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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 points)

Deadline expired on 10 January 2013.

Google Translation (Sorry, the solution is published in Hungarian only.)

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.

Statistics on problem C. 1149.
 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

•  Támogatóink: Morgan Stanley