Enigmatic Code

Programming Enigma Puzzles

Tag Archives: by: Susan Denham

Enigma 1071: Special occasion

From New Scientist #2227, 26th February 2000

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?



Enigma 439: Ten to twenty

From New Scientist #1589, 3rd December 1987 [link]

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


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In the above, digits have consistently been replaced by letters, different letters representing different digits.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


Enigma 1074: Changing times

From New Scientist #2230, 18th March 2000

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?


Enigma 435: An enigma to untangle

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.


Enigma 1079: Girls’ talk

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).


Enigma 1080: Magic product

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?


Enigma 430: Let your fingers do the walking

From New Scientist #1580, 1st October 1987 [link]

Enigma 430

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?


Enigma 426: Time and again

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.

Enigma 426

What is AGAIN?


Enigma 1085: Cut and run

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?


Enigma 1088: That’s torn it

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?


Enigma 422: Keeping fit by halves

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?


Enigma 1090: Week links

From New Scientist #2246, 8th July 2000 [link]

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


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:


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


Enigma 1091: One’s best years

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?


Enigma 417: Snooker triangle

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?


Enigma 1094: De-fence

From New Scientist #2250, 5th August 2000 [link]

In my garden there is a circular pond less than two metres across. Because my young nephew was coming to stay I asked a local handyman to erect a fence around it. He did this by taking three straight lengths of fencing, two of them equal, and each of them a whole number of metres long. He formed these into a triangle which fitted around the pond.

I complained that this took up too much space, so he adapted the construction to make a hexagonal fence around the pond. Opposite sides of the hexagon were parallel, three of the sides used bits of the original triangular fence without moving them, and all six sides touched the edge of the pond. The total perimeter of the new hexagonal fence was precisely half of that of the original triangular fence.

What were the lengths of the three original straight pieces of fencing?


Enigma 1098: Soccer heroes

From New Scientist #2254, 2nd September 2000 [link]

There are seven teams in our local football league. Each team plays each of the others once during the season. We are approaching the end of the season and I have constructed a table of the situation so far, with the teams in alphabetical order.

Here are the first two rows of the table, but with digits consistently replaced by letters, different letters being used for different digits.

What was the score when Albion played Borough?


Enigma 413: Quargerly dues

From New Scientist #1563, 4th June 1987 [link]

A native of Kipwarm had a gold necklace consisting of links joined together to form one long unbroken loop of chain.

He has fallen on hard times and to pay his gas bill he is going to give the gas board one link of his gold necklace every day.

He has broken just a certain number of links (thus forming that number of individual links and some other variously-sized pieces of chain). By giving away and taking back certain pieces he can ensure that, at the rate of one a day, his total of links decreases and the board’s increases. Furthermore, had his necklace had one more link, it would have been necessary to break one more in order to pay the board in this way.

His necklace will last him a whole number of quargers (a Kipwarmian period of a certain number of days, less than one year). But the board has offered him an alternative way of paying. His first quarger’s gas will be free, the next will cost him one link, the next two links, the next quarger’s will cost him four links, and so on, doubling each quarger. At that rate the necklace will pay for the same number of days’ gas.

How many days are there in a Kipwarmian quarger?


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 [link]

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 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?