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Sagot :
To determine which statements are true for all positive integers, let's analyze each one in detail.
### Statement 1: [tex]\( n^3 + 2n \)[/tex] is divisible by 3 for all positive integers [tex]\( n \)[/tex]
To check whether [tex]\( n^3 + 2n \)[/tex] is divisible by 3 for all positive integers [tex]\( n \)[/tex], we can investigate the divisibility properties:
1. Consider the expression modulo 3, [tex]\( n \equiv k \pmod{3} \)[/tex] where [tex]\( k \)[/tex] can be 0, 1, or 2 (as any integer modulo 3 is either 0, 1, or 2).
2. We need to check all possible values of [tex]\( k \)[/tex]:
- If [tex]\( n \equiv 0 \pmod{3} \)[/tex]:
[tex]\[ n = 3m \quad \text{for some integer } m \][/tex]
[tex]\[ n^3 + 2n = (3m)^3 + 2(3m) = 27m^3 + 6m = 3(9m^3 + 2m) \][/tex]
Clearly, this is divisible by 3.
- If [tex]\( n \equiv 1 \pmod{3} \)[/tex]:
[tex]\[ n = 3m + 1 \quad \text{for some integer } m \][/tex]
[tex]\[ n^3 + 2n = (3m+1)^3 + 2(3m+1) = 27m^3 + 27m^2 + 9m + 1 + 6m + 2 = 27m^3 + 27m^2 + 15m + 3 = 3(9m^3 + 9m^2 + 5m + 1) \][/tex]
Clearly, this is divisible by 3.
- If [tex]\( n \equiv 2 \pmod{3} \)[/tex]:
[tex]\[ n = 3m + 2 \quad \text{for some integer } m \][/tex]
[tex]\[ n^3 + 2n = (3m+2)^3 + 2(3m+2) = 27m^3 + 54m^2 + 36m + 8 + 6m + 4 = 27m^3 + 54m^2 + 42m + 12 = 3(9m^3 + 18m^2 + 14m + 4) \][/tex]
Clearly, this is divisible by 3.
Since, for all cases, [tex]\( n^3 + 2n \)[/tex] is divisible by 3, Statement 1 is true.
### Statement 2: [tex]\( 5^{2x} - 1 \)[/tex] is divisible by 24 for all positive integers [tex]\( x \)[/tex]
Next, let's analyze the expression [tex]\( 5^{2x} - 1 \)[/tex] for divisibility by 24:
1. Recognize that [tex]\( 5^{2x} \)[/tex] is an even power of 5:
- For [tex]\( x = 1 \)[/tex]:
[tex]\[ 5^2 - 1 = 25 - 1 = 24 \][/tex]
Clearly divisible by 24.
2. For higher positive integers [tex]\( x \)[/tex]:
- [tex]\( 5^{2x} - 1 \)[/tex] can be expressed using modular arithmetic:
[tex]\[ 5^{2x} - 1 = (5^2)^x - 1 = 25^x - 1 \][/tex]
- Any higher even power of 5 will also produce a result that can be expressed as a multiple of 24 plus one, making the original expression divisible by 24. This follows the same pattern since the factors grow larger but [tex]\( 5^2 = 25 \mod 24 \)[/tex] still holds generally for all [tex]\( x \)[/tex].
Thus, for all positive integers [tex]\( x \)[/tex], [tex]\( 5^{2x} - 1 \)[/tex] consistently yields a result divisible by 24. Statement 2 is true.
### Conclusion
Both statements are true for all positive integers:
1. [tex]\( n^3 + 2n \)[/tex] is divisible by 3.
2. [tex]\( 5^{2x} - 1 \)[/tex] is divisible by 24.
Therefore, the correct answer is:
Both statements are true.
### Statement 1: [tex]\( n^3 + 2n \)[/tex] is divisible by 3 for all positive integers [tex]\( n \)[/tex]
To check whether [tex]\( n^3 + 2n \)[/tex] is divisible by 3 for all positive integers [tex]\( n \)[/tex], we can investigate the divisibility properties:
1. Consider the expression modulo 3, [tex]\( n \equiv k \pmod{3} \)[/tex] where [tex]\( k \)[/tex] can be 0, 1, or 2 (as any integer modulo 3 is either 0, 1, or 2).
2. We need to check all possible values of [tex]\( k \)[/tex]:
- If [tex]\( n \equiv 0 \pmod{3} \)[/tex]:
[tex]\[ n = 3m \quad \text{for some integer } m \][/tex]
[tex]\[ n^3 + 2n = (3m)^3 + 2(3m) = 27m^3 + 6m = 3(9m^3 + 2m) \][/tex]
Clearly, this is divisible by 3.
- If [tex]\( n \equiv 1 \pmod{3} \)[/tex]:
[tex]\[ n = 3m + 1 \quad \text{for some integer } m \][/tex]
[tex]\[ n^3 + 2n = (3m+1)^3 + 2(3m+1) = 27m^3 + 27m^2 + 9m + 1 + 6m + 2 = 27m^3 + 27m^2 + 15m + 3 = 3(9m^3 + 9m^2 + 5m + 1) \][/tex]
Clearly, this is divisible by 3.
- If [tex]\( n \equiv 2 \pmod{3} \)[/tex]:
[tex]\[ n = 3m + 2 \quad \text{for some integer } m \][/tex]
[tex]\[ n^3 + 2n = (3m+2)^3 + 2(3m+2) = 27m^3 + 54m^2 + 36m + 8 + 6m + 4 = 27m^3 + 54m^2 + 42m + 12 = 3(9m^3 + 18m^2 + 14m + 4) \][/tex]
Clearly, this is divisible by 3.
Since, for all cases, [tex]\( n^3 + 2n \)[/tex] is divisible by 3, Statement 1 is true.
### Statement 2: [tex]\( 5^{2x} - 1 \)[/tex] is divisible by 24 for all positive integers [tex]\( x \)[/tex]
Next, let's analyze the expression [tex]\( 5^{2x} - 1 \)[/tex] for divisibility by 24:
1. Recognize that [tex]\( 5^{2x} \)[/tex] is an even power of 5:
- For [tex]\( x = 1 \)[/tex]:
[tex]\[ 5^2 - 1 = 25 - 1 = 24 \][/tex]
Clearly divisible by 24.
2. For higher positive integers [tex]\( x \)[/tex]:
- [tex]\( 5^{2x} - 1 \)[/tex] can be expressed using modular arithmetic:
[tex]\[ 5^{2x} - 1 = (5^2)^x - 1 = 25^x - 1 \][/tex]
- Any higher even power of 5 will also produce a result that can be expressed as a multiple of 24 plus one, making the original expression divisible by 24. This follows the same pattern since the factors grow larger but [tex]\( 5^2 = 25 \mod 24 \)[/tex] still holds generally for all [tex]\( x \)[/tex].
Thus, for all positive integers [tex]\( x \)[/tex], [tex]\( 5^{2x} - 1 \)[/tex] consistently yields a result divisible by 24. Statement 2 is true.
### Conclusion
Both statements are true for all positive integers:
1. [tex]\( n^3 + 2n \)[/tex] is divisible by 3.
2. [tex]\( 5^{2x} - 1 \)[/tex] is divisible by 24.
Therefore, the correct answer is:
Both statements are true.
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