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

Question

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.

You need the sum of the reciprocals, you are taking the sum of the divisors themselves. — Sarvesh Ravichandran Iyer, Mar 03 '17 at 09:51
yes астон вілла олоф мэллбэрг but how can i calculate it, thanks — DXT, Mar 03 '17 at 09:52
What about taking the reciprocals: $(1+2^{-1}+2^{-2}+2^{-3})\times(1+3^{-1}+3^{-2})\times(1+5^{-1}+5^{-2})$? — skyking, Mar 03 '17 at 09:54
Oh, just do geometric progression formula on the reciprocals instead. It's the same, basically. — Sarvesh Ravichandran Iyer, Mar 03 '17 at 09:55

score 0 · Accepted Answer · edited Apr 13 '17 at 12:21

0

HINT:

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

Now use this

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

edited Apr 13 '17 at 12:21

Community

answered Mar 03 '17 at 09:57

lab bhattacharjee

1 Answers1