Problem K. 611. (January 2019)

K. 611. Is it possible to arrange the integers 1 to 50 in pairs such that the sums of the numbers in the pairs are all distinct primes?

(6 pont)

Deadline expired on February 11, 2019.

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

Megoldás. Nem lehet, mert \(\displaystyle 1\)-től \(\displaystyle 50\)-ig az egész számokból \(\displaystyle 25\) párt alakíthatunk ki, melyekből a lehető legkisebb összeg \(\displaystyle 1+2 = 3\), a lehető legnagyobb pedig \(\displaystyle 49 + 50 = 99\), viszont \(\displaystyle 3\) és \(\displaystyle 99\) között csak \(\displaystyle 24\) prímszám szerepel: \(\displaystyle 3\), \(\displaystyle 5\), \(\displaystyle 7\), \(\displaystyle 11\), \(\displaystyle 13\), \(\displaystyle 17\), \(\displaystyle 19\), \(\displaystyle 23\), \(\displaystyle 29\), \(\displaystyle 31\), \(\displaystyle 37\), \(\displaystyle 41\), \(\displaystyle 43\), \(\displaystyle 47\), \(\displaystyle 53\), \(\displaystyle 59\), \(\displaystyle 61\), \(\displaystyle 67\), \(\displaystyle 71\), \(\displaystyle 73\), \(\displaystyle 79\), \(\displaystyle 83\), \(\displaystyle 89\), \(\displaystyle 97\).

Statistics:

158 students sent a solution.
6 points:	117 students.
5 points:	8 students.
4 points:	1 student.
3 points:	2 students.
2 points:	3 students.
0 point:	4 students.
Unfair, not evaluated:	8 solutionss.
Not shown because of missing birth date or parental permission:	15 solutions.

Problems in Mathematics of KöMaL, January 2019