The Alan Turing Cryptography Competition

2024 edition. From the people behind the MathsBombe Competition.
Home Archive

Solutions

Chapter 1

The message simply written downwards in the grid reading from left to right.

THE ANSWER REQUIRED IS SIMPLE TO SEE FIND THE PRIME FACTORS OF THREE HUNDRED AND THREE ADD THEM TOGETHER DIVIDE BY TWO THEN SQUARE THE RESULT THATS WHAT YOU MUST DO IF YOUR WORK IS CAREFUL AND IT MUST ALWAYS BE YOULL FIND OUR ANSWER ON DIVIDING BY THREE ENTER YOUR ANSWER TO THREE DECIMAL PLACES AND THEN YOU WILL SEE A BIG SMILE ON OUR FACES

The prime factors of 303 are 3 and 101. Adding them together gives 104, dividing by 2 and squaring the answer gives 2704. Finally, 2704รท3 is 9011/3, which is 901.333 to three decimal places.

Chapter 2

The message is a substitution cipher with each emoji representing a different letter. The original message is given below:

I WONDER IF YOU CAN ANSWER THIS QUESTION THE BASE OF A RIGHT ANGLED TRIANGLE IS TWELVE UNITS LONG THE HEIGHT OF THE TRIANGLE IS NINE UNITS HIGH USING THE PYTHAGORAS THEOREM WHAT IS THE LENGTH OF THE HYPOTENUSE OF THIS TRIANGLE.

The Pythagoras Theorem states that the square of the length of the hypotenuse of a right-angled triangle is equal to the sum of the squares of the lengths of the other two sides. If we let the length of the hypotenuse by H, then by the Pythagoras Theorem:

H^2 = 12^2 + 9^2 = 144 + 81 = 225

and so the length of the hypotenuse is 15.

Chapter 3

The message is written as a series of three words, each separated by a dot. This is the format used by what3words, a system for finding locations on the planet. If you type the three word combinations into the what3words web site, and you zoom the map out to an appropriate resolution you will find that the sets of words indicate the following places in the UK:

Whatley Isfield Forty Green Takeley Sevenoaks

If you now take the first part of each place name, you will spot the question:

WHAT IS FORTY TAKE SEVEN

The answer is, of course, 33.

Chapter 4

The code is based on a numerical representation of the letters of the alphabet, 1=A, 2=B, etc, but the numbers are not all written in base 10. If you are familar with binary, base 2, which uses only 0 and 1, you might suspect that a few of the numbers are binary, but what about the others? The binary numbers occur roughly every nine numbers, suggesting a repeating pattern of some sort.

In fact, the first number is binary, the second is in base 3, the third is in base 4 and so on up to base 10 and then the pattern repeats. Converting the numbers into base 10, we have

6-9-14-4-20-8-5-16-15-19-9-20-9-22-5-18-15-15-20-15-6-25-5- 17-21-1-12-19-20-8-18-5-5-25-3-21-2-5-4-26-26-26-26-26-26

and converting the numbers into letters reveals the message

FINDTHEPOSITIVEROOTOFYEQUALSTHREEYCUBEDZZZZZZ

The final Zs are just padding and so we need to solve the cubic equation

$$y = 3 y^3$$

One solution is y=0 and assuming that we don't have that solution, we can divide by y to obtain

$$1 = 3 y^2 \quad \text{or} \quad y^2 = 1/3$$

The answer required is the positive root which is $y = 1/\sqrt{3}$ and to five decimal places this is 0.57735.

Chapter 5

The code used is different in each row of the crossword. In the first row, the letters are shifted forwards one place in the alphabet. In the second row, the letters are shifted forwards two places in the alphabet and so on. The last, eighth, row contains numbers and each digit is also shifted forwards eight places in the set of digits, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 0, 1, 2, 3, ....

Applying the required shifts to each row, the original message is revealed:

WHAT IS THE AREA OF A REGULAR HEXAGON WITH PERIMETER 33CM

The formula for the area of a hexagon is $$\frac{3 \sqrt{3}}{2} l^2,$$ where $l$ is the length of one side of the hexagon. The formula can be easily found by decomposing the regular hexagon into six equilateral triangles. The length of each side is 33/6 = 5.5 cm. This means that the area is 78.59181 to five decimal places.

Chapter 6

Careful examination of the letters reveals that there aren't any vowels, but there are numbers above the letters, which take the values 1 to 5. Hmmm, could the numbers represent the vowels? Making the guess that the numbers correspond to the vowels in alphabetical order and putting the appropriate vowel(s) after each of the consonants with numbers above them gives:

    ASPAXORIHWOPIJIWARCO
    KXLEKIYEGEOQEXVIHWOPIJWZ
    LOWEKAHEWAVEAPPIJERXIHA
    PVEGUPAVSOPCGORWQEEXIRG
    AXXLEWAQEXLVEEJIQERWIORAP
    ARGPEWAVEGUPAVJOJEHALEJ
    VORIWOREOKXLESPAXORIH
    WOPIJWARJIXWKAHEWAVE
    VEGUPAVSERXAGORWLOZQARC
    WUHLSERXAGORWAVEREEJEJXO
    KOVQAVEGUPAVJOJEHALEJVOR 

That still doesn't make a lot of sense, but it now looks more like a substitution cipher. Using standard methods we can find the following plaintext:

A PLATONIC SOLID IS ANY OF THE FIVE GEOMETRIC SOLIDS WHOSE FACES ARE ALL IDENTICAL REGULAR POLYGONS MEETING AT THE SAME THREE DIMENSIONAL ANGLES. A REGULAR DODECAHEDRON IS ONE OF THE PLATONIC SOLIDS AND ITS FACES ARE REGULAR PENTAGONS. HOW MANY SUCH PENTAGONS ARE NEEDED TO FORM A REGULAR DODECAHEDRON.

In fact, twelve pentagonal faces are required to make a dodecahedron: dodeca- is a prefix meaning "having twelve". The required answer is 12.00 because we were asked to enter numerical values to two decimal places.

Alan Turing Cryptography Competition 2024 is organised by the The Department of Mathematics at The University of Manchester.
© The University of Manchester 2012–2024, All Rights Reserved
Contact us | Privacy notice