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Prove that if p is a perfect number then no multiple o f p is a perfect number.
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Answer to a math question Prove that if p is a perfect number then no multiple o f p is a perfect number.
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1. Assume \( p \) \( 1 + d_1 + d_2 + \ldots + d_k = 2p \)
2. Consider a multiple of\( p \) \( np \) \( n > 1 \)
3. The sum of divisors of\( np \) \( S(np) = S(n) \cdot (1 + d_1 + d_2 + \ldots + d_k) = S(n) \cdot 2p \).
4. Assume\( np \) \( S(np) - np = np \)
5. Substitute and simplify:\( S(n) \cdot 2p - np = np \) \( S(n) \cdot 2p = 2np \)
6. This implies\( S(n) = 2n \) \( n \)
7. Generally,\( n \) \( np \)
8. Therefore,\( np \neq \text{perfect number} \)
2. Consider a multiple of
3. The sum of divisors of
4. Assume
5. Substitute and simplify:
6. This implies
7. Generally,
8. Therefore,
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