This page gives you help for each of the questions. You can se this page when you are learning to play the game but not in the competiion.
This is the easiest question but you need to know a little bit about prime numbers. Prime numbers are numbers that do not have any factors (except 1 and the number itself). The first ten prime numbers are (2,3,5,7,11,13,17,23,29,31). There are an infinite number of primes.
Find two prime numbers that add up 32.
This question also uses prime numbers, but this time the numbers are multipled together instead of added. It looks difficult, but there sre things you can check to make the question easier. This example will show you how to do that.
Find two prime numbers that multiply together to make 437.
To solve the problem you need to divide 437 by smaller prime numbers until you find one which divides into 437 without any remainder. However you don't need to try every prime number less than 437. You do it like this.
This question uses square numbers. To do this question you need to find $x^2$ and then find the $\sqrt{x^2}$.
Find $x$ where $4x^2 + 1 = 677 $.
$4x^2 = 677 - 1 $
$4x^2 = 676 $
$x^2 = 676 \div 4 $
$x^2 = 169 $
$x = \sqrt{169} $
x = 13
To find the square root of a number use the list of square numbers. the first row has the squares from 1 - 9, the second row the squares from 10 - 19 and so on.
This question also uses square numbers. To do this question you need to find two square numbers that differ by n.
Example - Find x and y where $x^2 - y^2 = 39 $ .
The easiest way to. solve this problem is to use the equation $x^2 = 39 + y^2 $ and add a square number to 39 until you find one where $x^2 + 39 $ is also a square number.
To solve this puzzle you need to look at the letters that are before and after each of the code letters.
Example BBS
B B S
A A T
C C U
The word could be ACT or CAT and you need to try each one.With longer words there is usually only one choice.
To solve this puzzle you need to know about numbers raised to the power of 2, for example 1,2,4,8,18,32,64,128,256,512,1024 etc. are
$2^0, 2^1, 2^2,2^3,2^4,2^5,2^6,2^7,2^8,2^9,2^10 $
Find n where $2^n -1 = 4095 $
To find n you need to solve the equation
$2^n = 4096 $
To do that you need to keep dividing 4096 by 2 until you reach 1.
4096 = 2 x 2048
4096 = 2 x 2 x 1032
4096 = 2 x 2 x 2 x512
4096 = 2 x 2 x 2 x 2 x 256
4096 = 2 x 2 x 2 x 2 x 12 x 28
4096 = 2 x 2 x 2 x 2 x 2 x 2 x 64
4096 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 32
4096 = $2^7 x 2^5 $ (32 = $2^5$)
4096 = $2^{12}$
n = 12
For larger numbers however there is a quicker way.
Example $2^n -1 = 4194305 $
$2^n = 4194305 $
Divide 4194305 by $2^{10} $ ($2^{10} = 1024 $)
4194305 $ \div $ 1024 = 4096
4194305 = $ 4096 \times 2^{10} $
4096 = 1024 x 4 = $2^{10} \times 2^2 = 2^{12} $
4194305 = $2^{10} \times 2^{12} = 2^{(10 + 12)} = 2^{22}$
n = 22
To solve this puzzle you need to be able to solve 3 equations. To do that you need to add or subtract the equations or replace some of the variables. There are different ways to do it but for the equations in the puzzle the easiest way is to add the equations together and then replace A + B.
A + B = 12 (1)
2A + C = 14 (2)
C + 2B = 10 (3)
Addding them together gives
A + B + 2A + C + C + 2B = 12 + 14 + 10
3A + 3B + 2C = 36
3(A+B) + 2C = 36
we know that A + B = 12 (1)
3 x 12 + 2C = 36, 36 + 2C = 36
2C = 0 , C = 0
2A + C = 14 (2)
2A + 0 = 14, A = 7
A + B = 12 , 7 + B = 12 , B = 5
A = 7, B = 5, C = 0 , A + B + C = 12
In this puzzle you need to solve two equations by adding them together and solving the new equation.
$m^3 + n^3 = 189 $ (1)
$m^3 - n^3 = 61 $ (2)
Add the two equations together
$2m^3 = 189 + 61 $
$m^3 = 250 \div 2 $
$m^3 = 125 $
m = 5. (use the cubes table to find the $ \sqrt[3]{125} $
To find the cuberoot of a number use the list of cubes . the first row has the cubes from 1 - 9, the second row the cubes from 10 - 19 and so on.
This puzzle uses fractions that were used thousands of years ago in Egypt. The numerators in Egyptian fractions are always 1 and the denominators are different from one another.
Find a,b and c where a,b and c are all prime numbers and $ \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{103}{165} $
To solve the problem you need to find a,b,c where $a \times b \times c = 165 $. What makes it easy is that a,b and c are primes. So you need to find primes that divide into 165 without a remainder.
The prime numbers are 2,3,5,7,11,13,17,..
165 $ \div 2 $ = 82 + 1 remainder X
165 $ \div 3 $ = 53 + 0 remainder ✓
165 $ \div 5 $ = 33 + 0 remainder ✓
165 $ \div 7 $ = 23+ 4remainder X
165 $ \div 11 $ = 15 + 0 remainder ✓
a = 3, b = 5, c = 11
Check answer
$ \frac{1}{3} + \frac{1}{5} + \frac{1}{11} = \frac{5 \times 11 + 3 \times 11 + 3 \times 5 }{3 \times 5 \times 11}
= \frac{55 + 33 + 15}{165} = \frac{103}{165}$ ✓
This is the final puzzle in the game and the hardest. To solve the problem you need to use clock arithmetic when calculating the letter for each code.
Calculate $15^3 $ and then find the remainder when you divide the answer by 33 to find the remainder.
$15^3 = 3375 $
$ 3375 \div 33 $ = 102 + 9 remainder.
The 9th letter in the alphabet is I so 15 -> I
Use the same method for the other numbers that need to be decoded in the puzzle.