The sum of the proper factors of 945 is 975. Any number whose proper factors add up to more than the number itself is an
abundant number. 945 is the lowest odd abundant number. I don't know how many even abundant numbers there are below 945, but it looks like there are over 100.
Simple way of showing that there are more than 100 even abundant numbers under 945 (aside from the brute force method of exanmining each number below 945 and determining if it is both even and abundant):
ReplyDeleteFirst, note that any multiple of a perfect number (higher than that perfect number itself) is abundant. That's relatively easy to prove. Comment back if you want a proof.
Next, note that 6 is a perfect number.
Those two facts together imply that multiples of 6 higher than 6 itself are all abundant numbers. i.e., 12, 18, 24, 30, etc. are all abundant.
Since 945 is more than 101 × 6, we have more than 100 multiples of 6 (higher than 6 itself) that are less than 945. These are all abundant. In fact, that accounts for more than 150 abundant numbers less than 945.
If you want to go farther, 28 is also a perfect number. So you can also count all multiples of 28 other than 28 itself. But be careful not to doublecount multiples of 84, which are already included above by virtue of being mulitiples of 6.
You are right. And I think that any multiple of an abundant number is also abundant.
ReplyDeleteYes, a multiple of an abundant number is also abundant. The proof of that runs along the same lines as the proof that a multiple of a prfect number is abundant.
DeleteBefore continuing, let me define the function, R(n), as the ratio of the sum of the proper factors of n to n. For example, R(2) = 1/2 because the sum of the factors of 2 is 1. By definition, R(n) = 1 where n is a perfect number, R(n) > 1 where n is an abundant number, and R(n) < 1 where n is a deficient number.
That said, the proof is essentially showing that R(an) > R(n) where a and n are positive integers. Put another way, if a is a multiple of b, then R(a) > R(b). That multiples of perfect and abundant numbers are abundant follows.
Should have said that a > 1
Deleteim fascinated by this. I'd like to see the proof that any multiple of a perfect number (higher than that perfect number itself) is abundant, just because I'm interested. I'm confused about 28 being a perfect number. Isn't a perfect number one where the factors add up to the number itself, ie, 6's factors are 1, 2 and 3, which total to 6. I get 1, 2, 4, 7, 12, and 14 for 28. These add to 40. Am I missing something?
ReplyDelete28 is indeed a perfect number. You erroneously included 12 as a factor. But 28 is not divisible by 12.
DeleteAs for the proof, I shall try to present it here, but the limited formatting of blogspot comments will make it difficult. If I can't get it across, maybe you and I can correspond via email. I'm sure Amy can facilitate. I will try to present three things. First, I'll give a kind of handwaving description of the proof in nonprecise terms. That should help with following the proof when I present it. Since the proof is by construction, this is feasible. After this, I'll take a concrete example, and show how the construction works. Finally I'll go through a formal proof. Note that in this presentation it is necessary to repeatedly refer to "proper factors." In the interests of convenience, I will simply say "factors." From now on, whenever I say "factors" I mean "proper factors."
First a paraphrase of the proof:
If we start with a perfect number, P, then we know its factors add up to P. But we want to look at Q, which is a multiple of P (say it's zP). More specifically, let's look at the factors of Q. We know that, from each factor, f, of P, we can construct a factor of Q by multiplying f by z. Adding up these constructed factors, we get Q. But in addition, there is 1, which is also a factor of Q. Adding that to the list, we have a list of factors of Q whose sum is greater than Q. Thus we know that Q is abundant. Note that there can be more factors of Q that we haven't yet identified. But that doesn't matter.
Second, let's take a concrete example:
Let's consider a perfect number and a multiple of that perfect number. Let's choose 6 and 30 (Note that 30 = 5 × 6. The 5 is important). The factors of 6 are 1, 2, and 3. Form those three factors of 6, we can derive three factors of three factors of 30, simply by multiplying each of them by 5. so 5 (= 5 × 1), 10 (= 5 × 2) and 15 (= 5 × 3) are all factors of 30. Also, since they were derived by multiplying 5 by each of the factors of 6, we know that they add up to 30. Just to verify, we can see 5 + 10 + 15 = 30. But there's another factor of 30. Namely, 1. So we know that 1, 5, 10 and 15 are all factors of 30 and they add up to 31 which is greater than 30. Thus 30 is abundant. I note thyat there are other factors of 30 that are not listed. Namely 3 and 6. But that's irrelevant to the question at hand. Identuifying more factors just raises the sum of the factors. Once we know the sum of the factors of 30 is greater than 30, we are done with the question.
Finally, the formal proof.
Consider the perfect number, P. The factors of P are P(1), P(2), ...P(n), where n is the number of factors. We know that P(1) + P(2) + ... + P(n) = P.
Given z, a positive integer greater than 1, we need to show that the sum of the factors of zP is greater than zP.
Because we know that P(1), P(2), ... P(n) are factors of P, we also know that zP(1), zP(2), ... zP(n) are factors of zP. Further, because we know P(1) + P(2) + ... + P(n) = P, we also know that zP(1) + zP(2) + ... +zP(n) =zP.
Now, consider the number 1. 1 is a factor of zP. But it is not in the list, zP(1), zP(2), ... zP(n). We4 know this because z > 1 and P(i) >= 1 for all i. So now consider the list 1, zP(1), zP(2), ... zP(n). That is a list of factors of zP that adds up to zP + 1, which is greater than zP. Thus zP must be abundant.
Let me note that I didn't actually use the ratio R that I mentioned in my Januatry 12 comment above. It could be restated using R if I wanted to demonstarte the more general result that if A is a multiple of B, then R(A) > R(B).