Problems
Does there exist a quadrilateral which can be cut into six parts with two straight lines?
Cut a square into \(3\) parts which you can use to construct a triangle with angles less than \(90^{\circ}\) and three different sides.
Find all rectangles that can be cut into \(13\) equal squares.
Cut a square into two equal:
1. Triangles.
2. Pentagons
3. Hexagons.
Find one way to encrypt letters of Latin alphabet as sequences of \(0\)s and \(1\)s, each letter corresponds to a sequence of five symbols.
Using the representation of Latin alphabet as sequences of \(0\)s and \(1\)s five symbols long, encrypt your first and last name.
Decipher the quote from "Alice in Wonderland" from the following matrix:
\[\begin{array}{@{}*{26}{c}@{}}
Y&q&o&l&u&e&c&d&a&i&n \\
w&a&r&l&a&w&e&a&t&y&k \\
s&n&t&c&a&e&k&c&e&a&m \\
t&o&d&r&w&e&a&t&a&h&r \\
a&c&n&t&n&e&o&d&t&r&h \\
n&i&d&n&l&g&m&e&x&s&z
\end{array}\]
Decipher the following quote from Alice in Wonderland:
Lw zrxog eh vr qlfh li vrphwklqj pdgh vhqvh iru d fkdqjh.
The same letters correspond to the same in the phrase, different letters correspond to different. We know that no original letters stayed in place, meaning that in places of e,r,h there was surely something else.
After mastering the Caesar shift cypher one may wonder how to generalize it. One possible way is to use Affine cypher. The difference between these two methods can be described as follows:
In case of Caesar cypher we took a letter with position \(n\) from \(1\) to \(26\) and added to its position a number \(d\) obtaining the number \(n+d\), then we compute its residue modulo \(26\).
In case of affine cypher we take a letter with position \(n\) and consider a number \(nx + d\) modulo \(26\).
To decipher such code we need to know values \(x\) and \(d\), then if we have a letter in the code with position \(m\), we can find \(n\) as \(n= (m-d)x^{-1}\) modulo \(26\). Here we have to explain what is \(x^{-1}\): for a number \(x < 26\) we are looking for such a number \(y\), that \(26\) divides \(xy-1\).
Does there always exist a number \(x^{-1}\) modulo \(26\) for any \(x\)?
Using data \(x=3\), \(d=8\) encrypt the word "SOLUTION".