Mathematical and Physical Journal
for High Schools
Issued by the MATFUND Foundation
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# Problem C. 1159. (February 2013)

C. 1159. Consider all non-congruent rectangles whose side lengths are two distinct elements selected from the set . What is the total area of all such rectangles?

(5 pont)

Deadline expired on March 11, 2013.

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

Megoldás. A feladatot tetszőleges $\displaystyle n$ pozitív egész esetére oldjuk meg. A területösszegben az 1, 2, ..., $\displaystyle n$ számok egymással vett szorzatai szerepelnek, kivéve a négyzetszámokat. Vagyis a keresett összeg:

$\displaystyle \frac{(1+2+\dots+n)^2-(1^2+2^2+\dots+n^2)}{2}=\frac12\cdot\left(\left(\frac{(n+1)n}{2}\right)^2-\frac{n(n+1)(2n+1)}{6}\right)=$

$\displaystyle =\frac{(n+1)^2n^2}{8}-\frac{n(n+1)(2n+1)}{12}.$

Ha $\displaystyle n$ helyébe behelyettesítjük a 100-at, akkor az eredmény: 12582075.

### Statistics:

 164 students sent a solution. 5 points: 75 students. 4 points: 22 students. 3 points: 23 students. 2 points: 15 students. 1 point: 12 students. 0 point: 7 students. Unfair, not evaluated: 10 solutions.

Problems in Mathematics of KöMaL, February 2013