Let $ X$, $ Y$, $ Z$ be distinct positive integers having exactly two digits in such a way that:

$ X\equal{}10a \plus{}b$
$ Y\equal{}10b \plus{}c$
$ Z\equal{}10c \plus{}a$

($ a,b,c$ are digits)

Find all posible values of $ gcd(X,Y,Z)$
1. Given the expressions for \(X\), \(Y\), and \(Z\):
   \[
   X = 10a + b, \quad Y = 10b + c, \quad Z = 10c + a
   \]
   where \(a\), \(b\), and \(c\) are digits (i.e., integers from 0 to 9).

2. We need to find \(\gcd(X, Y, Z)\). By the properties of the greatest common divisor, we know:
   \[
   \gcd(X, Y, Z) \mid \alpha X + \beta Y + \gamma Z
   \]
   for some integers \(\alpha\), \(\beta\), and \(\gamma\).

3. Choosing \((\alpha, \beta, \gamma) = (1, 1, 1)\), we get:
   \[
   \gcd(X, Y, Z) \mid X + Y + Z
   \]
   Substituting the values of \(X\), \(Y\), and \(Z\):
   \[
   X + Y + Z = (10a + b) + (10b + c) + (10c + a) = 11a + 11b + 11c = 11(a + b + c)
   \]
   Therefore:
   \[
   \gcd(X, Y, Z) \mid 11(a + b + c)
   \]
   Since \(a\), \(b\), and \(c\) are digits, \(a + b + c\) ranges from 0 to 27. Hence:
   \[
   \gcd(X, Y, Z) \mid 11 \times (0 \text{ to } 27)
   \]
   This implies:
   \[
   \gcd(X, Y, Z) \mid 11
   \]
   However, since \(a\), \(b\), and \(c\) are distinct digits, \(a + b + c\) cannot be a multiple of 11. Thus, \(\gcd(X, Y, Z) \mid 11\) is not possible.

4. Next, choosing \((\alpha, \beta, \gamma) = (1, 100, -10)\), we get:
   \[
   \gcd(X, Y, Z) \mid X + 100Y - 10Z
   \]
   Substituting the values of \(X\), \(Y\), and \(Z\):
   \[
   X + 100Y - 10Z = (10a + b) + 100(10b + c) - 10(10c + a) = 10a + b + 1000b + 100c - 100c - 10a = 1001b
   \]
   Therefore:
   \[
   \gcd(X, Y, Z) \mid 1001b
   \]
   Since \(1001 = 7 \times 11 \times 13\), we have:
   \[
   \gcd(X, Y, Z) \mid 7 \times 11 \times 13 \times b
   \]
   Similarly, we can derive:
   \[
   \gcd(X, Y, Z) \mid 7 \times 11 \times 13 \times a
   \]
   \[
   \gcd(X, Y, Z) \mid 7 \times 11 \times 13 \times c
   \]

5. Since \(a\), \(b\), and \(c\) are digits, the possible values of \(\gcd(X, Y, Z)\) are the divisors of \(7 \times 13 = 91\) that are less than or equal to 27. These divisors are:
   \[
   1, 7, 13
   \]

6. We can rule out some cases because one of the numbers \(5, 6, 8, 9\) cannot divide \(a, b, c\) simultaneously. Thus, the possible values of \(\gcd(X, Y, Z)\) are:
   \[
   \gcd(X, Y, Z) \in \{1, 2, 3, 4, 7, 13, 14\}
   \]

7. Each of these values, except \(21\) and \(26\), gives a solution indeed.

The final answer is \( \boxed{ \gcd(X, Y, Z) \in \{1, 2, 3, 4, 7, 13, 14\} } \).