A community for students.
Here's the question you clicked on:
 0 viewing
AravindG
 4 years ago
find the number of divisors of 8400 excluding 1 and 8400 also find sum of all these divisors
AravindG
 4 years ago
find the number of divisors of 8400 excluding 1 and 8400 also find sum of all these divisors

This Question is Closed

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh hold on we can do this , i just have to recall how. but the first thing we need is to factor

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0first find the prime factorization of the number 8400

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0exponents are 4,1,2,1 respectively, and if remember correctly that means number of factors is \[5\times 2\times 3\times 2\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0but don't take my word for it. ok take my word for it

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[(2^0 + 2^1+2^2+2^3+2^4) \times (3^0+3^1) \times (5^0+5^1+5^2) \times (7^0+7^1)\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0the old counting principle rears its head again

AravindG
 4 years ago
Best ResponseYou've already chosen the best response.2can u relate this to permutation and combinations??

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.1you'll just need to subtract to 2 from that answer to take out 1 and 8400

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0what i had above is the sum of all divisors

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0For number of divisors is, (4+1) (1+1) (2+1)(1+1)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0For 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
 4 years ago
Best ResponseYou've already chosen the best response.2k hw we find sum of them

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.02^0 * 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
 4 years ago
Best ResponseYou've already chosen the best response.0sum is found by \[\sigma(n)\] where \[\sigma(p^k)=\frac{p^{k+1}}{p1}\] and \[\sigma(n)\] is multiplicative

AravindG
 4 years ago
Best ResponseYou've already chosen the best response.2hw to find sum using permutation and combination?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Aravind needs a prof?

AravindG
 4 years ago
Best ResponseYou've already chosen the best response.2i want to relate to permuation and combination

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Check this out: http://www.artofproblemsolving.com/Wiki/index.php/Divisor_function

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i write it wrong! \[\sigma(p^k)=\frac{p^{k+1}1}{p1}\] so for this example you would have \[\sigma(8400)=\sigma(2^5)\sigma(3)\sigma(5^2)\sigma(7)\] \[=(2^61)\times\frac{3^21}{2}\times \frac{5^31}{4}\times \frac{7^21}{6}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0The answer should be 30752 for the sum of all factors of 8400

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0A solution using Mathematica is attached.
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.