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- False, as proven by phil1882. A smaller example is 40 < 1+2+4+5+8+10+20.
- True for k>1. For each divisor di of n, consider the larger number kdi which is a divisor of kn. Sum(kdi) = k*Sum(di) = 2kn, but 1 is also a divisor of kn which has not yet been accounted for. So the sum of kn's divisors is at least 2kn+1, which makes kn abundant.
- Not sure what you mean. Should it be p1*p2? In that case, 2*3 = 6 is perfect, all others are deficient. If p1 = p2, the divisors of p1*p2 are 1, p1, and p12. The sum 1+p1+p12 < p1+p1+p12 = 2*p1+p12 </= p1*p1+p12 = 2*p12 = 2*p1*p2, so p1*p2 is deficient. If p1 < p2, the divisors of p1*p2 are 1, p1, p2, and p1*p2. The sum 1+p1+p2+p1*p2 </= p2+p2+p1*p2 = 2p2+p1*p2 </= p1*p2+p1*p2 = 2*p1*p2, with equality iff p1 = 2 and p2 = 3.
- True. The negation is equivalent to "there is a non-deficient positive integer n and a positive integer k such that kn is deficient", which is easily disproven using the method in my proof for conjecture 2.
- False. 40 and 12 are abundant, but 40+12 = 52 > 1+2+4+13+26 is deficient.
- True. If n is abundant, then nm is abundant for any m.
- False. Every n>1, abundant or otherwise, has a prime factor p, and therefore is of the form pm for some natural m.