## AravindG 4 years ago find the number of divisors of 8400 excluding 1 and 8400 also find sum of all these divisors

1. anonymous

really?

2. anonymous

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

3. AravindG

but till where?

4. AravindG

there are many terms

5. anonymous

first find the prime factorization of the number 8400

6. anonymous

$2^4×3×5^2×7$

7. anonymous

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

8. JamesJ

Right ... In general:

9. anonymous

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

10. anonymous

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

11. anonymous

the old counting principle rears its head again

12. AravindG

can u relate this to permutation and combinations??

13. JamesJ

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

14. anonymous

next part...

15. anonymous

what i had above is the sum of all divisors

16. anonymous

For number of divisors is, (4+1) (1+1) (2+1)(1+1)

17. anonymous

it's permutation

18. AravindG

cnfused

19. 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

20. AravindG

k hw we find sum of them

21. 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)

22. anonymous

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

23. AravindG

hw to find sum using permutation and combination?

24. anonymous

Aravind needs a prof?

25. anonymous

proof*

26. AravindG

i want to relate to permuation and combination

27. anonymous
28. anonymous

i write it wrong! $\sigma(p^k)=\frac{p^{k+1}-1}{p-1}$ so for this example you would have $\sigma(8400)=\sigma(2^5)\sigma(3)\sigma(5^2)\sigma(7)$ $=(2^6-1)\times\frac{3^2-1}{2}\times \frac{5^3-1}{4}\times \frac{7^2-1}{6}$

29. AravindG

k thx all

30. anonymous

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

31. anonymous

A solution using Mathematica is attached.