Päevaülesanne

Juhuslik

Harjutuskomplekt

Balti Tee 1996 · Ülesanne 8

Arvuteooria

Consider the sequence

x1=19,x2=95,xn+2=lcm⁡(xn+1,xn)+xn,\begin{aligned} x_{1} & =19, \\ x_{2} & =95, \\ x_{n+2} & =\operatorname{lcm}\left(x_{n+1}, x_{n}\right)+x_{n}, \end{aligned}

for n>1n>1, where lcm⁡(a,b)\operatorname{lcm}(a, b) means the least common multiple of aa and bb. Find the greatest common divisor of x1995x_{1995} and x1996x_{1996}.

Muuda valikut

Kui oled valmis

Ülevaatematerjal muutub kättesaadavaks järgmise päevaülesannete komplektiga.

Ülevaade

Teemad

SÜT ja VÜK · Jaguvus ja tegurdamine

Lahendused

Lahendus

Solution:

Let d=gcd⁡(xk,xk+1)d = \operatorname{gcd}\left(x_{k}, x_{k+1}\right). Then lcm⁡(xk,xk+1)=xkxk+1/d\operatorname{lcm}\left(x_{k}, x_{k+1}\right) = x_{k} x_{k+1} / d, and

gcd⁡(xk+1,xk+2)=gcd⁡(xk+1,xkxk+1d+xk)=gcd⁡(xk+1,xkd(xk+1+d)).\operatorname{gcd}\left(x_{k+1}, x_{k+2}\right) = \operatorname{gcd}\left(x_{k+1}, \frac{x_{k} x_{k+1}}{d} + x_{k}\right) = \operatorname{gcd}\left(x_{k+1}, \frac{x_{k}}{d}\left(x_{k+1} + d\right)\right).

Since xk+1x_{k+1} and xk/dx_{k} / d are relatively prime, this equals gcd⁡(xk+1,xk+1+d)=d\operatorname{gcd}\left(x_{k+1}, x_{k+1} + d\right) = d. It follows by induction that gcd⁡(xn,xn+1)=gcd⁡(x1,x2)=19\operatorname{gcd}\left(x_{n}, x_{n+1}\right) = \operatorname{gcd}\left(x_{1}, x_{2}\right) = 19 for all n≥1n \geq 1. Hence gcd⁡(x1995,x1996)=19\operatorname{gcd}\left(x_{1995}, x_{1996}\right) = 19.

Võistluse kontekst

Balti Tee tulemused 1996

10 võistkonda

Keskmine tulemus
4,0 / 5
4 või 5 punkti
8 / 10
Eesti
5 / 5

Punktijaotus

02
10
20
30
40
58
Kõigi võistkondade punktid
VõistkondPunktid
Poland5 / 5
Latvia5 / 5
Sweden5 / 5
Denmark5 / 5
St. Petersburg5 / 5
Finland5 / 5
Norway0 / 5
Lithuania5 / 5
Estonia5 / 5
Iceland0 / 5