Mathematical and Physical Journal
for High Schools
Issued by the MATFUND Foundation
Already signed up?
New to KöMaL?
I want the old design back!!! :-)

Problem C. 1078. (April 2011)

C. 1078. Define ``words'' as strings of at most ten characters made out of the 26 letters of the English alphabet. (A ``word'' may as well consist of a single letter, and two ``words'' are considered different if they differ in at least one letter.) Prove that the number of all possible ``words'' obtained in this way is divisible by 27.

(5 pont)

Deadline expired on May 10, 2011.

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

Megoldás. Ha egy "szó" \(\displaystyle k\) karakterből áll, akkor az ilyen "szavak" száma \(\displaystyle 26^k\), mert ez egyes helyekre egymástól függetlenül bármely betűt választhatjuk az abc-ből. Összesen \(\displaystyle N=26+26^2+\dots 26^{10}=26\cdot \frac{26^{10}-1}{25}\) "szót" alkottunk. Ha 27 osztja \(\displaystyle 26^{10}-1\)-t, akkor \(\displaystyle N\)-t is, mert 27 és 25 relatív prímek. \(\displaystyle 26^{10}-1=(26^5-1)(26^5+1)=(26^5-1)(26+1)(26^4-26^3+26^2-26+1)\), mely szorzatnak 27 osztója.


141 students sent a solution.
5 points:98 students.
4 points:31 students.
3 points:6 students.
2 points:1 student.
0 point:5 students.

Problems in Mathematics of KöMaL, April 2011