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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?
 one year ago
 one year ago
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?
 one year ago
 one year ago

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experimentXBest ResponseYou've already chosen the best response.2
hmm... i never seen this ... could you elaborate?
 one year ago

CalcmathleteBest ResponseYou've already chosen the best response.0
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.
 one year ago

CalcmathleteBest ResponseYou've already chosen the best response.0
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.
 one year ago

experimentXBest ResponseYou've already chosen the best response.2
so it would be dw:1344116637049:dw
 one year ago

CalcmathleteBest ResponseYou've already chosen the best response.0
I believe so...I don't get why it doesn't work though...I'm pretty sure that's the formula...hold on...
 one year ago

CalcmathleteBest ResponseYou've already chosen the best response.0
Oh wait, it does work! I'm guessing that means that the formula is correct?
 one year ago

experimentXBest ResponseYou've already chosen the best response.2
dw:1344117005528:dw 4 is missing
 one year ago

CalcmathleteBest ResponseYou've already chosen the best response.0
But 7 x 4 = 28 1 + 2 + 3 + 4 + 6 + 12 = 28
 one year ago

experimentXBest ResponseYou've already chosen the best response.2
hmm ... try this for any other number let it for 9
 one year ago

CalcmathleteBest ResponseYou've already chosen the best response.0
It should be 13. Using the formula.
 one year ago

experimentXBest ResponseYou've already chosen the best response.2
dw:1344117256625:dw interesting ...do you know what does this work?
 one year ago

CalcmathleteBest ResponseYou've already chosen the best response.0
I have no idea how this works...it's number theory, but I don't know why it works...
 one year ago

CalcmathleteBest ResponseYou've already chosen the best response.0
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.
 one year ago

experimentXBest ResponseYou've already chosen the best response.2
dw:1344118722255:dw what't the deal with this? seems the sum of factors is never a prime no
 one year ago

experimentXBest ResponseYou've already chosen the best response.2
dw:1344119038064:dw
 one year ago

CalcmathleteBest ResponseYou've already chosen the best response.0
What? I'm not really following...
 one year ago

experimentXBest ResponseYou've already chosen the best response.2
the sum of factors is equal to the ... = product of geometric sum of prime divisors.
 one year ago

CalcmathleteBest ResponseYou've already chosen the best response.0
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 :)
 one year ago

experimentXBest ResponseYou've already chosen the best response.2
lol ... this is easy.
 one year ago

experimentXBest ResponseYou've already chosen the best response.2
Not sure this workds let a number \( X \) has prime factors \( a^n \) and \( b^m\)
 one year ago

experimentXBest ResponseYou've already chosen the best response.2
let the divisors of X be \( 1 + a+ b + c +d +f + ... +X = (1 + a +a^2 + a^3 + ... +a^n) \times \\ (1 +b +b^2 + b^3 + ... +b^m) \)
 one year ago

experimentXBest ResponseYou've already chosen the best response.2
eg ... X = a^n*b^m a = 1*a b = 1*b
 one year ago

experimentXBest ResponseYou've already chosen the best response.2
from the left (divisors of X) ... choose any number ... you should be able to get it ... by one of the combination of the right.
 one year ago

CalcmathleteBest ResponseYou've already chosen the best response.0
I see what is going on from there...then you just apply the formula for the sum of a geometric series?
 one year ago

experimentXBest ResponseYou've already chosen the best response.2
yep ... who want's to calculate that long?? I wouldn't even ask wolf ..
 one year ago

CalcmathleteBest ResponseYou've already chosen the best response.0
lol...me either...thanks again :)
 one year ago

experimentXBest ResponseYou've already chosen the best response.2
yw .. and thank you too for ... this Q
 one year ago

CalcmathleteBest ResponseYou've already chosen the best response.0
lol np :) KG just watching from a distance XD
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.1
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 \(n\).
 one year ago

CalcmathleteBest ResponseYou've already chosen the best response.0
I am not familiar with the totient formula, so I'm clueless...
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.1
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 \[\text{sopd}(n)=2(4)+5(2)+11(5)=73\]
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.1
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.
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.1
So back to \(12=2^2\cdot3\). According to the formula, the sum of divisors should be \[\frac{81}{21}\cdot\frac{91}{31}=28\]When we look at the divisors and add them manually, \[1+2+3+4+6+12=28\]
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.1
Why this works, is actually pretty neat, although a little messy to prove.
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.1
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.
 one year ago
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