Mathematical and Physical Journal
for High Schools
Issued by the MATFUND Foundation
Already signed up?
New to KöMaL?
I want the old design back!!! :-)

Problem C. 1149. (December 2012)

C. 1149. The number of one-element subsets, the number of two-element subsets and the number of three-element subsets of a set of more than three elements are consecutive terms of an arithmetic progression. How many elements does the set have?

(5 pont)

Deadline expired on January 10, 2013.

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

Megoldás. Legyen a halmaz \(\displaystyle n\) elemű. Ekkor felírható: \(\displaystyle \binom n1 +\binom n3 =2\binom n2\), ahol \(\displaystyle n\geq3\). Ebből rendezéssel:

\(\displaystyle n+\frac{n(n-1)(n-2)}{6}=2\cdot\frac{n(n-1)}{2},\)

\(\displaystyle 6n+n(n-1)(n-2)=6n(n-1).\)

Osztva \(\displaystyle n\neq0\)-val:

\(\displaystyle 6+(n-1)(n-2)=6(n-1),\)

\(\displaystyle n^2-9n+14=(n-2)(n-7)=0.\)

Mivel \(\displaystyle n\geq3\), ezért az egyetlen megoldás \(\displaystyle n=7\), a halmaz 7 elemű.


284 students sent a solution.
5 points:132 students.
4 points:96 students.
3 points:23 students.
2 points:11 students.
1 point:9 students.
0 point:9 students.
Unfair, not evaluated:4 solutions.

Problems in Mathematics of KöMaL, December 2012