Problem K. 516. (October 2016)

K. 516. Consider the sets \(\displaystyle A = \{a, 2a+1, a^{2}+1\}\), \(\displaystyle B =\{b+3, 10, b-1\}\). Find appropriate positive integers \(\displaystyle a\) and \(\displaystyle b\) such that the two sets have

\(\displaystyle a)\) no element in common;

\(\displaystyle b)\) exactly 1 element in common;

\(\displaystyle c)\) exactly 2 elements in common;

\(\displaystyle d)\) the same elements.

(6 pont)

Deadline expired on November 10, 2016.

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

Megoldás. a) Legyen \(\displaystyle a = 1\), így \(\displaystyle A = \{1; 3; 2\}\), ha pl. \(\displaystyle b = 10\), akkor \(\displaystyle B = \{13; 10; 9\}\).

b) Legyen \(\displaystyle a = 10\), így \(\displaystyle A = \{10; 21; 101\}\), ha pl. \(\displaystyle b = 10\), akkor \(\displaystyle B = \{13; 10; 9\}\).

c) Legyen \(\displaystyle a = 10\), így \(\displaystyle A = \{10; 21; 101\}\), ha \(\displaystyle b = 18\), akkor \(\displaystyle B = \{21; 10; 17\}.\)

d) Keressük meg, melyik elem lehet az \(\displaystyle A\) halmazból 10. Ha \(\displaystyle a = 10\), akkor az \(\displaystyle A\) halmaz elemei: 10, 21, 101, viszont a \(\displaystyle B\) halmazban \(\displaystyle b+3\) és \(\displaystyle b–1\) különbsége 4, így azok nem lehetnek semmilyen \(\displaystyle b\) érték mellett 21 és 101. \(\displaystyle 2a+1\) nem lehet 10, mert \(\displaystyle a\) egész szám. Ha \(\displaystyle a^2+1=10\), akkor \(\displaystyle a = 3\), az \(\displaystyle A\) halmaz elemei pedig: 3, 7, 10. Ha \(\displaystyle b = 4\), akkor a \(\displaystyle B\) halmaz elemei is ezek.

Statistics:

121 students sent a solution.
6 points:	77 students.
5 points:	26 students.
4 points:	7 students.
3 points:	8 students.
2 points:	1 student.
Unfair, not evaluated:	2 solutionss.

Problems in Mathematics of KöMaL, October 2016