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

Problems in Mathematics of KöMaL, October 2016