helpers some gifts from a big bag of different trinkets, and they each got a different number and none were left.

To make it fairer I gave each helper 10p for each gift that he

didn’tget and deducted 40p for each gift that hedidget, but that still gave each of them some 10p coins as well as some gifts. It cost me £12.60 in addition to the gifts.What was the highest number of gifts received by any helper (that little fellow got less than 50p cash)?

What part was I playing?

This puzzle completes the archive of *Enigma* puzzles from 1987. There is now a complete archive from the start of *Enigma* in February 1979 to the end of 1987, and also from February 2000 to the final *Enigma* puzzle in December 2013. Making 1162 *Enigma* puzzles posted so far, which means there are about 626 left to post.

[enigma442b] [enigma442]

26 March 2018

Posted by on **From New Scientist #2225, 12th February 2000** [link]

I have a cube. On each of its faces is a digit, the style of writing being rather like the display on a calculator. I hold the cube to look at one of its faces from the front and then, keeping the upper and lower faces horizontal, I swivel the cube around and note the digits which I see (all seemingly the right way up) and hence I read off a four-figure number which is divisible by eleven.

Now I repeat the process starting this time looking from the front at one of the faces which was horizontal in the previous manoeuvre. Once again I read off a four-figure number which is divisible by eleven.

Now I start again with the cube in exactly the same position as it was at the start of the first process. This time I keep the left-hand and right-hand faces vertical and I swivel the cube around. Once again I read a four-figure number which is divisible by eleven and also by three odd integers less than eleven.

What was that last four figure number?

[enigma1069]

12 March 2018

Posted by on **From New Scientist #2227, 26th February 2000** [link]

Your task this week is to find the day and date of my birthday this year in the form:

(for example, Monday / 8 / May).

If I told you the DAY and the NUMBER you could also work out the MONTH.

So now if I told you the first letter in the spelling of the MONTH you could work out the MONTH.

So now if I told you how many Es there are in the spelling of the MONTH you could work out the MONTH.

So now if I told you the NUMBER you could work out the DAY and MONTH.

What are the DAY, NUMBER and MONTH of my birthday this year?

[enigma1071]

9 March 2018

Posted by on **From New Scientist #1589, 3rd December 1987** [link]

“How many perfect squares are there between TEN and TWENTY?”

“ONE.”

“Right. And are TWO, TEN, TWELVE and TWENTY even?”

“Of course. In fact the first and last digits of TWENTY are both even.”

“Right. And is TEN divisible by 3?”

“Of course not.”

In the above, digits have consistently been replaced by letters, different letters representing different digits.

Find NOW.

[enigma439]

16 February 2018

Posted by on **From New Scientist #2230, 18th March 2000** [link]

In Enigmaland they work in the usual decimal arithmetic using the usual +, × and =, but they have different symbols for the digits zero to nine. Or, to be more accurate, they use the same symbols 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 but in an entirely different order. In fact, for each of the digits, their symbol differs from the usual one.

Here are some correct sums which were done in Enigmaland (which surprisingly also work in our conventional system):

4 × 7 = 28

5 × 7 = 35

4 × 6 = 24

1 + 4 + 6 + 7 + 7 = 25

In Enigmaland if they wrote 2 × 302, what would their answer be?

[enigma1074]

9 February 2018

Posted by on **From New Scientist #1585, 5th November 1987** [link]

Here is an addition sum with some occurrences of digits replaced consistently by letters, different letters being used for different digits, with gaps being left in the remaining places, as shown:

By some cunning logic you can untangle this and … find the genuine number GENUINE.

[enigma435]

15 January 2018

Posted by on **From New Scientist #2235, 22nd April 2000** [link]

One of the three girls Angie, Bianca and Cindy always tells the truth, one always lies, and the other is unreliable in the sense that a true statement is always followed by a false one and vice versa. Here are some things they just said about themselves:

Angie:The eldest is dishonest. The tallest is unreliable.

Bianca:The youngest is honest. The shortest is unreliable. Angie is taller than me.

Cindy:The youngest is unreliable. The tallest is honest.What are Cindy’s characteristics? (For example – honest, youngest and mid-height).

[enigma1079]

8 January 2018

Posted by on **From New Scientist #2236, 29th April 2000** [link]

A magic square consists of a 3 × 3 array of nine different integers, nought or more, such that the sum of each row, column and main diagonal is the same. The most familiar one is:

If you form the product of each row and add them up you get 8×1×6 + 3×5×7 + 4×9×2 = 225. Similarly if you form the product of each column and add them up you get 8×3×4 + 1×5×9 + 6×7×2 = 225. Remarkably the two numbers will be equal for any magic square, and it is known as the “magic product”.

There is a magic square whose middle entry is a single digit number which equals the sum of the three digits in its magic product.

What is that magic product?

[enigma1080]

5 January 2018

Posted by on **From New Scientist #1580, 1st October 1987** [link]

My Welsh friend, Dai the dial, has a telephone number consisting of nine different digits and, as you telephone him on my push-button phone illustrated above, you push a sequence of buttons each adjacent (across or down) to the one before.

The digit not used in his number is odd, the last digit of the number is larger than the first, and (ignoring the leading digit if it is zero) the number is divisible by 21.

What is Dai’s number?

[enigma430]

11 December 2017

Posted by on **From New Scientist #1576, 3rd September 1987** [link]

Time and again, you’ve been asked to sort out letters-for-digits puzzles, where digits are consistently replaced by letters, different letters being used for different digits. Today, that recurring theme is used in a truly recurring way. The fraction on the left (which is in its simplest form) represents the recurring decimal on the right. Should you want an extra optional clue, I can also tell you that the last two digits of the numerator of the fraction are equal.

What is AGAIN?

[enigma426]

4 December 2017

Posted by on **From New Scientist #2241, 3rd June 2000** [link]

Place your finger in the starting box in the grid and follow each instruction. You will find that you visit each square before finishing in the appropriate box.

Now cut up the board into six rectangles each consisting of two adjacent squares. Then put them back together to form a new three-by-four grid with the starting square one place lower than it was before. Do all this in such a way that, once again, if you follow the instructions you visit each square.

In your new grid, what instruction is in the top left-hand corner, and what is the instruction in the bottom right-hand corner?

[enigma1085]

13 November 2017

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

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]

30 October 2017

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

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]

23 October 2017

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

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]

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