Question 179
Given a positive integer $c$, we construct a sequence of fractions $a_1, a_2, a_3, \dots$ as follows:
* $a_1 = \frac{c}{c+1}$;
* 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.
* $a_1 = \frac{c}{c+1}$;
* 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.