But âandâ implies all.

February 22, 2026

But âandâ implies all.

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Unless the number is not required to be three-digit, but the context says âthree-digit integer $ z $â.
Thus, the only resolution is that the **least common multiple is 1001**, and since no three-digit number is divisible by all three, the problem may be flawed.
But in educational context, perhaps they meant: divisible by **7** and **11**, or **13**âbut not **and** both.
Alternatively, perhaps the number is divisible by **7**, and the sum $ p+q $ involves 11 and 13? But not specified.
Given the difficulty, and to align with olympiad style, suppose the problem meant: the number is divisible by **7** and **by 11**, and also by **by 13**âsame issue.
But waitâperhaps âdivisible by 7, 11, and 13â is a red herring, but no.

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