Mathematical and Physical Journal
for High Schools
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Problem K. 244. (February 2010)

K. 244. Find the largest prime factor of 11!+13!\,. (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