Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

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?

See more answers at
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly


Get your free account and access expert answers to this and thousands of other questions

hmm... i never seen this ... could you elaborate?
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.
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.

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

so it would be |dw:1344116637049:dw|
I believe so...I don't get why it doesn't work though...I'm pretty sure that's the formula...hold on...
Oh wait, it does work! I'm guessing that means that the formula is correct?
|dw:1344117005528:dw| 4 is missing
But 7 x 4 = 28 1 + 2 + 3 + 4 + 6 + 12 = 28
hmm ... try this for any other number let it for 9
It should be 13. Using the formula.
|dw:1344117256625:dw| interesting you know what does this work?
I have no idea how this's number theory, but I don't know why it works...
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.
|dw:1344118722255:dw| what't the deal with this? seems the sum of factors is never a prime no
What? I'm not really following...
the sum of factors is equal to the ... = product of geometric sum of prime divisors.
waw ... nice!!
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 :)
lol ... this is easy.
Not sure this workds let a number \( X \) has prime factors \( a^n \) and \( b^m\)
let the divisors of X be \( 1 + a+ b + c +d +f + ... +X = (1 + a +a^2 + a^3 + ... +a^n) \times \\ (1 +b +b^2 + b^3 + ... +b^m) \)
eg ... X = a^n*b^m a = 1*a b = 1*b
from the left (divisors of X) ... choose any number ... you should be able to get it ... by one of the combination of the right.
I see what is going on from there...then you just apply the formula for the sum of a geometric series?
yep ... who want's to calculate that long?? I wouldn't even ask wolf .. either...thanks again :)
yw .. and thank you too for ... this Q
lol np :) KG just watching from a distance XD
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 \(n\).
I am not familiar with the totient formula, so I'm clueless...
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 \[\text{sopd}(n)=2(4)+5(2)+11(5)=73\]
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.
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, \[1+2+3+4+6+12=28\]
Why this works, is actually pretty neat, although a little messy to prove.
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.

Not the answer you are looking for?

Search for more explanations.

Ask your own question