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Sagot :
To address the question, let's analyze both the assertion and the reason provided:
### Assertion:
[tex]\( 12^n \)[/tex] ends with the digit zero, where [tex]\( n \)[/tex] is a natural number.
### Reason:
Any number ends with zero if its prime factor is of the form [tex]\( 2^m \times 5^n \)[/tex], where [tex]\( m \)[/tex] and [tex]\( n \)[/tex] are natural numbers.
### Step-by-Step Solution:
1. Analyzing the Reason:
- For a number to end with the digit zero, it must be divisible by 10.
- The number 10 has the prime factorization [tex]\( 2 \times 5 \)[/tex].
- Therefore, a number can only end with zero if it has both 2 and 5 in its prime factorization.
The given reason is correct. Any number ending in zero must include at least one factor of 2 and one factor of 5 in its prime factorization.
2. Analyzing the Assertion:
- We need to check if [tex]\( 12^n \)[/tex] can end with the digit zero.
- Let's start with the prime factorization of 12:
[tex]\[ 12 = 2^2 \times 3 \][/tex]
- Raising 12 to any power [tex]\( n \)[/tex]:
[tex]\[ 12^n = (2^2 \times 3)^n = 2^{2n} \times 3^n \][/tex]
- For [tex]\( 12^n \)[/tex] to end with the digit zero, it must have both a factor of 2 and a factor of 5.
- From the prime factorization [tex]\( 2^{2n} \times 3^n \)[/tex], we see that while 2 is present (in the form [tex]\( 2^{2n} \)[/tex]), there is no factor of 5.
- Since [tex]\( 12^n \)[/tex] does not include a factor of 5, it cannot end with the digit zero regardless of the value of [tex]\( n \)[/tex].
Thus, the assertion [tex]\( 12^n \)[/tex] ends with the digit zero is not correct.
### Conclusion:
- The assertion [tex]\( 12^n \)[/tex] ends with the digit zero is false.
- The reason given that any number ends with zero if its prime factor is of the form [tex]\( 2^m \times 5^n \)[/tex] is true.
Therefore, the correct detailed solution is:
1. Assertion: False
2. Reason: True
### Assertion:
[tex]\( 12^n \)[/tex] ends with the digit zero, where [tex]\( n \)[/tex] is a natural number.
### Reason:
Any number ends with zero if its prime factor is of the form [tex]\( 2^m \times 5^n \)[/tex], where [tex]\( m \)[/tex] and [tex]\( n \)[/tex] are natural numbers.
### Step-by-Step Solution:
1. Analyzing the Reason:
- For a number to end with the digit zero, it must be divisible by 10.
- The number 10 has the prime factorization [tex]\( 2 \times 5 \)[/tex].
- Therefore, a number can only end with zero if it has both 2 and 5 in its prime factorization.
The given reason is correct. Any number ending in zero must include at least one factor of 2 and one factor of 5 in its prime factorization.
2. Analyzing the Assertion:
- We need to check if [tex]\( 12^n \)[/tex] can end with the digit zero.
- Let's start with the prime factorization of 12:
[tex]\[ 12 = 2^2 \times 3 \][/tex]
- Raising 12 to any power [tex]\( n \)[/tex]:
[tex]\[ 12^n = (2^2 \times 3)^n = 2^{2n} \times 3^n \][/tex]
- For [tex]\( 12^n \)[/tex] to end with the digit zero, it must have both a factor of 2 and a factor of 5.
- From the prime factorization [tex]\( 2^{2n} \times 3^n \)[/tex], we see that while 2 is present (in the form [tex]\( 2^{2n} \)[/tex]), there is no factor of 5.
- Since [tex]\( 12^n \)[/tex] does not include a factor of 5, it cannot end with the digit zero regardless of the value of [tex]\( n \)[/tex].
Thus, the assertion [tex]\( 12^n \)[/tex] ends with the digit zero is not correct.
### Conclusion:
- The assertion [tex]\( 12^n \)[/tex] ends with the digit zero is false.
- The reason given that any number ends with zero if its prime factor is of the form [tex]\( 2^m \times 5^n \)[/tex] is true.
Therefore, the correct detailed solution is:
1. Assertion: False
2. Reason: True
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