Question

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

Answer 1

really?

Answer 2

oh hold on we can do this , i just have to recall how. but the first thing we need is to factor

Answer 3

but till where?

Answer 4

there are many terms

Answer 5

first find the prime factorization of the number 8400

Answer 6

\[2^4×3×5^2×7 \]

Answer 7

exponents are 4,1,2,1 respectively, and if remember correctly that means number of factors is
\[5\times 2\times 3\times 2\]

Answer 8

Right ... In general:

Answer 9

but don't take my word for it.
ok take my word for it

Answer 10

\[(2^0 + 2^1+2^2+2^3+2^4) \times (3^0+3^1) \times (5^0+5^1+5^2) \times (7^0+7^1)\]

Answer 11

the old counting principle rears its head again

Answer 12

can u relate this to permutation and combinations??

Answer 13

you'll just need to subtract to 2 from that answer to take out 1 and 8400

Answer 14

next part...

Answer 15

what i had above is the sum of all divisors

Answer 16

For number of divisors is,

(4+1) (1+1) (2+1)(1+1)

Answer 17

it's permutation

Answer 18

cnfused

Answer 19

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

Answer 20

k hw we find sum of them

Answer 21

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)

Answer 22

sum is found by
\[\sigma(n)\] where
\[\sigma(p^k)=\frac{p^{k+1}}{p-1}\] and
\[\sigma(n)\] is multiplicative

Answer 23

hw to find sum using permutation and combination?

Answer 24

Aravind needs a prof?

Answer 25

proof*

Answer 26

i want to relate to permuation and combination

Answer 27

Check this out: http://www.artofproblemsolving.com/Wiki/index.php/Divisor_function

Answer 28

i write it wrong!
\[\sigma(p^k)=\frac{p^{k+1}-1}{p-1}\]

so for this example you would have
\[\sigma(8400)=\sigma(2^5)\sigma(3)\sigma(5^2)\sigma(7)\]
\[=(2^6-1)\times\frac{3^2-1}{2}\times \frac{5^3-1}{4}\times \frac{7^2-1}{6}\]

Answer 29

k thx all

Answer 30

The answer should be 30752 for the sum of all factors of 8400

Answer 31

A solution using Mathematica is attached.