This code calculated the remainder if we divide 5 into 11, which is one.

这段代码计算的是11除以5的余数，结果是1。

And then the remainder, 358, I'll just say the remainder.

然后余数，358，我就说余数。

When you divide them by two, there's nothing left over.

当你把它们除以二时，没有余数。

If there's no remainder, then okay, two is a factor.

果没有余数，二就是一个因数。

'69 is 5 twelves, remainder 9, which is 2 fours.

'69 为 5 个 12，余数 9，即 2 个 4。

If the number of steps in the distance between the cabinet-door and the arm-chair will divide by three without a remainder, then the new regimen will cure him; but if it does not so divide, then it will not.

果柜门到扶手椅之间的步数除以三没有余数，的治疗方法就会治愈；果它不分裂，它就不会分裂。

The second is that many of these residue classes contain zero prime numbers or sometimes just one so they won't show up.

第二个原因是，许多这样的余数类包含零个素数，或者有时只有一个素数，所以它们不会显示出来。

" 10 divided by 4" gives you 2 with a remainder of 2, so: " 2 remainder 2" .

所以：“10 除以 4” 得到 2，余数为 2， 所以：“2 余数 2” 。

So too, at the present time there is a remnant chosen by grace.

今也是这样，照着拣选的恩典，还有所留的余数。

The hard part is figuring out the exponent given only the result and the base.

重点是，果只给余数和基数，很难得知指数是多少。

Remainder. The next word is remainder.

余。 下一个词是余数。

There's no " remainder, " we would say.

我们不会说“余数”。

Finally, how many times does 4 go into the remainder?

最后，余数中 4 除以多少？

Some of the recent breakthroughs on small gaps between primes, edging towards that ever elusive twin prime conjecture, have their basis in understanding how primes split up among these kinds of residue classes.

最近关于素数之间小间隙的一些突破性进展，逐渐接近个一直难以捉摸的孪生素数猜想，其基础在于理解素数何分布在这些余数类中。

This means taking one number, the base, to the power of another number, the exponent, and taking the remainder when dividing by a third number, the modulus.

这意味着将一个数（底数）乘以自身另一个数（指数）次，然后除以第三个数（模数）取余数。

76 is 6 twelves, remainder 4, and there's 1 four in the remainder.

76 是 6 个十二，余数为 4，余数中有 1 个四。

So, for example, if we wanted to calculate 3 to the 5th power, modulo 31, we would calculate 3 to the 5th, which is 243, then take the remainder when divided by 31, which is 26.

所以，例，果我们想要计算3的5次方，模31，我们会先计算3的5次方，也就是243，然后求除以31的余数，即26。

And this isn't just antiquated technology either Understanding the distribution of primes in residue classes like this Continues to be relevant in modern research too!

而这不仅仅是过时的技术，理解素数在余数类中的分布，在现代研究中仍然具有相关性！

If I tell you I raised 3 to some secret number, modulo 31, and got 7 as the remainder, you'd have to test a lot of exponents to know which one I picked.

果我告诉你我把3的某个秘密数次方，然后对31取模，得到的余数是7，你得测试很多指数才能知道我选的是哪一个。

Everything whose last digit is 0 is a residue class, everything whose last digit is 1 is another residue class and so on.

所有个位数是0的数构成一个余数类，所有个位数是1的数构成另一个余数类，依此类推。