MathDB
a sequence of fractions construction

Source: Dutch NMO 2022 p3

November 17, 2022
algebrasimplifysimplificationnumber theory

Problem Statement

Given a positive integer cc, we construct a sequence of fractions a1,a2,a3,...a_1, a_2, a_3,... as follows: \bullet a1=cc+1a_1 =\frac{c}{c+1} \bullet to get ana_n, we take an1a_{n-1} (in its most simplified form, with both the numerator and denominator chosen to be positive) and we add 22 to the numerator and 33 to the denominator. Then we simplify the result again as much as possible, with positive numerator and denominator. For example, if we take c=20c = 20, then a1=2021a_1 =\frac{20}{21} and a2=2224=1112a_2 =\frac{22}{24} = \frac{11}{12} . Then we find that a3=1315a_3 =\frac{13}{15} (which is already simplified) and a4=1518=56a_4 =\frac{15}{18} =\frac{5}{6}. (a) Let c=10c = 10, hence a1=1011a_1 =\frac{10}{11} . Determine the largest nn for which a simplification is needed in the construction of ana_n. (b) Let c=99c = 99, hence a1=99100a_1 =\frac{99}{100} . Determine whether a simplification is needed somewhere in the sequence. (c) Find two values of cc for which in the first step of the construction of a5a_5 (before simplification) the numerator and denominator are divisible by 55.