Mathematical and Physical Journal
for High Schools
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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 (n3). 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 solutions.

Problems in Mathematics of KöMaL, October 2012