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Calcmathlete Group Title

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?

  • 2 years ago
  • 2 years ago

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  1. experimentX Group Title
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    hmm... i never seen this ... could you elaborate?

    • 2 years ago
  2. Calcmathlete Group Title
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    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.

    • 2 years ago
  3. Calcmathlete Group Title
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    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.

    • 2 years ago
  4. experimentX Group Title
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    so it would be |dw:1344116637049:dw|

    • 2 years ago
  5. Calcmathlete Group Title
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    I believe so...I don't get why it doesn't work though...I'm pretty sure that's the formula...hold on...

    • 2 years ago
  6. Calcmathlete Group Title
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    Oh wait, it does work! I'm guessing that means that the formula is correct?

    • 2 years ago
  7. experimentX Group Title
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    |dw:1344117005528:dw| 4 is missing

    • 2 years ago
  8. Calcmathlete Group Title
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    But 7 x 4 = 28 1 + 2 + 3 + 4 + 6 + 12 = 28

    • 2 years ago
  9. experimentX Group Title
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    hmm ... try this for any other number let it for 9

    • 2 years ago
  10. Calcmathlete Group Title
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    It should be 13. Using the formula.

    • 2 years ago
  11. experimentX Group Title
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    |dw:1344117256625:dw| interesting ...do you know what does this work?

    • 2 years ago
  12. Calcmathlete Group Title
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    I have no idea how this works...it's number theory, but I don't know why it works...

    • 2 years ago
  13. Calcmathlete Group Title
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    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.

    • 2 years ago
  14. experimentX Group Title
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    |dw:1344118722255:dw| what't the deal with this? seems the sum of factors is never a prime no

    • 2 years ago
  15. experimentX Group Title
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    |dw:1344119038064:dw|

    • 2 years ago
  16. Calcmathlete Group Title
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    What? I'm not really following...

    • 2 years ago
  17. experimentX Group Title
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    the sum of factors is equal to the ... = product of geometric sum of prime divisors.

    • 2 years ago
  18. experimentX Group Title
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    waw ... nice!!

    • 2 years ago
  19. Calcmathlete Group Title
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    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 :)

    • 2 years ago
  20. experimentX Group Title
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    lol ... this is easy.

    • 2 years ago
  21. experimentX Group Title
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    Not sure this workds let a number \( X \) has prime factors \( a^n \) and \( b^m\)

    • 2 years ago
  22. experimentX Group Title
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    let the divisors of X be \( 1 + a+ b + c +d +f + ... +X = (1 + a +a^2 + a^3 + ... +a^n) \times \\ (1 +b +b^2 + b^3 + ... +b^m) \)

    • 2 years ago
  23. experimentX Group Title
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    eg ... X = a^n*b^m a = 1*a b = 1*b

    • 2 years ago
  24. experimentX Group Title
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    from the left (divisors of X) ... choose any number ... you should be able to get it ... by one of the combination of the right.

    • 2 years ago
  25. Calcmathlete Group Title
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    I see what is going on from there...then you just apply the formula for the sum of a geometric series?

    • 2 years ago
  26. experimentX Group Title
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    yep ... who want's to calculate that long?? I wouldn't even ask wolf ..

    • 2 years ago
  27. Calcmathlete Group Title
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    lol...me either...thanks again :)

    • 2 years ago
  28. experimentX Group Title
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    yw .. and thank you too for ... this Q

    • 2 years ago
  29. Calcmathlete Group Title
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    lol np :) KG just watching from a distance XD

    • 2 years ago
  30. KingGeorge Group Title
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    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 \(n\).

    • 2 years ago
  31. Calcmathlete Group Title
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    I am not familiar with the totient formula, so I'm clueless...

    • 2 years ago
  32. KingGeorge Group Title
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    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 \[\text{sopd}(n)=2(4)+5(2)+11(5)=73\]

    • 2 years ago
  33. KingGeorge Group Title
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    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.

    • 2 years ago
  34. KingGeorge Group Title
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    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, \[1+2+3+4+6+12=28\]

    • 2 years ago
  35. KingGeorge Group Title
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    Why this works, is actually pretty neat, although a little messy to prove.

    • 2 years ago
  36. KingGeorge Group Title
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    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.

    • 2 years ago
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