# 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?

• @ColinMcLarty What about, say, 8? Sep 12, 2016 at 17:40
• @GoodDeeds The divisors correspond, but are not identical. Each number 2*k that divides 1000 corresponds to the number k that divides 500. Sep 12, 2016 at 17:55
• @chepner I had misunderstood, thank you. Sep 12, 2016 at 17:56

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.

• Interesting visual approach, but this would only work for smaller numbers, yes? If I take i.e. $1778700=2^2\cdot3^1\cdot5^2\cdot7^2\cdot11^2$, you would have 5 "dimensions" for your table, which would not be easy to get the even factors out? Or is there another way? Sep 13, 2016 at 11:44
• @hamena314 All odd factors will lie in the 4-dimensional 'row' which corresponds to the factor $2^0$ and all even ones will lie in the row corresponding to $2^1$ and $2^2$. So again, the sum of positive even divisors will be $(2^1 + 2^2) \times (3^0 + 3^1) \times (5^0 + 5^1 + 5^2) \times (7^0 + 7^1 + 7^2) \times (11^0 + 11^1 + 11^2).$ Sep 13, 2016 at 16:04

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.

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.

• The product of two evens is even and also product of an odd and even is also even. The only odd terms are product of odd and an odd .
– user356774
Sep 13, 2016 at 8:58

$n$ 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 $2 \times 1092 = 2184$

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*}