If x, y, and z are different positive prime numbers, then the product xyz is divisible by how many different positive integers?
The product xyz is divisible by eight different positive integers.
The product of three different positive prime numbers, xyz, has a total of eight positive divisors. This is calculated using the formula for the number of divisors, which states that if a number is expressed in its prime factorization form as p1^e1 * p2^e2 * ... * pn^en, then the number of divisors is (e1 + 1)(e2 + 1)...(en + 1). In the case of xyz, each prime is raised to the power of one.
The number one only has one positive divisor, which is itself. Since xyz consists of three distinct prime factors, it has more than one divisor; hence, this option is incorrect.
Three divisors would imply that there are only two distinct prime factors, as a prime number p has exactly two positive divisors: 1 and p. In this case, with three different primes, xyz must have more than three divisors.
Four divisors would suggest a structure like p^3 or p^1 * q^1, where p and q are primes. However, since we have three distinct primes, xyz is not limited to four divisors; this option is therefore incorrect.
Six divisors would imply a combination of prime factors that is not possible with three distinct primes. The only way to achieve six divisors would require a prime raised to the second power, which contradicts the requirement for distinct primes.
The product xyz, where x, y, and z are distinct primes, has each exponent equal to 1. The total number of divisors can be calculated as (1+1)(1+1)(1+1) = 2 * 2 * 2 = 8, confirming that this option is correct.
The product of three different positive prime numbers yields eight distinct positive divisors. By applying the divisor-counting formula, we see that each unique prime contributes to the overall count of divisors, leading to the conclusion that the number of different positive integers by which the product xyz is divisible is eight.
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