Mathematical and Physical Journal
for High Schools
Issued by the MATFUND Foundation
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Problem K. 414. (February 2014)

K. 414. In a chess tournament, every participant plays with every other participant exactly once. Two players withdrew their application, so the actual number of games played were 17 less than planned. How many participants remained?

(6 pont)

Deadline expired on March 10, 2014.

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

Megoldás. Ha az összes versenyző részt vett volna, akkor \(\displaystyle \binom n2=\frac{n(n-1)}{2}\) mérkőzést játszottak volna le.

Mivel ketten lemondták, ezért csak \(\displaystyle \binom{n-2}{2}=\frac{(n-2)(n-3)}{2}\) mérkőzés volt. Mivel ez 17-tel kevesebb, mint eredtileg tervezték, felírható a következő egyenlet:

\(\displaystyle \frac{n(n-1)}{2}-\frac{(n-2)(n-3)}{2}=17,\)


\(\displaystyle n^{2}-n-(n^{2}-5n+6)=34,\)


\(\displaystyle 4n=40\)

és így

\(\displaystyle n=10\)


Tehát a lemondás után 8 résztvevő lesz.


186 students sent a solution.
6 points:89 students.
5 points:33 students.
4 points:17 students.
3 points:16 students.
2 points:27 students.
1 point:2 students.
Unfair, not evaluated:2 solutions.

Problems in Mathematics of KöMaL, February 2014