Mathematical and Physical Journal
for High Schools
Issued by the MATFUND Foundation
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Problem K. 506. (September 2016)

K. 506. Out of the set of integers from 1 to 1000, remove all multiples of 2. Then remove all multiples of 3 from the remaining set of numbers. Then continue to remove the multiples of 5, 7, 11, 13, 15, 17, up to 19. At the end of this procedure, select all composite numbers remaining, and add them together. What is the total obtained?

(6 pont)

Deadline expired on October 10, 2016.

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

Megoldás. A megmaradt összetett számok legkisebb prímtényezője a \(\displaystyle 23\). Három prímszám szorzata, melyek mindegyike legalább \(\displaystyle 23\), meghaladja az \(\displaystyle 1000\)-et, így a keresett összetett számok két prímszám szorzataként állíthatók elő, melyek mindegyike legalább \(\displaystyle 23\). A megfelelő számok: \(\displaystyle 23\cdot23\), \(\displaystyle 23\cdot29\), \(\displaystyle 23\cdot31\), \(\displaystyle 23\cdot37\), \(\displaystyle 23\cdot41\), \(\displaystyle 23\cdot43\), \(\displaystyle 29\cdot29\), \(\displaystyle 29\cdot31\), \(\displaystyle 31\cdot31\). Ezek összege \(\displaystyle 7393\).


119 students sent a solution.
6 points:63 students.
5 points:14 students.
4 points:17 students.
3 points:6 students.
2 points:7 students.
1 point:3 students.
0 point:8 students.
Unfair, not evaluated:1 solutions.

Problems in Mathematics of KöMaL, September 2016