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12 Sep 2006, 12:16
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What is the sum of all the factors of 12972960?
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12 Sep 2006, 15:07
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Yezz,
Based on your problem I went out researching and found some interesting things. Suffice to say I second your statement about this being useful...
Ok.. given any number n we define the following functions:
a. D(n) := the total number of +ve divisors of n
b. σk(n) := the divisor function, which is the sum of the k-th powers of all the positive divisors of n
Specifically by definition :
* σ0(n) = d(n) (k = 0)
* σ1(n) = σ(n), (k = 1)
Both these functions are multiplicative i.e.
D(a x b) = D(a) x D(b) &
σk(a x b) = σk(a) x σk(b)
For example:
144 = 12 x 12
= 2x2x3 x 2x2x3
= 2^4 x 3^2
Therefore D(n) = # of +ve divisors of 144
= D(144) = σ0(2^4 x 3^2)
= σ0(2^4) x σ0(3^2)
= (1^0 + 2^0+4^0+8^0+16^0) x (1^0 + 3^0 +9^0)
= 5 x 3
= 15
Therefore total number of divisors of 144 = 15.
Similarly sum of all the divisors of 144 = σ1(144)
= σ1(2^4 x 3^2)
= σ1(2^4) x σ1(3^2)
= (1^1 + 2^1+4^1+8^1+16^1) x (1^1 + 3^1 + 9^1)
= 31 x 13 = 403
For the given problem the number 1297296 = 2^4 x 3^4 x 7 x 11 x 13
Therefore sum of divisors = σ1(2^4 x 3^4 x 7x11x13)
= σ1(2^4) x σ1(3^4) x σ1(7) x σ1(11)xσ1(13)
= (1^1+2^1+4^1+8+1+16^1) x (1^1+3^1+9^1+27^1+81^1)
x (1^1+7^1) x (1^1 +11^1)x(1^1+13^1)
= 31 x 121 x 8x 12x14
= 5041344
FYI the number of divisors of 1297296 = σ0(2^4 x 3^4 x 7x11x13) = 5 x 5 x 2 x 2 x 2= 200.
FYI: https://en.wikipedia.org/wiki/Multiplicative_function
Based on your problem I went out researching and found some interesting things. Suffice to say I second your statement about this being useful...
Ok.. given any number n we define the following functions:
a. D(n) := the total number of +ve divisors of n
b. σk(n) := the divisor function, which is the sum of the k-th powers of all the positive divisors of n
Specifically by definition :
* σ0(n) = d(n) (k = 0)
* σ1(n) = σ(n), (k = 1)
Both these functions are multiplicative i.e.
D(a x b) = D(a) x D(b) &
σk(a x b) = σk(a) x σk(b)
For example:
144 = 12 x 12
= 2x2x3 x 2x2x3
= 2^4 x 3^2
Therefore D(n) = # of +ve divisors of 144
= D(144) = σ0(2^4 x 3^2)
= σ0(2^4) x σ0(3^2)
= (1^0 + 2^0+4^0+8^0+16^0) x (1^0 + 3^0 +9^0)
= 5 x 3
= 15
Therefore total number of divisors of 144 = 15.
Similarly sum of all the divisors of 144 = σ1(144)
= σ1(2^4 x 3^2)
= σ1(2^4) x σ1(3^2)
= (1^1 + 2^1+4^1+8^1+16^1) x (1^1 + 3^1 + 9^1)
= 31 x 13 = 403
For the given problem the number 1297296 = 2^4 x 3^4 x 7 x 11 x 13
Therefore sum of divisors = σ1(2^4 x 3^4 x 7x11x13)
= σ1(2^4) x σ1(3^4) x σ1(7) x σ1(11)xσ1(13)
= (1^1+2^1+4^1+8+1+16^1) x (1^1+3^1+9^1+27^1+81^1)
x (1^1+7^1) x (1^1 +11^1)x(1^1+13^1)
= 31 x 121 x 8x 12x14
= 5041344
FYI the number of divisors of 1297296 = σ0(2^4 x 3^4 x 7x11x13) = 5 x 5 x 2 x 2 x 2= 200.
FYI: https://en.wikipedia.org/wiki/Multiplicative_function
