Sum of reciprocal of the positive divisors of $1800.$

Attempt: divisors of $1800 = 2^3\times 3^2 \times 5^2$

so sum of divisors of $1800$ is $\displaystyle (1+2+2^2+2^3)\times (1+3+3^2)\times (1+5+5^2)$

$ \displaystyle = \bigg(\frac{2^4-1}{2-1}\bigg)\times \bigg(\frac{3^3-1}{3-1}\bigg)\times \bigg(\frac{5^3-1}{5-1}\bigg)$

but answer is different, could some help me to solve it. Thanks.

  • $\begingroup$ You need the sum of the reciprocals, you are taking the sum of the divisors themselves. $\endgroup$ – Teresa Lisbon Mar 3 '17 at 9:51
  • $\begingroup$ yes астон вілла олоф мэллбэрг but how can i calculate it, thanks $\endgroup$ – DXT Mar 3 '17 at 9:52
  • 1
    $\begingroup$ What about taking the reciprocals: $(1+2^{-1}+2^{-2}+2^{-3})\times(1+3^{-1}+3^{-2})\times(1+5^{-1}+5^{-2})$? $\endgroup$ – skyking Mar 3 '17 at 9:54
  • $\begingroup$ Oh, just do geometric progression formula on the reciprocals instead. It's the same, basically. $\endgroup$ – Teresa Lisbon Mar 3 '17 at 9:55
  • $\begingroup$ means divided it by $1800$ $\endgroup$ – DXT Mar 3 '17 at 9:57


$$\sum_{d=1,d|n}^n\dfrac1d=\dfrac1n\sum_{d=1,d|n}^n\dfrac nd=\dfrac1n\sum_{d=1,d|n}^nd$$

Now use this

See also: Is there a formula to calculate the sum of all proper divisors of a number?

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