I know that because two is divisible into eight.

我知道这个 因为8能被2整除。

I will switch every light bulb that's divisible by two.

我会按下编号能被2整除的那些灯泡。

This covers every number which divides evenly into 45, once, and only once.

这涵盖了所有能被45整除一次且仅一次的数。

These major aspects are those that are divisible by 10 and evenly divided in relation to the 360 degree wheel.

这些主要相位指那些能被10整除，并且360盘中均匀分布的相位。

Negative 20 over 84, which is the same thing, they're both divisible by 4, the numerator divided by 4 is negative 5, over 21.

负20除以84， 也样的意思，它们都能被4整除， 分子除以4负5， 分母21。

We started with this question asking us to count the number of subsets of 1 up  to 2 000 whose sum is divisible by five.

我们从这个问题开始，要求计算从1到2000的集合中，元素和能被5整除的子集数量。

The factors of 710 are 71, 5 and 2. So if the remainder or residue is divisible by any of those then so is the number.

710的因数71、5和2。所以如果余数或剩余部分能被其中任何一个整除，那么这个数也能被整除。

The factors or the numbers that evenly divided into twenty eight are one, two, four, seven and 14 and those happen to add up to twenty eight numbers.

能被28整除的因数有1、2、4、7和14，这些数字恰好加起来等于28。

Again, thinking back to what these coefficients  encode for us, that will be answering our final question that will be counting the total number  of subsets whose sum is divisible by five.

再次回想这些系数为我们编码了什么，这将回答我们最后一个问题， 即计算所有元素和能被5整除的子集的总数。

So in the 16th Century, Pope Gregory XIII tweak the system again so that leap ears would occur in any year that's divisible by four, but only in centuries that are evenly divided by 400, confusing, I know.

所以16世纪，教皇格里高利十三世再次修改了系统，使得闰年发生每四年一次的年份中，但只有能被400整除的年份才闰年，令人困惑，我知道。

It also can't be 3 above a multiple of 6 unless it's the number 3 itself since all of those are divisible by 3. So at least at the smaller-scale nothing magical is going on.

它也不能比6的倍数大3的数字，除非这个数字本身3，因为所有这样的数字都能被3整除。所以较小的范围内并没有什么神奇的事情发生。

And all in all the ones that we care about, the subsets with a sum divisible by five, have been put over here on the left and it looks  like there's a total of eight of them.

总的来说， 我们关心的那些子集，即元素和能被5整除的子集，都被放了左边， 看起来总共有8个。

And if we turn to chapter 2 advanced problems, problem number 10 asks this seemingly innocent question " find the number of subsets of the set  one up to two thousand, the sum of whose elements is divisible by five" .

而如果我们转向第二章高级问题，第10题提出了这个看似无辜的问题：“找出从1到2000的集合中，元素和能被5整除的子集的数量” 。