Mathematical and Physical Journal
for High Schools
Issued by the MATFUND Foundation

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Problem C. 835. (January 2006)

C. 835. How many solutions does the equation x+y+z=100 have on the set of positive integers?

(5 pont)

Deadline expired on February 15, 2006.

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

Megoldás: Meghatározzuk az x+y<100 egyenlőtlenség megoldásainak a számát. Ez lesz a feladat kérdésére is a válasz. Az (1;y) alakú párokból 98 darab van, a (2;y) alakú párokból 97 darab van, $\ldots$ , a (98;y) alakú párokból 1 darab van. Vagyis összesen $98+97+\ldots+2+1=4851$ darab.

Statistics:

391 students sent a solution.

5 points: 286 students.

4 points: 4 students.

3 points: 12 students.

2 points: 18 students.

1 point: 13 students.

0 point: 48 students.

Unfair, not evaluated: 10 solutionss.

Problems in Mathematics of KöMaL, January 2006

391 students sent a solution.
5 points:	286 students.
4 points:	4 students.
3 points:	12 students.
2 points:	18 students.
1 point:	13 students.
0 point:	48 students.
Unfair, not evaluated:	10 solutionss.