anonymous
  • anonymous
Just testing to see if I've memorized this formula again: \[\text{Sum of the factors of a number: }\frac{a^{p + 1} - 1}{a - 1} \times \frac{b^{q + 1} - 1}{b - 1} \times \frac{c^{r + 1} - 1}{c - 1}...\]where a, b, and c are prime factors of the given number and where p, q, and r are exponents relating to the multiplicity of the prime factors. Is it correct?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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experimentX
  • experimentX
hmm... i never seen this ... could you elaborate?
anonymous
  • anonymous
For instance, if we were given the number 12, and you prime factor it: \[12 = 2^2 \times 3\]Then a would be 2, p would be 2, b would be 3, and q would be 1.
anonymous
  • anonymous
Then the sum of the factors would be 2 + 2 + 3 = 7. I think the formula is a spinoff of the number of factors formula, but I'm trying to see if I have completely got this formula down.

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experimentX
  • experimentX
so it would be |dw:1344116637049:dw|
anonymous
  • anonymous
I believe so...I don't get why it doesn't work though...I'm pretty sure that's the formula...hold on...
anonymous
  • anonymous
Oh wait, it does work! I'm guessing that means that the formula is correct?
experimentX
  • experimentX
|dw:1344117005528:dw| 4 is missing
anonymous
  • anonymous
But 7 x 4 = 28 1 + 2 + 3 + 4 + 6 + 12 = 28
experimentX
  • experimentX
hmm ... try this for any other number let it for 9
anonymous
  • anonymous
It should be 13. Using the formula.
experimentX
  • experimentX
|dw:1344117256625:dw| interesting ...do you know what does this work?
anonymous
  • anonymous
I have no idea how this works...it's number theory, but I don't know why it works...
anonymous
  • anonymous
But does it seem to work as a formula? lol The original question was to see if my formula was correct since I am in the midst of memorizing a whole bunch of miscellaneous formulas.
experimentX
  • experimentX
|dw:1344118722255:dw| what't the deal with this? seems the sum of factors is never a prime no
experimentX
  • experimentX
|dw:1344119038064:dw|
anonymous
  • anonymous
What? I'm not really following...
experimentX
  • experimentX
the sum of factors is equal to the ... = product of geometric sum of prime divisors.
experimentX
  • experimentX
waw ... nice!!
anonymous
  • anonymous
THat does make sense somewhat. lol I didn't come up with the formula, I'm just trying to memorize it? Thanks for helping me verify :)
experimentX
  • experimentX
lol ... this is easy.
experimentX
  • experimentX
Not sure this workds let a number \( X \) has prime factors \( a^n \) and \( b^m\)
experimentX
  • experimentX
let the divisors of X be \( 1 + a+ b + c +d +f + ... +X = (1 + a +a^2 + a^3 + ... +a^n) \times \\ (1 +b +b^2 + b^3 + ... +b^m) \)
experimentX
  • experimentX
eg ... X = a^n*b^m a = 1*a b = 1*b
experimentX
  • experimentX
from the left (divisors of X) ... choose any number ... you should be able to get it ... by one of the combination of the right.
anonymous
  • anonymous
I see what is going on from there...then you just apply the formula for the sum of a geometric series?
experimentX
  • experimentX
yep ... who want's to calculate that long?? I wouldn't even ask wolf ..
anonymous
  • anonymous
lol...me either...thanks again :)
experimentX
  • experimentX
yw .. and thank you too for ... this Q
anonymous
  • anonymous
lol np :) KG just watching from a distance XD
KingGeorge
  • KingGeorge
Watching...And slowly coming to the decision that there may not be a formula that counts the sum of prime factors. Looking at a graph of this function, it appears to be fairly random. On another note, the formula you originally wrote in the question strongly resembles the totient function which counts how many numbers less than \(n\) are coprime to \(n\).
anonymous
  • anonymous
I am not familiar with the totient formula, so I'm clueless...
KingGeorge
  • KingGeorge
Actually, ignore what I said about there not being a formula. Suppose you have a number \(n=2^4\cdot5^2\cdot11^5\), and call this sum of prime divisors function "sopd," then \[\text{sopd}(n)=2(4)+5(2)+11(5)=73\]
KingGeorge
  • KingGeorge
From what I can tell, you're mixing up this function with the sum of divisors function. The sum of divisors function is exactly what you wrote above.
KingGeorge
  • KingGeorge
So back to \(12=2^2\cdot3\). According to the formula, the sum of divisors should be \[\frac{8-1}{2-1}\cdot\frac{9-1}{3-1}=28\]When we look at the divisors and add them manually, \[1+2+3+4+6+12=28\]
KingGeorge
  • KingGeorge
Why this works, is actually pretty neat, although a little messy to prove.
KingGeorge
  • KingGeorge
And @experimentX nailed it when he said "the sum of factors is equal to the ... = product of geometric sum of prime divisors." That's one of the key points of this when you want to prove it.

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