1973 AMC 12 Problem 18

Factor $p^2-1=(p-1)(p+1)$

Among the three consecutive integers $p-1,$ $p,$ $p+1,$ locate factors of $3$ and $8$

A prime $p\ge5$ is odd, so $p-1$ and $p+1$ are consecutive even integers. One is divisible by $4,$ making their product divisible by $8.$ Among the three consecutive integers $p-1,$ $p,$ $p+1,$ one is divisible by $3.$ It cannot be $p,$ since $p\ge5$ is prime, so $3$ also divides $(p-1)(p+1).$ Because $3$ and $8$ are relatively prime, $$(p-1)(p+1)=p^2-1$$ is always divisible by $24.$

Therefore, the correct answer is C.