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AravindG

  • 4 years ago

find the number of divisors of 8400 excluding 1 and 8400 also find sum of all these divisors

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  1. anonymous
    • 4 years ago
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    really?

  2. anonymous
    • 4 years ago
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    oh hold on we can do this , i just have to recall how. but the first thing we need is to factor

  3. AravindG
    • 4 years ago
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    but till where?

  4. AravindG
    • 4 years ago
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    there are many terms

  5. anonymous
    • 4 years ago
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    first find the prime factorization of the number 8400

  6. anonymous
    • 4 years ago
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    \[2^4×3×5^2×7 \]

  7. anonymous
    • 4 years ago
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    exponents are 4,1,2,1 respectively, and if remember correctly that means number of factors is \[5\times 2\times 3\times 2\]

  8. JamesJ
    • 4 years ago
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    Right ... In general:

  9. anonymous
    • 4 years ago
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    but don't take my word for it. ok take my word for it

  10. anonymous
    • 4 years ago
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    \[(2^0 + 2^1+2^2+2^3+2^4) \times (3^0+3^1) \times (5^0+5^1+5^2) \times (7^0+7^1)\]

  11. anonymous
    • 4 years ago
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    the old counting principle rears its head again

  12. AravindG
    • 4 years ago
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    can u relate this to permutation and combinations??

  13. JamesJ
    • 4 years ago
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    you'll just need to subtract to 2 from that answer to take out 1 and 8400

  14. anonymous
    • 4 years ago
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    next part...

  15. anonymous
    • 4 years ago
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    what i had above is the sum of all divisors

  16. anonymous
    • 4 years ago
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    For number of divisors is, (4+1) (1+1) (2+1)(1+1)

  17. anonymous
    • 4 years ago
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    it's permutation

  18. AravindG
    • 4 years ago
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    cnfused

  19. anonymous
    • 4 years ago
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    For example, if the number is 12. then it's prime factorization will be 2^2 * 3 factors are: 2^0* 3^0 , 2^1 * 3^0, 2^2 * 3^0, 2^0 * 3^1, 2^1 * 3^1, 2^2 * 3^1

  20. AravindG
    • 4 years ago
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    k hw we find sum of them

  21. anonymous
    • 4 years ago
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    2^0 * 3^0 + 2^1 * 3^0 + 2^2 * 3^0 + 2^0 * 3^1 + 2^1 * 3^1 + 2^2 * 3^1 3^0 (2^0 + 2^1 + 2^2) + 3^1 (2^0 + 2^1 + 2^2) = (3^0 + 3^1)(2^0 + 2^1 + 2^2)

  22. anonymous
    • 4 years ago
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    sum is found by \[\sigma(n)\] where \[\sigma(p^k)=\frac{p^{k+1}}{p-1}\] and \[\sigma(n)\] is multiplicative

  23. AravindG
    • 4 years ago
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    hw to find sum using permutation and combination?

  24. anonymous
    • 4 years ago
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    Aravind needs a prof?

  25. anonymous
    • 4 years ago
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    proof*

  26. AravindG
    • 4 years ago
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    i want to relate to permuation and combination

  27. anonymous
    • 4 years ago
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    Check this out: http://www.artofproblemsolving.com/Wiki/index.php/Divisor_function

  28. anonymous
    • 4 years ago
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    i write it wrong! \[\sigma(p^k)=\frac{p^{k+1}-1}{p-1}\] so for this example you would have \[\sigma(8400)=\sigma(2^5)\sigma(3)\sigma(5^2)\sigma(7)\] \[=(2^6-1)\times\frac{3^2-1}{2}\times \frac{5^3-1}{4}\times \frac{7^2-1}{6}\]

  29. AravindG
    • 4 years ago
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    k thx all

  30. anonymous
    • 4 years ago
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    The answer should be 30752 for the sum of all factors of 8400

  31. anonymous
    • 4 years ago
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    A solution using Mathematica is attached.

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