The message is written in a clockwise spiral starting from the top row and working around the grid.

The plaintext is: WHAT IS THE UNIQUE SOLUTION TO THE EQUATION X SQUARED MINUS 2X PLUS 1 EQUALS 0?

This is a quadratic equation

$$ X^2 -2X +1 = (X-1)(X-1) = 0.$$

Therefore, the equation has two repeated roots X = 1, so the answer is 1.

Looking at the vertical axis (the ordinate), you can see that there are 26 marks. This suggests that each distinct vertical position corresponds to a different letter. Making the guess that this is a standard cipher 1=A, 2=B, and so on; and that the letters run from left to right then the message is revealed. The additional circle around each point corresponds to the start of a word.

The plaintext is: WHAT IS THE INTERIOR ANGLE OF A REGULAR DODECAGON.

A dodecagon is a polygon with 12 sides. The interior angle in degrees of a regular polygon with $N$ sides is given by the formula $180 - 360/N$. Hence, the answer is $180 - 360/12 = 180 - 30 = 150$.

Looking at the image, it appears that each symbol could represent a letter. Start with the symbols that appear on their own, could the zebra be I or A? A is more likely. Frequency analysis can also help. Looks like the dragon appears quite frequently, could it be the E? Working through the rest of the symbols reveals the message.

The plaintext is: A STRAIGHT LADDER IS LEANING AGAINST A VERTICAL WALL. THE TOP OF THE LADDER TOUCHES THE WALL AT A HEIGHT OF TWELVE METRES AND THE BOTTOM OF THE LADDER IS FIVE METRES AWAY FROM THE BASE OF THE WALL ON FLAT GROUND. THE BOTTOM OF THE LADDER IS NOW MOVED SO THAT IT IS SEVEN METRES FROM THE WALL. HOW HIGH UP THE WALL TO THE NEAREST CENTIMETRE WILL THE LADDER REACH?

In order to solve this problem, we need to work out the length of the ladder, which we can do using the Pythagoras Theorem based on the information about the initial position. We have the base length and the height and the length of the ladder is the hypotenuse, $\sqrt{12^2 + 5^2} = 13m$.

Once the ladder has been moved, we can use Pythagoras again, but now we have the base length and the hypotenuse and need to work out the height. This is

$$ \sqrt{13^2 - 7^2} = \sqrt{120} = 10.95m.$$

We are asked for the answer in centimetres, so the required answer is 1095.

This looks like slightly jumbled English. There are some words in there, but also things that don't make sense. From the clue, there seems to be something significant about the number three. Could this be a permutation cipher?

The answer is yes. Each group of three characters (including spaces) has been permuted by moving the first character to the back of the group. To unscramble the message, simply move every third character to the front of the previous two characters.

The plaintext is: USING ONLY INTEGERS BETWEEN ONE AND ONE HUNDRED WHAT IS THE LARGEST PROPER FRACTION THAT CAN BE WRITTEN WITHOUT USING THE LETTER E?

A proper fraction has a smaller numerator than a denominator, so it is less than one. In order to get the largest value we need to make the numerator as close to the denominator as possible. This suggests choosing a large denominator.

Clearly anything in the nineties, eighties or seventies is out, so that leaves the sixties and the highest number without an "e" is sixty-six. The largest proper fraction without an "e" with 66 as the denominator is $\frac{64}{66}$, so that is a plausible answer. However, although sixty-five contains an "e" sixty-fifths does not and $\frac{64}{65} > \frac{64}{66}$. Therefore, the answer is sixty-four sixty-fifths or $64/65$.

The clue mentions an evening stroll or night walk; and the $8 \times 8$ grid looks like a chessboard. This is a reference to a knight's tour : a classic problem in which the challenge is to visit every square of the chessboard using the move that a knight would make.

There are a large number of possible knight's tours, so it's helpful to know the starting point. The clue here is in the paper sizes, which indicate the squares using standard chess notation: the columns are labelled from A to H and the rows are labelled from 1 to 8.

Starting from A5 (W) and then moving to B7 (H) and following knight's moves after lots of trial and error reveals the message. It may be quicker to use a computer program to help with this....

The plaintext is: WHAT IS THE PRODUCT OF THE NUMBERS OF CITIES IN THE UK AND INCHES IN A MILE VWXYZ.

The final five characters are padding to make the message exactly 64 letters long.

The number of cities in the UK is 76 and the number of inches in a mile is 63360, giving a final answer of $76 \times 63360 = 4815360$.

What is going on here? The numbers could correspond to letters, but why are some of the times AM and some PM? The answer is that this is a polyalphabetic cipher and the AM and PM indicate how to shift the letters that are represented by the numbers in the time.

Each time corresponds to two letters. One letter represented by the hour and one by the minute. The numbers without AM or PM simply represent letters in the usual way A=1, B=2 and so on. For minutes greater than 26 the letters just cycle around again, so 27=A. In other words, the first time 23:08 corresponds to the two letters WH

The AM or PM indicates that the original letter is shifted forward in the alphabet by 1(A), 13(M) or 16(P) places. The first letter corresponds to the shift of the letter represented by the hours and the second by the minutes. Looking at the second time 11:07PM an applying these shifts gives 27:20, corresponding to the letters AT. Using the fact that the first word is likely to be WHAT gives a clue to help work out what is going on with the shifts.

Applying this approach to all the times reveals the plaintext.

The plaintext is: WHAT IS THE AREA OF A TRIANGLE WITH SIDES OF LENGTHS PI E AND THE SQUARE ROOT OF TWO TO THREE DECIMAL PLACES.

In order to find the area of triangle from all three sides, Heron's formula can be used. For three sides of length $a$, $b$ and $c$ the formula gives the area as:

$$ A = \sqrt{s(s-a)(s-b)(s-c)},$$

where $s = (a + b + c)/2$.

Using this formula the required area is found to be $1.918$.

Alan Turing Cryptography Competition 2025 is organised by the The Department of Mathematics at The University of Manchester.
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