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Sagot :
Let's carefully examine the prime factorization of the number 60 provided by Darius, which is expressed as [tex]\(2 \times 3 \times 5\)[/tex].
### Step-by-Step Solution:
1. Identify the Prime Factors of 60:
To correctly factorize 60 into its prime components, we perform the following steps:
- Divide 60 by the smallest prime number, which is 2:
[tex]\[ 60 \div 2 = 30 \][/tex]
- Next, divide 30 by 2 again:
[tex]\[ 30 \div 2 = 15 \][/tex]
- Finally, divide 15 by the next smallest prime, which is 3:
[tex]\[ 15 \div 3 = 5 \][/tex]
- At this point, 5 remains, which is a prime number itself.
2. List all Prime Factors:
Collecting all the prime factors from each division step, we have:
[tex]\[ 60 = 2 \times 2 \times 3 \times 5 \][/tex]
3. Compare with Darius's Factorization:
Darius's factorization is:
[tex]\[ 60 = 2 \times 3 \times 5 \][/tex]
4. Analyze the Possible Errors:
- Option 1: Darius should have started the factor tree with [tex]\(2 \times 30\)[/tex].
This is not a crucial error as starting the factor tree at different points can still lead to the correct prime factorization.
- Option 2: Darius did not complete the factorization; 5 is a composite number.
This statement is factually incorrect because 5 is a prime number, not a composite number. Therefore, this is an incorrect error.
- Option 3: Darius should have expressed 15 as [tex]\(5 \times 5\)[/tex], not [tex]\(3 \times 5\)[/tex].
This is incorrect because 15 is indeed [tex]\(3 \times 5\)[/tex]; expressing it as [tex]\(5 \times 5\)[/tex] would be an error.
- Option 4: Darius should have expressed 60 as [tex]\(2^2 \times 3 \times 5\)[/tex], not [tex]\(2 \times 3 \times 5\)[/tex].
Indeed, Darius has missed one prime factor of 2. The correct factorization includes two 2s:
[tex]\[ 60 = 2^2 \times 3 \times 5 \][/tex]
Therefore, Darius's error is:
Corrected Statement:
- Darius should have expressed 60 as [tex]\(2^2 \times 3 \times 5\)[/tex], not [tex]\(2 \times 3 \times 5\)[/tex].
Darius missed one of the prime factors 2, leading to an incomplete factorization. The correct prime factorization fully describes 60 as [tex]\(2^2 \times 3 \times 5\)[/tex].
### Step-by-Step Solution:
1. Identify the Prime Factors of 60:
To correctly factorize 60 into its prime components, we perform the following steps:
- Divide 60 by the smallest prime number, which is 2:
[tex]\[ 60 \div 2 = 30 \][/tex]
- Next, divide 30 by 2 again:
[tex]\[ 30 \div 2 = 15 \][/tex]
- Finally, divide 15 by the next smallest prime, which is 3:
[tex]\[ 15 \div 3 = 5 \][/tex]
- At this point, 5 remains, which is a prime number itself.
2. List all Prime Factors:
Collecting all the prime factors from each division step, we have:
[tex]\[ 60 = 2 \times 2 \times 3 \times 5 \][/tex]
3. Compare with Darius's Factorization:
Darius's factorization is:
[tex]\[ 60 = 2 \times 3 \times 5 \][/tex]
4. Analyze the Possible Errors:
- Option 1: Darius should have started the factor tree with [tex]\(2 \times 30\)[/tex].
This is not a crucial error as starting the factor tree at different points can still lead to the correct prime factorization.
- Option 2: Darius did not complete the factorization; 5 is a composite number.
This statement is factually incorrect because 5 is a prime number, not a composite number. Therefore, this is an incorrect error.
- Option 3: Darius should have expressed 15 as [tex]\(5 \times 5\)[/tex], not [tex]\(3 \times 5\)[/tex].
This is incorrect because 15 is indeed [tex]\(3 \times 5\)[/tex]; expressing it as [tex]\(5 \times 5\)[/tex] would be an error.
- Option 4: Darius should have expressed 60 as [tex]\(2^2 \times 3 \times 5\)[/tex], not [tex]\(2 \times 3 \times 5\)[/tex].
Indeed, Darius has missed one prime factor of 2. The correct factorization includes two 2s:
[tex]\[ 60 = 2^2 \times 3 \times 5 \][/tex]
Therefore, Darius's error is:
Corrected Statement:
- Darius should have expressed 60 as [tex]\(2^2 \times 3 \times 5\)[/tex], not [tex]\(2 \times 3 \times 5\)[/tex].
Darius missed one of the prime factors 2, leading to an incomplete factorization. The correct prime factorization fully describes 60 as [tex]\(2^2 \times 3 \times 5\)[/tex].
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