AoPS - Problem

p1. Consider a number of two or more digits, none of them being zero $(0)$. Such a number is called a thirteen if every pair of consecutive digits forms a number divisible by $13$. For example: $139$ is thirteen since $13 = 13\times 1$ and $39 = 13\times 3$. How many thirteen five-digit numbers are there? p2. Katia and Mariela play the following game: In each of their three turns, Katia replaces one of the stars in the expression $\star \star\star\star\star\star\star\star\star$ for some a digit of between $1$, $2$, $3$, $4$, $5$, $6$, $7$, $8$, $9$ that has not been used before in the game. In shifts of Mariela, she replaces two of the stars with two different digits that have not been used. Katia starts playing and plays alternately. Mariela wins if the resulting number at the end of the game is divisible by $27$. Does Mariela have any to ensure the triumph? p3. A rectangle has been divided, using three segments, into four sections, as illustrated in the figure. If the areas of the indicated sectors are $3$, $4$ and $5$ respectively, find the area of the sector in gray. p4. On the surface of a cylinder of height $12$ meters and base circle of perimeter $4$ meters, a rope is wound whose initial end is on the circle upper, makes a total of four full turns of the cylinder and its lower end is at the lower circle. What is the shortest length that this string can have? PS. Juniors P2 was also Seniors P1

p1 Firstly let's list the two-digit multiples of $13$ which are: $13,26,39,52,65,78,91$ Let $\overline{abcde}$ be five-digit $thirteen$ number, after a little bit of playing around we find that we can only plug in values of $a$ and then the values of $b,c,d,e$ follow from the conditions of the problem. So we need to take the first digit of these two-digit multiples of 13 and make them equal to $a$. Example: When $a=1$, we have: $\overline{1bcde}$,now $b$ must be $3$ (because we need to have $13\vert ab$), $\overline{13cde}$,now $c=9$ (for the same reason), $\overline{139de}$,now $d=1$, $\overline{1391e}$,now $e=3$, Finally our number is $13913$ This same procedure can be done for when $a=2,3,5,6,9$ and for every case we have exactly one $thirteen$ number. (We're not considering the case when $a=7$ because that would imply $b=8$ and since there's no 2-digit multiple of $13$ with tenths digit $8$ we cannot have such a number) So in total there are $\boxed{6}$ $thirteen$ five-digit numbers. (They are $13913, 26526,39136,52652,65265,91365$)