What is the sum of all positive even divisors of 1000? I know similar questions and answers have been posted here, but I don't understand the answers. Can anyone show me how to solve this problem in a simple way? This is a math problem for 8th grade students.Thank you very much!

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

 A: 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.
A: We know that product of two odds is always odd and since $1000=2^3\cdot5^3$, the only odd terms are $1$, $5^1$, $5^2$, $5^3$, and their sum is 1 + 5 + 25 + 125 = 156. 
Also sum of divisors of 1000 = σ($2^3$.$5^3$) = [($2^4$-1)/ (2-1)].[($5^4$-1)/(5-1)] = 15.156 = 2340. 
Subtracting the sum of odd divisors gives the sum of even divisors, 2340-156 = 2184.
I know the function for the summation of divisors of a number, σ ,maybe a bit new for the 8th grade but it is easy to grasp and worthwhile to know.
A: $n$ is a positive even divisor of $1000$ if and only if $n = 2m$ where $m$ is a divisor of $500$. Since $500 = 2^2 \times 5^3$, there are $(2+1)(3+1) = 12$ divisors of $500$. Those divisors are
\begin{array}{rr}
    1, & 500, \\
    2, & 250, \\
    4, & 125, \\
    5, & 100, \\
   10, &  50, \\
   20, &  25 \\
\end{array}
so there are $12$ positive even divisors of $1000$.
Those divisors are
\begin{array}{cc}
    2, & 1000, \\
    4, &  500, \\
    8, &  250, \\
   10, &  200, \\
   20, &  100, \\
   40, &   50 \\
\end{array}
The sum of the positive divisors of $500 = 2^2 \times 5^3$ equals 
$\dfrac{2^3 - 1}{2 - 1} \times \dfrac{5^4 - 1}{5 - 1} = 1092$
So the sum of the even divisors of $1000$ is $2 \times 1092 = 2184$
A: 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 $\sum_{i=1}^3\sum_{j=0}^32^i5^j=(\sum_{i=1}^3 2^i)(\sum_{j=0}^3 5^j)=\frac{2^4-2}{2-1}\cdot\frac{5^4-1}{5-1}=14\cdot156=2184$.
A: We want to exclude the odd divisors, and the odd divisors of $1000$ are exactly the divisors of $125$. Therefore your answer is the sum of divisors of $1000$, minus the sum of divisors of $125$, or using $\sigma$ for the sum of divisors,
$$
\sigma(1000) - \sigma(125).
$$
Now, to compute $\sigma(n)$, you split $n$ into its relatively prime parts and then sum the divisors of each part. So we get
\begin{align*}
\sigma(1000) - \sigma(125)
&= \sigma(125) \sigma(8) - \sigma(125) \\
&= (1 + 5 + 25 + 125)(1 + 2 + 4 + 8) - (1 + 5 + 25 + 125) \\
&= 156 \cdot15 - 156 \\
&= 156 \cdot 14 \\
&= 2184.
\end{align*}
A: 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.
