Mathematical and Physical Journal
for High Schools
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Problem K. 602. (November 2018)

K. 602. Andrew and Paul are playing a game. The winner is always awarded \(\displaystyle x\) points and the loser always gets \(\displaystyle y\) points (where \(\displaystyle x>y\) are integers). There is no draw. After a few rounds, we observe that Andrew has 30 points and Paul has 25 points since Paul has only won twice. How many points are awarded to the winner?

(6 pont)

Deadline expired on December 10, 2018.

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

Megoldás. Ha \(\displaystyle n\)-szer játszottak összesen, akkor Andrásnak \(\displaystyle (n–2)x+2y = 30\), Palinak \(\displaystyle 2x+(n–2)y=25\) pontja van. Vonjuk ki az első egyenletből a másodikat: \(\displaystyle (n-2)x+2y – (2x+(n–2)y) = 5\). A bal oldalt rendezve:

\(\displaystyle (n–2)x–(n–2)y–2x+2y = (n–2)(x–y)–2(x–y) = (n–2–2)(x–y) = (n–4)(x–y) = 5.\)

Mivel \(\displaystyle x>y\) egész számok, így \(\displaystyle n–4 = 5\) és \(\displaystyle x–y = 1\) vagy \(\displaystyle n–4 = 1\) és \(\displaystyle x–y = 5\), azaz \(\displaystyle n = 9\) és \(\displaystyle y = x–1\) vagy \(\displaystyle n = 5\) és \(\displaystyle y = x–5\). Az első esetben András pontjai \(\displaystyle 7x+2(x–1)=30\), azaz \(\displaystyle 9x = 32\), ami nem lehet, ha \(\displaystyle x\) egész. A második esetben \(\displaystyle 3x+2(x–5)=30\), azaz \(\displaystyle 5x=40\), ahonnan \(\displaystyle x=8\), \(\displaystyle y=3\), ami mindhárom egyenletnek megfelel. Tehát a nyertes \(\displaystyle 8\) pontot kap körönként.


172 students sent a solution.
6 points:58 students.
5 points:31 students.
4 points:12 students.
3 points:18 students.
2 points:12 students.
1 point:15 students.
0 point:6 students.
Not shown because of missing birth date or parental permission:20 solutions.

Problems in Mathematics of KöMaL, November 2018