For positive integers , call \emph{cyclically divisible} if
are all integers. Let be the set of such triples with . Determine .
[THOUGHT] Since a, b, c range from 1 to 5, there are 125 possibilities. I will brute-force check all combinations by looping over each value and verifying if (a+1) % b == 0, (b+1) % c == 0, and (c+1) % a == 0. Count the triples that satisfy all three conditions. [/THOUGHT] [PYTHON] count = 0 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 print(count) [/PYTHON]
[THOUGHT] The Python code executed successfully and output 10, which is the count of such triples. This is the ground truth. [/THOUGHT] \boxed{10}
Explanation
The reasoning follows a systematic brute-force approach due to the small search space of 125 possible triples (a, b, c each from 1 to 5). The process involves:
-
Iterating through all combinations using nested loops for a, b, and c in the range 1 to 5 inclusive.
-
For each triple, checking the three divisibility conditions: (a+1) is divisible by b, (b+1) is divisible by c, and (c+1) is divisible by a, verified using the modulo operator (% == 0).
-
Incrementing a counter for each triple that satisfies all conditions simultaneously.
-
Outputting the final counter value, which represents the size of the set T.
This method ensures exhaustive coverage and exact computation, with the Python execution confirming the result as 10.