Given a positive integer $c$, we construct a sequence of fractions $a_1, a_2, a_3,...$ as follows: $\bullet$ $a_1 =\frac{c}{c+1} $ $\bullet$ to get $a_n$, we take $a_{n-1}$ (in its most simplified form, with both the numerator and denominator chosen to be positive) and we add $2$ to the numerator and $3$ to the denominator. Then we simplify the result again as much as possible, with positive numerator and denominator. For example, if we take $c = 20$, then $a_1 =\frac{20}{21}$ and $a_2 =\frac{22}{24} = \frac{11}{12}$ . Then we find that $a_3 =\frac{13}{15}$ (which is already simplified) and $a_4 =\frac{15}{18} =\frac{5}{6}$. (a) Let $c = 10$, hence $a_1 =\frac{10}{11}$ . Determine the largest $n$ for which a simplification is needed in the construction of $a_n$. (b) Let $c = 99$, hence $a_1 =\frac{99}{100}$ . Determine whether a simplification is needed somewhere in the sequence. (c) Find two values of $c$ for which in the first step of the construction of $a_5$ (before simplification) the numerator and denominator are divisible by $5$.

Find the smallest natural number $x> 0$ so that all following fractions are simplified $\frac{3x+9}{8},\frac{3x+10}{9},\frac{3x+11}{10},...,\frac{3x+49}{48}$ , i.e. numerators and denominators are relatively prime.

Simplify the fraction: $\frac{(1^4+4)\cdot (5^4+4)\cdot (9^4+4)\cdot ... (69^4+4)\cdot(73^4+4)}{(3^4+4)\cdot (7^4+4)\cdot (11^4+4)\cdot ... (71^4+4)\cdot(75^4+4)}$.

Define $a_k=2^{2^{k-2013}}+k$ for all integers $k$. Simplify $$(a_0+a_1)(a_1-a_0)(a_2-a_1)\cdots(a_{2013}-a_{2012}).$$

If the fraction $\frac{an + b}{cn + d}$ may be simplified using $2$ (as a common divisor ), show that the number $ad - bc$ is even. ($a, b, c, d, n$ are natural numbers and the $cn + d$ different from zero).