For positive integers , call \emph{cyclically divisible} if
are all integers. Let be the set of such triples with . Determine .
[THOUGHT] We need to find all integer triples (a,b,c) with 1 ≤ a,b,c ≤ 5 such that:
- (a+1) divisible by b → (a+1) % b == 0
- (b+1) divisible by c → (b+1) % c == 0
- (c+1) divisible by a → (c+1) % a == 0
We can brute-force all 5^3 = 125 possibilities and count those satisfying all three conditions.
Let's implement in Python. [/PYTHON] [PYTHON] count = 0 valid_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 valid_triples.append((a,b,c))
print("Number of valid triples:", count) print("Valid triples:", valid_triples) [/PYTHON]
[THOUGHT] The Python code computed the number of valid triples as 10. The output shows the count directly, so the answer is 10.
\boxed{10}