The Alan Turing Cryptography Competition

2023 edition. From the people behind the MathsBombe Competition.
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Solutions

Chapter 1

The images represent letters from the NATO phonetic alphabet. The images represent:
  • A glass of WHISKEY
  • A HOTEL from Monopoly
  • The Greek letter ALPHA (in the NATO phonetic alphabet, A is represented by Alfa - an alternative spelling of Alpha to ensure that it is pronounced as intended)
  • A can of TANGO
  • The flag of INDIA
  • The flag of SIERRA Leone
  • An Amazon ECHO
  • A map of INDIA
  • A woman playing GOLF
  • A HOTEL resort
  • A can of TANGO
  • A can of TANGO
  • The flag of INDIA
  • MIKE from Monsters Inc
  • A diagram of an ECHO
  • The macOS SIERRA desktop
  • Two people dancing the FOXTROT
  • OSCAR the Grouch from Sesame Street
  • School UNIFORMs
  • A silhouette of ROMEO from Romeo and Juliet
These spell: WHATISEIGHTTIMESFOUR. Eight times four is 32, which is the solution.

Chapter 2

The grid of letters are coloured in a way to highlight the word 'PRIMES'. If one takes the letters occurring in the prime places (so the H in place 2, the O in place 3, the W in place 5, etc), then one obtains the plaintext: how many prime numbers are less than 150?

There are 35 primes less than 150.

Chapter 3

This message is encoded using a substitution cipher, but the slight catch is that the letter e has been removed from all the words in the message. There is still enough information to use frequency analysis, and the plaintext is:

a mixd class at at primary school has a numbr of childrn of diffrnt ags th man of thir ags is nin and a half th youngst child is fiv and th ldst child is lvn th answr that w sk is th minimum numbr of childrn that could b prsnt in th class multiplid by minus on point two thr you should assum that th ags ar intgrs or whol numbrs

which, once the e's have been added back in, reads

a mixed class at at primary school has a number of children of different ages the mean of their ages is nine and a half the youngest child is five and the eldest child is eleven the answer that we seek is the minimum number of children that could be present in the class multiplied by minus one point two three you should assume that the ages are integers or whole numbers

We now need to solve the problem. There must be at least two children because we are told that one child is five and one is eleven. If there are only these two children in the class then the mean would be (5 + 11)/2, which is eight, but we are told that the mean is nine. This means that there must be more than two children in the class.

If we let the total number of children be n and the sum of the ages of the remaining n-2 children be x, then we have that

\[ 9.5n = 16 + x. \]

We are also told that the children's ages are all integers, which means that 16 + x is an integer and so n must be even. We can, therefore, write that n = 2m, where m is greater than 1, and so

\[ 19m = 16 + x.\]

The maximum possible age of each child is 11 and so the maximum value that x can take is $11(n-2) = 11(2m -2) = 22(m-1)$, so if we assume that x does take its maximum then,

\[19m = 16 + 22(m-1) \Rightarrow -3m = -6 \Rightarrow m = 2. \]

Hence, the minimum number of children is four (11, 11, 11, 5) and so multiplying that by $-1.23$ gives $-4.92$.

Chapter 4

The numbers in the dates tell you which letters/punctuation from the corresponding message to take. Spaces are excluded from the count. For example, for '05/22/23 Theres something out of equilibrium here', take the 5th, 22nd and 23rd letters to obtain EQU. The plaintext is 'Which triangle has all sides equal?'. The answer is 'Equilateral'.

Chapter 5

The messages is encoded using a variant of a Vigenere cipher, which uses different shifts of the alphabet to encode each letter. For example, if we have a shift of one then ABC would becomed BCD; but, if we have a shift of ten then ABC becomes KLM. In this case Mike has used his own name as a codeword with a negative shift corresponding to the numerical value of each letter, i.e. M would be a shift of 13 places backwards in the alphabet. The decoded message reads:

THE ANSWER WE SEEK IS A COMMA SEPARATED LIST OF THE THREE ROOTS OF THE EQUATION X CUBED PLUS TWO X SQUARED MINUS FOUR X PLUS ONE EQUALS ZERO YOU MUST LIST THE ROOTS IN ORDER OF INCREASING MAGNITUDE YOU SHOULD WRITE EACH ROOT TO TWO DECIMAL PLACES

This means that we need to solve the cubic equation

\[ x^3 + 2x^2 -4x + 1 = 0.\]

Hopefully, you can spot that x=1 is one solution. We can then factorise the cubic to obtain a quadratic that can be solved by the quadratic formula. The three roots are \[ x = 1,\; -3/2 - \sqrt{13/2} \quad \mbox{and}\quad -3/2 + \sqrt{13/2}. \] The question asks us to enter the roots as a comma separated list in increasing magnitude. The magnitude of a real number is its distance from zero, i.e. the magnitude of 2 is 2 and the magnitude of -2 is also 2. The question also asks us to work to two decimal places. Thus, the required answer is "0.30, 1.00, -3.30"

Chapter 6

The first hidden question is simply given by reading across the grid of letters on the left "What is the area of a circle of radius nine m divided by pi?" The answer is of course, 81 metres squared, so the required number is 81.

The second hidden question is harder, and relies on spotting that each pair of numbers in brackets represents a type of polar coordinate: the first number is the distance (squared) from the bottom left-hand corner of the letter grid, the second number is the direction in degrees with zero along the horizontal direction and 90 in the vertical. For example the first pair (25,90), corresponds to the letter at the Cartesian coordinate (5,0), which is W. Once you've work this out then after some calculation, you can find the second hidden message: What is the product of the first ten primes?

The first ten primes are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29 and their product is 6469693230. This is the second number required and it is clearly the larger.

The final answer is then 6469693230/81 to three decimal places, which gives: 79872755.926

Alan Turing Cryptography Competition 2023 is organised by the The Department of Mathematics at The University of Manchester.
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