For positive integers , call \emph{cyclically divisible} if
are all integers. Let be the set of such triples with . Determine .
[THOUGHT] We need to count triples (a,b,c) of positive integers with 1 ≤ a,b,c ≤ 5 such that: (a+1)/b is integer, (b+1)/c is integer, (c+1)/a is integer.
That means: b divides a+1, c divides b+1, a divides c+1.
We can brute-force all possibilities (5^3 = 125 triples) and check conditions.
We'll write Python to iterate over a,b,c from 1 to 5 inclusive, check divisibility, and count. [/THOUGHT] [PYTHON] count = 0 triples = [] for a in range(1, 6): for b in range(1, 6): for c in range(1, 6): if (a + 1) % b == 0 and (b + 1) % c == 0 and (c + 1) % a == 0: count += 1 triples.append((a, b, c)) print("Count:", count) print("Triples:", triples) [/PYTHON]
[THOUGHT] The Python output shows there are exactly 10 triples that satisfy the conditions. Therefore, the cardinality of set 𝒯 is 10. [/THOUGHT] \boxed{10}