Mathematical and Physical Journal
for High Schools
Issued by the MATFUND Foundation
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Problem K. 699. (October 2021)

K. 699. We have six discs. Each disc has a letter on one side (A, B, C, D, E, F), and a number on the other side (1, 2, 3, 4, 5, 6, in some order). The discs are placed on the table with their letter side up. Given that the sum of the numbers on the discs marked A, B and C is 14, and the sum of the numbers on discs A, D and E is 12, what is the minimum number of discs to be turned over in order to know which number is on which disc?

(5 pont)

Deadline expired on November 10, 2021.

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

Megoldás. Az \(\displaystyle A\), \(\displaystyle B\), \(\displaystyle C\), \(\displaystyle D\), \(\displaystyle E\) és \(\displaystyle F\) betű jelölje a megfelelő korong túloldalán lévő számot.

Mivel \(\displaystyle A+B+C+D+E+F= 1+2+3+4+5+6=21\) és \(\displaystyle A+B+C = 14\), így \(\displaystyle D+E+F= 21-14=7\), ahonnan \(\displaystyle D+E\) legfeljebb \(\displaystyle 7-1=6\) lehet. \(\displaystyle A+D+E = 12\) miatt (mivel bármely \(\displaystyle A\) legfeljebb 6 ) \(\displaystyle D + E\) legalább 6.

Így tehát \(\displaystyle D + E\) éppen 6, illetve így \(\displaystyle A = 6\). Mivel \(\displaystyle D + E = 6\), így \(\displaystyle F = 1\).

Tehát eddig biztosan tudjuk, hogy \(\displaystyle A=6\) és \(\displaystyle F=1\).

\(\displaystyle B + C = 8\) miatt így \(\displaystyle B\) és \(\displaystyle C\) csak \(\displaystyle 5\) és \(\displaystyle 3\) lehet, \(\displaystyle D + E = 6\) miatt pedig \(\displaystyle D\) és \(\displaystyle E\) értéke csak 2 és 4 lehet valamilyen sorrendben. Ahhoz, hogy megtudjuk, melyik melyik, egy-egy korongot meg kell fordítanunk a \(\displaystyle B\) és a \(\displaystyle C\), illetve a \(\displaystyle D\) és az \(\displaystyle E\) közül is.

Tehát legalább két korongot kell megfodítanunk.


141 students sent a solution.
5 points:64 students.
4 points:20 students.
3 points:12 students.
2 points:5 students.
1 point:4 students.
0 point:17 students.
Unfair, not evaluated:16 solutionss.
Not shown because of missing birth date or parental permission:3 solutions.

Problems in Mathematics of KöMaL, October 2021