Posted March 3, 2014 (edited) All the numbers below should be assumed to be positive integers. Definition. An abundant number is an integer n whose divisors add up to more than In. Definition. A perfect number is an integer n whose divisors add up to exactly In. Definition. A deficient number is an integer n whose divisors add up to less than In. Example. 12 is an abundant number, because 1 + 2 + 3+ 4 + 6+12 = 28 and 28 > 2x12. However, 14 is a deficient number, because 1 + 2 + 7 + 14 = 24, and 24 < 2 x 14. Your task is to consider the following conjectures and determine, with proofs, whether they are true or false. Conjecture 1. A number is abundant if and only if it is a multiple of 6. Conjecture 2. If n is perfect, then kn is abundant for any k in N. Conjecture 3. If p1 and p2 are primes, then p1/p2 is abundant. Conjecture 4. If n is deficient, then every divisor of n is deficient. Conjecture 5. If n and m are abundant, then n + m is abundant. Conjecture 6. If n and m are abundant, then nm is abundant. Conjecture 7. If n is abundant, then n is not of the form pm for some natural m and prime p. Edited March 3, 2014 by BMAD 0 Share this post Link to post Share on other sites

0 Posted March 3, 2014 The definitions should be in comparison of 2n not in. All the numbers below should be assumed to be positive integers. Definition. An abundant number is an integer n whose divisors add up to more than In. Definition. A perfect number is an integer n whose divisors add up to exactly In. Definition. A deficient number is an integer n whose divisors add up to less than In. Example. 12 is an abundant number, because 1 + 2 + 3+ 4 + 6+12 = 28 and 28 > 2x12. However, 14 is a deficient number, because 1 + 2 + 7 + 14 = 24, and 24 < 2 x 14. Your task is to consider the following conjectures and determine, with proofs, whether they are true or false. Conjecture 1. A number is abundant if and only if it is a multiple of 6. Conjecture 2. If n is perfect, then kn is abundant for any k in N. Conjecture 3. If p1 and p2 are primes, then p1/p2 is abundant. Conjecture 4. If n is deficient, then every divisor of n is deficient. Conjecture 5. If n and m are abundant, then n + m is abundant. Conjecture 6. If n and m are abundant, then nm is abundant. Conjecture 7. If n is abundant, then n is not of the form pm for some natural m and prime p. 0 Share this post Link to post Share on other sites

0 Posted March 3, 2014 If p1 and p2 are primes then p1/p2 cannot be an integer and therefore not abundant (Definition 1). 0 Share this post Link to post Share on other sites

0 Posted March 4, 2014 i think he ment multiplication not division there. 1) false. 28000 :: 1 +2 +4 +5 +7 +10 +14 +16 +25 +28 +32 +125 +224 875 +1000+1120+1750+2000+2800 +4000 +5600 +7000 +14000 = 40638, definitely abundant. 2) true, though i have no idea how to prove it at this point 3) true, with the exception of 2 and 3. multiplication of two primes will produce a larger result than addition. 4) true, follows from 2 assuming that can be proven. 0 Share this post Link to post Share on other sites

0 Posted March 4, 2014 False, as proven by phil1882. A smaller example is 40 < 1+2+4+5+8+10+20. True for k>1. For each divisor d_{i} of n, consider the larger number kd_{i} which is a divisor of kn. Sum(kd_{i}) = k*Sum(d_{i}) = 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 p_{1}*p_{2}? In that case, 2*3 = 6 is perfect, all others are deficient. If p_{1} = p_{2}, the divisors of p_{1}*p_{2} are 1, p_{1}, and p_{1}^{2}. The sum 1+p_{1}+p_{1}^{2} < p_{1}+p_{1}+p_{1}^{2} = 2*p_{1}+p_{1}^{2} </= p_{1}*p_{1}+p_{1}^{2} = 2*p_{1}^{2} = 2*p_{1}*p_{2}, so p_{1}*p_{2} is deficient. If p_{1} < p_{2}, the divisors of p_{1}*p_{2} are 1, p_{1}, p_{2}, and p_{1}*p_{2}. The sum 1+p_{1}+p_{2}+p_{1}*p_{2} </= p_{2}+p_{2}+p_{1}*p_{2} = 2p_{2}+p_{1}*p_{2} </= p_{1}*p_{2}+p_{1}*p_{2} = 2*p_{1}*p_{2}, with equality iff p_{1} = 2 and p_{2} = 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. 0 Share this post Link to post Share on other sites

0 Posted March 4, 2014 nicely done for 2.for seven i'm not sure if thats what hes asking. i think he means that if n is abundant, then it must not have one prime raised to some power. that is if n is abundant, it must be something like 2^4*3^2*5^3.... but i agree based on 2 its definitely false. 0 Share this post Link to post Share on other sites

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