ext{lcm}(7, 11, 13) = 7 imes 11 imes 13 = 1001

February 22, 2026

ext{lcm}(7, 11, 13) = 7 imes 11 imes 13 = 1001

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Since 75 divides 150 and \( k + l = 2 \) satisfies coprimality, the largest possible \(\gcd(p, q)\) is \(oxed{75}\).
An entomologist is examining the cycles of a rare insect species, where the cycle length is represented by a three-digit integer \( z \) such that \( z \) is divisible by 7, 11, and 13. What is the smallest such \( z \)?
Since \( z \) is divisible by 7, 11, and 13, compute the least common multiple:
But 1001 is a four-digit number. The condition specifies a three-digit number, and no three-digit number divisible by all three exists. However, reconsider: the smallest three-digit number divisible by \( 7 imes 11 imes 13 = 1001 \) is impossible.
But note: the problem asks for the smallest three-digit number divisible by **each** of 7, 11, and 13. Since their product is 1001 > 999, no three-digit number satisfies this.
But perhaps a misinterpretation: maybe the number is divisible by **at least one**, but the context implies "divisible by" the product. Given the constraints, no such three-digit number exists. But rechecking, the smallest number divisible by 7, 11, and 13 is 1001âexceeding three digits.

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