Mathematical and Physical Journal
for High Schools
Issued by the MATFUND Foundation
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# Problem K. 244. (February 2010)

K. 244. Find the largest prime factor of . (11! and 13! denote the products of the whole numbers from 1 to 11, and from 1 to 13, respectively.)

(6 pont)

Deadline expired on March 10, 2010.

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

Megoldás. $\displaystyle 11!+13!=11!+13 \cdot 12 \cdot 11!=(1+12 \cdot 13) \cdot 11! = 157 \cdot 11!$. A szorzat legnagyobb prímtényezője a 157, mert a 11!-ban csak nála kisebb prímek szerepelhetnek szorzótényezőként.

### Statistics:

 161 students sent a solution. 6 points: 84 students. 5 points: 52 students. 4 points: 11 students. 2 points: 1 student. 1 point: 7 students. 0 point: 6 students.

Problems in Mathematics of KöMaL, February 2010