Problem K/C. 883. (December 2025)

K/C. 883. A sequence is obtained with the following recursive method: if member \(\displaystyle t\) is a positive odd number, then the next member is \(\displaystyle 3t-9\), if it's a positive even number, then the next member is \(\displaystyle 2t-7\). Suppose that the sequence alternates between two positive values: \(\displaystyle a\), \(\displaystyle b\), \(\displaystyle a\), \(\displaystyle b\), \(\displaystyle \ldots\) Find the possible values of \(\displaystyle a\) and \(\displaystyle b\).

(5 pont)

Deadline expired on January 12, 2026.

Sorry, the solution is available only in Hungarian. Google translation

Megoldás. Ha \(\displaystyle a\) páros, akkor \(\displaystyle b = 2a – 7\), ami páratlan, ezért \(\displaystyle a = 3(2a – 7) – 9\), ahonnan \(\displaystyle a = 6a – 30\), azaz \(\displaystyle a = 6\), és \(\displaystyle b = 12 – 7 = 5\). Ha \(\displaystyle a\) páratlan, akkor \(\displaystyle b = 3a – 9\), ami páros, ezért \(\displaystyle a = 2(3a – 9 ) – 7\), ahonnan \(\displaystyle a = 6a – 25\), azaz \(\displaystyle a = 5\), és \(\displaystyle b = 15 – 9 = 6\). Tehát két megfelelő sorozat van: 5, 6, 5, 6, ... és 6, 5, 6, 5, ...

Statistics:

263 students sent a solution.
5 points:	172 students.
4 points:	50 students.
3 points:	19 students.
2 points:	13 students.
1 point:	2 students.
0 point:	1 student.
Not shown because of missing birth date or parental permission:	5 solutions.

Problems in Mathematics of KöMaL, December 2025