<s>[INST] 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?

 [/INST] $n$ 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 $2 \times 1092 = 2184$
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