Which is the six-digit number (abcdef) in the decimal system, whose 2, 3, 4, 5, 6 times multiples are also six-digit and their digits are formed by cyclic permutations of the digits of the above number and start with $c, b, e, f, d$ respectively?
I. Solution: The number is certainly 1, because its sextuple is less than 10, $3 f$ can only be 1, 11, 21, because $3 f \leq 27$. Of these, only 21 is divisible by 3, so $f=7$. The double of $f$ ends with $b$, so $b=4$, its quadruple ends with $d$, so $d=8$, its quintuple ends with $e$, so $e=5$, and its sextuple ends with $c$, so $c=2$. Therefore, the sought number can only be 142857, and this indeed meets the conditions.

Tamás Kóvári (Budapest, Evangelical Gymnasium VII. grade).

Solved by: Bakonyi Kornélia, Bánó Klára, Bendzsák Z., Bécsy Cecília, Bognár J., Csík M., Csuhássy Edit, Dobó F., Fülöp M., ifj. Gacsályi S., Gehér L., Hegedűs L., Hosszú M., Inkovics Gabriella, Izsák I., Királyfalvi I., Kovács G., Kovács J., László F., Márton J., Miskolczi Ida, Németh R., Reiner Éva, Reppert F., Réthy Eszter, Sós Vera, Spitz Vera, Szabó Á., Szathmári D., Szúcs L., Turczi Gy., Ungár P., Varróh Enikő, Vermes R., Vékony Mária, Vizi Mária, Vörös M.

II. Solution: According to the condition: $3 \cdot a b c d e f=b c d e f a=10 \cdot b c d e f+a$. From this, $7 \cdot b c d e f=299999 a, b c d e f=42857 \cdot a$, so either $a=1, b c d e f=42857$, or $a=2, b c d e f=85714$. The latter solution (which is twice the first) does not fit, because its quadruple is already a 7-digit number, so the sought number can only be 142857, and this indeed meets the conditions.

Similar solution: by Tamás I.

In the solution, they only used the triple of the number: Bodonyi J., Csordás L., Horváth M., Magyarosi B., Tarnóczi T., Tarnóczi Z., Vata L.

Other solutions: Berendik S., Blaskó F., Erdősy Gy., Korányi Á.

Partially solved: Magyar Á. Sz., Párkány M., Szépfalussy P.

Partially correct: Róna P.