Mathematical and Physical Journal
for High Schools
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Problem K. 224. (December 2009)

K. 224. There are dice of six and four faces (cubes and tetrahedra) on a table. Their faces are numbered with dots, 1 to 6 and 1 to 4, respectively. The number of all dots on the dice is 323. If we had as many six-sided dice as we have of the four-sided dice and vice versa, the number of dots would be 185. How many dice of each kind are there on the table?

(6 pont)

Deadline expired on January 11, 2010.

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

Megoldás. Egy dobókockán összesen \(\displaystyle 21\) pötty van és \(\displaystyle h\) darab dobókockánk van. A ,,dobótetraéderek'' mindegyikén \(\displaystyle 10\) pötty van, a számuk \(\displaystyle t\). A pöttyök száma összesen \(\displaystyle 21h+10t=314\), a megfordított szituációban \(\displaystyle 10h+21t=182\). A két egyenlet különbségének 11-ed részéből \(\displaystyle h-t=12\), azaz \(\displaystyle h=12+t\). Pl. az első összefüggésbe visszahelyettesítve \(\displaystyle 252+21t+10t=314\), amiből \(\displaystyle t=2\) és \(\displaystyle h=14\). \(\displaystyle \mathbf{14}\) dobókockánk és \(\displaystyle \mathbf{2}\) dobótetraéderünk van.


133 students sent a solution.
6 points:93 students.
5 points:21 students.
4 points:6 students.
3 points:3 students.
2 points:2 students.
1 point:1 student.
0 point:1 student.
Unfair, not evaluated:6 solutions.

Problems in Mathematics of KöMaL, December 2009