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A set of 9-digit numbers is composed, with all numbers are formed by using each of the digits 1 to 9 one and only once. How many of these numbers are prime numbers? Explaining your answer

A set of 9-digit numbers is composed, with all numbers are formed by using each of the digits 1 to 9 one and only once. How many of these numbers are prime numbers? Explaining your answer
 
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To determine how many 9-digit numbers can be formed using each of the digits from 1 to 9 exactly once that are also prime, we need to consider a few key points about the properties of prime numbers and the constraints given by the digits.

1. **Understanding Primality**: A prime number is a natural number greater than 1 that cannot be formed by multiplying two smaller natural numbers. In other words, a prime number has exactly two distinct positive divisors: 1 and itself.

2. **Sum of Digits**: The sum of the digits from 1 to 9 is calculated as follows:
\[
1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = \frac{9 \times 10}{2} = 45
\]
Since 45 is divisible by 3, any number formed using all these digits will also be divisible by 3. This is because of the property of divisibility: If the sum of the digits of a number is divisible by 3, then the number itself is also divisible by 3.

3. **Consequences for Primality**: Since a prime number greater than 3 cannot be divisible by 3, and since all 9-digit numbers formed from the digits 1 to 9 will be divisible by 3 (as we've seen in point 2), it follows that none of these numbers can be prime.

In conclusion, **there are no 9-digit numbers formed using all the digits from 1 to 9 exactly once that are prime**. Therefore, the answer is:

\[
\text{Number of prime 9-digit numbers} = 0
\]

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