AravindG
  • AravindG
find the number of divisors of 8400 excluding 1 and 8400 also find sum of all these divisors
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
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.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
really?
anonymous
  • anonymous
oh hold on we can do this , i just have to recall how. but the first thing we need is to factor
AravindG
  • AravindG
but till where?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

AravindG
  • AravindG
there are many terms
anonymous
  • anonymous
first find the prime factorization of the number 8400
anonymous
  • anonymous
\[2^4×3×5^2×7 \]
anonymous
  • anonymous
exponents are 4,1,2,1 respectively, and if remember correctly that means number of factors is \[5\times 2\times 3\times 2\]
JamesJ
  • JamesJ
Right ... In general:
anonymous
  • anonymous
but don't take my word for it. ok take my word for it
anonymous
  • anonymous
\[(2^0 + 2^1+2^2+2^3+2^4) \times (3^0+3^1) \times (5^0+5^1+5^2) \times (7^0+7^1)\]
anonymous
  • anonymous
the old counting principle rears its head again
AravindG
  • AravindG
can u relate this to permutation and combinations??
JamesJ
  • JamesJ
you'll just need to subtract to 2 from that answer to take out 1 and 8400
anonymous
  • anonymous
next part...
anonymous
  • anonymous
what i had above is the sum of all divisors
anonymous
  • anonymous
For number of divisors is, (4+1) (1+1) (2+1)(1+1)
anonymous
  • anonymous
it's permutation
AravindG
  • AravindG
cnfused
anonymous
  • anonymous
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
AravindG
  • AravindG
k hw we find sum of them
anonymous
  • anonymous
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)
anonymous
  • anonymous
sum is found by \[\sigma(n)\] where \[\sigma(p^k)=\frac{p^{k+1}}{p-1}\] and \[\sigma(n)\] is multiplicative
AravindG
  • AravindG
hw to find sum using permutation and combination?
anonymous
  • anonymous
Aravind needs a prof?
anonymous
  • anonymous
proof*
AravindG
  • AravindG
i want to relate to permuation and combination
anonymous
  • anonymous
Check this out:http://www.artofproblemsolving.com/Wiki/index.php/Divisor_function
anonymous
  • anonymous
i write it wrong! \[\sigma(p^k)=\frac{p^{k+1}-1}{p-1}\] so for this example you would have \[\sigma(8400)=\sigma(2^5)\sigma(3)\sigma(5^2)\sigma(7)\] \[=(2^6-1)\times\frac{3^2-1}{2}\times \frac{5^3-1}{4}\times \frac{7^2-1}{6}\]
AravindG
  • AravindG
k thx all
anonymous
  • anonymous
The answer should be 30752 for the sum of all factors of 8400
anonymous
  • anonymous
A solution using Mathematica is attached.
1 Attachment

Looking for something else?

Not the answer you are looking for? Search for more explanations.