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. 1137. (October 2012)

C. 1137. The first two terms of the Fibonacci sequence are a1=1, a2=1, and every further term equals the sum of the two preceding terms, that is, an=an-2+an-1 (n\ge3). Prove that the sequence has no term that leaves a remainder of 4 when divided by 13.

(5 pont)

Deadline expired on November 12, 2012.


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

Megoldás. A 13-mal való osztási maradékok sorozatát jelölje \(\displaystyle b_1\), \(\displaystyle b_2\) stb. Nyilván \(\displaystyle b_{n}=b_{n-2}+b_{n-1}\) is teljesül.

Írjuk fel a \(\displaystyle b_n\) sorozatot:

\(\displaystyle 1,~1,~2,~3,~5,~8,~0,~8,~8,~3,~11,~1,~12,~0,~12,~12,~11,~10,~8,~5,~0,~5,~5,~10,~2,~12,~1,~0,~1,~1,\ldots\)

Innentől kezdve a maradékok sorozata ismétlődik. Látható, hogy egyik maradék sem 4, vagyis valóban nincs a Fibonacci sorozatnak olyan tagja, ami 13-mal osztva 4 maradékot ad.


Statistics:

293 students sent a solution.
5 points:243 students.
4 points:12 students.
3 points:11 students.
2 points:4 students.
1 point:6 students.
0 point:16 students.
Unfair, not evaluated:1 solution.

Problems in Mathematics of KöMaL, October 2012