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

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

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

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Other answers:

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