minor fixes

exams/fermat-primality/index.md

@@ -26,13 +26,13 @@ respective number, are avoided when testing your implementation of this task.
 To generate pseudorandom numbers in a given interval, use 
 the *Linear Congruential Generator (LCG)* 
 $$
-  x_{n+1} = (A x_n + C) \ \texttt{mod }M,
+  x_{n+1} = (A x_n + C) \ \texttt{mod}\, M,
 $$
 where $A$, $C$ and $M$ are constants. This equation generates the next pseudorandom number $x_{n+1}$ from the previous $x_n$. The number $x_0$ is the seed.
 
 The number $b$ drawn from $(1)$ can be transformed to the interval $b^\text{lower} \leq b' < b^\text{upper}$ as 
 $$
-  b' = (b \ \texttt{mod } (b^\text{upper} - b^\text{lower})) + b^\text{lower}. 
+  b' = (b \ \texttt{mod}\, (b^\text{upper} - b^\text{lower})) + b^\text{lower}.
 $$
 
 

exams/minimum-spanning-tree/index.md

@@ -13,7 +13,7 @@ $E'\subseteq E$, $T$ is a tree (i.e., a connected graph without cycles) and $\su
 minimum possible among such trees. The figure shows an example of a connected weighted
 graph and its minimum spanning tree.
 
-![Left: A connected, weighted graph. Right: Its minimum spanning tree of weight 16.](/img/minimum-spanning-tree-graph.svg){class="inverting-image"}
+![Left: A connected, weighted graph. Right: Its minimum spanning tree of weight 16.](/img/minimum-spanning-tree-graph.svg){style="width: 100%; margin: auto", class="inverting-image"}
 
 Your task is to implement an algorithm computing the minimum spanning tree, i.e.,
 a function returning for a given connected weighted graph $(V,E)$ the subset $E'$

exams/sierpinski-carpet/index.md

@@ -10,7 +10,7 @@ outline: deep
 The *Sierpiński carpet* is a plane fractal first described by Wacław Sierpiński in 1916.
 Your task is to generate this fractal in a text format represented as a list of strings. 
 Each string represent a single row in the picture. The picture $f(n)$ is defined recursively.
-For $n=0$, we define $f(0)="\#"$. For $n>0$, we define $f(n)$ as the picture depicted below.
+For $n=0$, we define $f(0) = \texttt{"\#"}$. For $n>0$, we define $f(n)$ as the picture depicted below.
 In other words, $f(n)$ consists of eigth copies of $f(n-1)$ and 
 the middle box of the same size as $f(n-1)$ filled with spaces.