Sum of factors of 27000

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• 27000

[konopong]

To determine the sum of the factors for any natural number just take the product of the sum of the choices for each prime factor of the number (choices when determining each factor).

Since 27,000 = 2**3 x 3**3 x 5**3

Sum of factors = (1 + 2 +4 +8) x (1 + 3 + 9 + 27) x (1 + 5 + 25 +125)

Sum of factors = 15 x 40 x 156 = 93,600

Also if you multiply the above expression out without combining terms you will generate all 64 factors of 27,000.

# of factors = 4 x 4 x 4

[Thymonico]

Sweet!

This is the type of maths wake me you from your dreams grasping for pen and paper in an entertaining effort to still your mind. Or is it just me? >.>

[Vindex27]

@Thymonico No, it's not just you and thank goodness for that!

[PoketoMtg]

@anfedorov (1+2+4+8)=(2^0+2^1+2^2+2^3); (1+3+9+27)=(3^0+3^1+3^2+3^3); (1+5+25+125)=(5^0+5^1+5^2+5^3)­;

Now you have every multiple of 27000 if you distribute this enough times.

A easy, but good, example is 36=2^2*3^2 the sum of all the values can be looked at as

(2^0+2^1+2^2)(3^0+3^1+3^2)=(1+­2+4)(1+3+9)=(7)(13)=91 and if you find every single factor:

(1+2+4)+(3+6+12)+(9+18+36)=91

This is an easy way to save paper.

[WildChildftw]

@anfedorov

It follows rather trivially from the sum of a geometric series, and the fact that the sum of all positive divisors of n can be written as the product of (p_i + (p_i)^2 + ... + (p_i)^k_i) for i=1,2,3, ..., n, (since each positive divisors of n appears once and only once in the expansion of this product.)

[anfedorov]

@91jgphonecall Start with the Calculus lectures, then. Quantum physics isn't about anything you actually can see/feel/touch, but about abstract models for some phenomena observed, and a strong foundation in math (several years of it, most likely) will be necessary.

[anfedorov]

@konopong Proof (or citation) required.

[91jgphonecall]

off topic but, khan every time you say you will eventually teach us string theory and higher stuff like that i get all excited. im from a small town and pretty much a bum but someday ill be a bum who noes string thoery.

[cssftw100]

@tomEwylde so he actually writes with his hand?

[MalaTemporaCurrunt]

you could use gnu/maxima function 'divsum'