What is the sum of all positive even divisors of 1000?

First consider the prime factorization of $1000$. We have:

$$1000=2^3\times 5^3$$

Now, how can we list all the factors of $1000$? We see that we can try listing them in a table:

$$\begin{array}{c|c|c|} & \text{$5^0$} & \text{$5^1$} & \text{$5^2$} & \text{$5^3$} \\ \hline \text{$2^0$} & 1 & 5 & 25 & 125 \\ \hline \text{$2^1$} & 2 & 10 & 50 & 250 \\ \hline \text{$2^2$} & 4 & 20 & 100 & 500 \\ \hline \text{$2^3$} & 8 & 40 & 200 & 1000 \\ \hline \end{array}$$

We see that we can take $(2^1+2^2+2^3) \times (5^0 + 5^1 + 5^2 + 5^3) = 2184$. To get the sum of all factors, we would also include $2^0$ on the left side of the multiplication. We exclude $2^0$ because those would be odd factors.

Since $1000=2^3\cdot5^3$, the even divisors of $1000$ have the form $2^i5^j$, where $1\leq i\leq 3$ and $0\leq j\leq 3$. There are only 12 of them, so you can do this calculation directly.

Alternatively, it is $\sum_{i=1}^3\sum_{j=0}^32^i5^j=(\sum_{i=1}^3 2^i)(\sum_{j=0}^3 5^j)=\frac{2^4-2}{2-1}\cdot\frac{5^4-1}{5-1}=14\cdot156=2184$.

First, factor $1000=2^3\cdot 5^3$. A divisor of $1000$ has to be of the form $2^a\cdot 5^b$. If you want it to be even, you need $a \ge 1$. How many choices do you have? You can simplify the calculation (though it is not worth it for this small a case) by making it the product of two geometric series. If you wanted the sum of even divisors of $10^{32}$ it would be worthwhile, and is worth understanding.