# 6. Variant 1.

Vasya cut out a triangle from cardboard and numbered its vertices with the digits 1, 2, and 3. It turned out that if Vasya's triangle is rotated 15 times clockwise around its vertex numbered 1 by an angle equal to the angle at this vertex, the triangle will return to its original position. If Vasya's triangle is rotated 6 times clockwise around its vertex numbered 2 by an angle equal to the angle at this vertex, the triangle will return to its original position. Vasya claims that if his triangle is rotated \( n \) times around the vertex numbered 3 by an angle equal to the angle at this vertex, the triangle will return to its original position. What is the smallest \( n \) that Vasya could have named so that his claim is true for at least some cardboard triangle?
Answer: 5.

Solution. Let the angle at the first vertex be $\alpha$, at the second $\beta$, and at the third $\gamma$. When the triangle is rotated around its vertices, it may complete several full rotations before returning to its initial position. Suppose that during the rotation around the first vertex, it makes $k$ full rotations, around the second vertex $m$ full rotations, and around the third vertex $l$ full rotations. Then from the problem's conditions, we have $15 \alpha = 360 k$, $6 \beta = 360 m$, and $n \gamma = 360 l$. From the first two equations, we find that $\alpha = 24 k$, $\beta = 60 m$, which means $\alpha \geq 24$, $\beta \geq 60$. Note that $n \geq 4$, otherwise $\gamma > 120$, which together with $\beta$ would exceed 180. $n \neq 4$, because otherwise $\gamma = 90$, so $\alpha + \beta = 90$, which is impossible given $\alpha = 24 k$, $\beta = 60 m$.

$n = 5$ works: take a triangle with angles $\alpha = 48$, $\beta = 60$, $\gamma = 72$.