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\sqrt[3]{8000} = 20
Since \( p \) must be a multiple of 6, we consider multiples of 6 that are less than 20. These multiples are 6, 12, and 18. We check which of these satisfy \( p^3 < 8000 \):
For \( p = 18 \), \( 18^3 = 5832 \) which is less than 8000.
For \( p = 24 \), \( 24^3 = 13824 \) which exceeds 8000, but 24 is not less than 20.
Thus, the largest multiple of 6 less than 20 is 18, and it satisfies \( 18^3 = 5832 < 8000 \). Therefore, the largest possible value of \( p \) is \(\boxed{18}\).
Question:** A virologist is studying viral loads and notes that a particular parameter \( v \) is a positive integer less than or equal to 30. If the probability that \( v \) is a factor of 360 is to be calculated, what is this probability?
First, we determine the number of positive integers \( v \) that are factors of 360 and are less than or equal to 30. Begin by finding the prime factorization of 360:
= 2^3 \times 3^2 \times 5^1
The total number of positive divisors of 360 is:
Now, list the divisors of 360 that are less than or equal to 30: