How many 5-digit prime numbers can be formed using the digits 1, 2, 3, 4, 5 if the repetition of digits is not allowed?
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Understanding the Problem: Forming 5-Digit Prime Numbers
The question asks us to determine how many 5-digit prime numbers can be created using the specific digits 1, 2, 3, 4, and 5, with the condition that each digit must be used exactly once (repetition is not allowed). We need to form numbers using these five distinct digits.
Constructing 5-Digit Numbers from Given Digits
We are given the set of digits {1, 2, 3, 4, 5}. To form a 5-digit number using these digits without repetition, we must use each digit exactly once. The number of such distinct 5-digit numbers that can be formed is the number of permutations of 5 distinct items, which is \(5!\) (5 factorial).
Calculation of total possible numbers:
- The first digit can be any of the 5 digits.
- The second digit can be any of the remaining 4 digits.
- The third digit can be any of the remaining 3 digits.
- The fourth digit can be any of the remaining 2 digits.
- The fifth digit must be the last remaining digit.
So, the total number of distinct 5-digit numbers possible is \(5 \times 4 \times 3 \times 2 \times 1 = 120\).
We need to find out how many of these 120 numbers are prime.
Applying the Divisibility Rule of 3
A crucial rule in number theory is the divisibility rule for 3. A number is divisible by 3 if the sum of its digits is divisible by 3.
Let's calculate the sum of the digits available to us: 1, 2, 3, 4, 5.
Sum of digits \(= 1 + 2 + 3 + 4 + 5 = 15\).
Since the sum of the digits (15) is divisible by 3 (15 ÷ 3 = 5), any number formed using these digits will be divisible by 3.
Analyzing Prime Numbers and Divisibility by 3
A prime number is a natural number greater than 1 that has exactly two distinct positive divisors: 1 and itself.
We have established that any 5-digit number formed using the digits 1, 2, 3, 4, 5 exactly once will be divisible by 3.
Consider a number, \(N\), that is divisible by 3. If \(N\) is prime, its only positive divisors can be 1 and \(N\). For \(N\) to be divisible by 3, 3 must be one of its divisors. The only prime number that is divisible by 3 is 3 itself.
However, the numbers we are forming are 5-digit numbers. The smallest 5-digit number is 10,000, and the smallest number we can form using the digits 1, 2, 3, 4, 5 is 12,345. All such numbers are significantly greater than 3.
Therefore, any 5-digit number formed using these digits will be:
- Greater than 3.
- Divisible by 3 (because the sum of digits is 15, which is divisible by 3).
A number greater than 3 that is divisible by 3 will have divisors 1, 3, itself, and potentially others. This means it has at least three distinct positive divisors (1, 3, and the number itself), which contradicts the definition of a prime number (having exactly two distinct positive divisors: 1 and itself).
Thus, no 5-digit number formed using the digits 1, 2, 3, 4, 5 exactly once can be a prime number.
Conclusion on 5-Digit Prime Numbers
Since every 5-digit number formed by rearranging the digits 1, 2, 3, 4, and 5 is divisible by 3 and is greater than 3, none of these numbers can be prime. The count of such 5-digit prime numbers is 0.
| Property | Value/Observation |
|---|---|
| Digits used | 1, 2, 3, 4, 5 (no repetition) |
| Number of digits | 5 |
| Sum of digits | \(1 + 2 + 3 + 4 + 5 = 15\) |
| Divisibility of sum by 3 | 15 is divisible by 3 |
| Divisibility of formed number by 3 | Any number formed is divisible by 3 |
| Range of formed numbers | 5-digit numbers (all > 3) |
| Prime number definition | Number > 1 with exactly two divisors (1 and itself) |
| Conclusion | No formed number is prime |
Revision Table: Key Concepts
| Concept | Explanation |
|---|---|
| Prime Number | A natural number greater than 1 that has no positive divisors other than 1 and itself. Examples: 2, 3, 5, 7, 11. |
| Composite Number | A natural number greater than 1 that is not prime. It has at least one positive divisor other than 1 and itself. Examples: 4, 6, 8, 9, 10. |
| Divisibility Rule for 3 | A number is divisible by 3 if the sum of its digits is divisible by 3. |
| Numbers formed by 1,2,3,4,5 | Using each digit once forms a 5-digit number. The sum of digits is always 15, which is divisible by 3. Thus, all such numbers are divisible by 3. |
Additional Information: Number Properties and Combinatorics
This problem combines concepts from number theory (prime numbers, divisibility rules) and combinatorics (permutations). Understanding the properties of numbers, like divisibility rules, can often provide shortcuts in problems involving primes or factors without needing to test each number individually.
In this specific case, the divisibility rule of 3 is a powerful tool. Any number formed by permuting a set of digits whose sum is divisible by 3 will itself be divisible by 3. If the number formed is greater than 3 and divisible by 3, it cannot be prime.
The total number of permutations of 5 distinct items is \(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\). These are all the possible 5-digit numbers we could form. However, we didn't need to list or check any of these 120 numbers individually because the divisibility rule gave us a definitive property that applies to all of them.
Knowing common divisibility rules (for 2, 3, 4, 5, 6, 9, 10, 11) is very helpful in number theory problems.
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