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Smuggling yourself aboard the rogue submarine was the easy part.
偷搭上叛逃的潜水艇还算是简单的部分。
Hacking into the nuclear missile launch override -- a little harder.
骇入核弹发射撤消系统这就稍微难一点了。
But now you've got a problem: you don't have the override code.
但现在你遇到问题了：你没有撤消密码。
You know you need the same two numbers that the agents of chaos just used to authorize the launch. But one wrong answer will lock you out.
你知道你需要的两个数字，就是混乱特务刚才用来授权发射的数字。但弄错一个数字就会让系统锁住。
From your hiding spot, you've been able to learn the following:
从你躲藏的位置，你得到了下列信息：
The big boss didn't trust any minion with the full information to launch nuclear missiles on their own.
大头目不相信任何喽啰，不会把全部的信息都给同一个人，以免楼啰自己发射核弹。
So he gave one launch code to Minion A, the other to minion B, and forbade them to share the numbers with each other.
所以他把一个发射密码给喽啰A，另一个给喽啰B，并禁止他们告诉对方自己的数字。
When the order came, each entered their own number and activated the countdown.
命令下来时，他们各自输入他们的数字，启动发射倒数。
That was 50 minutes ago, and there's only 10 minutes left before the missiles launch.
那是五十分钟前的事，只剩下十分钟核弹就要发射了。
Suddenly, the boss says, "Funny story -- your launch codes were actually related.
突然，头目说：“说件趣事--你们的发射密码其实是相关的。
I chose a set of distinct positive integers with at least two elements, each less than 7, and told their sum to you, A, and their product to you, B."
我选了一组正整数，至少有两个数字，每个都小于7，把它们的加总告诉A，乘积告诉B。”
After a moment of awkward silence, A says to B, "I don't know whether you know my number."
在短暂的尴尬沉默后，A对B说：“我不知道你是否知道我的数字。”
B thinks this over, then responds, "I know your number, and now I know you know my number too."
B想过之后，响应说：“我知道你的数字，且现在我知道你也知道我的数字。”
That's all you've got. What numbers do you enter to override the launch?
你就只听到这些。你要输入什么数字来撤销飞弹发射？
Ignorance-based puzzles like this are notoriously difficult to work through.
这种以“不知情”为基础的谜题是出名了难解。
The trick is to put yourself in the heads of both characters and narrow down the possibilities based on what they know or don't know.
技巧在于从两位角色的角度来思考，并根据他们知道什么或不知道什么，来找到更接近的答案。
So let's start with A's first statement.
咱们先从A的第一句陈述开始。
It means that B could conceivably have something with the potential to reveal A's number, but isn't guaranteed to.
意即，B可能有信息可以揭露出A的数字，但不保证一定可以。
That doesn't sound very definitive, but it can lead us to a major insight.
听起来无法确定什么，但能让我们窥见一件重要的事。
The only scenarios where B could know A's number are when there's exactly one valid way to factor B's number.
只有一种情况下，B会知道A的数字，就是B的数字只有一种因子拆解的方式。
Try factoring a few and you'll find the pattern.
试着拆解几个数字，你就会找到模式。
It could be prime -- where the product must be of 1 and itself -- or it could be the product of 1 and the square of a prime, such as 4.
可能是质数--乘积即1和质数本身--或是一和质数的平方，比如4。
In both cases, there is exactly one sum. For a number like 8, factoring it into 2 and 4, or 1, 2, and 4, creates too many options.
在这两种情况下，都刚好只有一个加总值。如果是八这类数字，因子可能是2与4，或者1、2、4，会有太多加总的可能性。
Because the boss's numbers must be less than 7, A's list of B's possibilities only has these 4 numbers.
因为头目说每个数字都小于七，A把B可能的数字列成列表，上面就只会有这四个数字。
Here's where we can conclude a major clue.
这里我们就能得到一条重要线索。
To think B could have these numbers, A's number must be a sum of their factors -- so 3, 4, 5, or 6.
A认为B可能有这些数字，所以A的数字一定是其因子的总和--3、4、5，或6。
We can eliminate 3 and 4, because if the sum was either,
我们可以拿掉3和4，因为如果总和是3或4，
the product could only be 2 or 3, in which case A would know that B already knows A's number, contradicting A's statement.
那乘积就只有可能是2或3，不论是2或3，A都能知道B已经知道A的数字，这和A的说法矛盾。
5 and 6, however, are in play, because they can become sums in multiple ways.
不过，5和6都有可能，因为有多种方式加总出这两个数字。
The need to consider this is one of the most difficult parts of this puzzle.
知道要考虑这一点是这类谜题最困难的地方。
The crucial thing to remember is that there's no guarantee that B's number is on A's list
重点是要记住，B的数字并不保证一定在A的列表，
those are just the possibilities from A's perspective that would allow B to deduce A's number.
列表上的只是从A的角度认为可能的数字，A认为这样B就能推理出A的数字。
That ambiguity forces us to go through unintuitive multi-step processes like:
这种模糊性，迫使我们要进行非直觉式的多步骤过程，比如：
consider a product, see what sums can result from its factors, then break those apart and see what products can result.
先想一个乘积，看看它的因子能加出哪些总和，接着再把这些总和拆开，看能乘出哪些乘积。
We'll soon have to do something similar going from sums to products and back to sums.
我们很快就得要采用类似的过程，从总和到乘积，再回到总和。
But now we know -- when A made his first statement, he must have been holding either 5 or 6.
但现在我们知道，当A说出第一句陈述时，他的数字一定是五或六。
B has access to the same information we do, so he knows this too.
我们能得到的信息B也都能得到，所以他也知道这一点。
Let's review what's in each brain at this point: everyone knows a lot about the sum, but only B knows the product.
咱们来回顾一下每个人的脑中有什么：大家都对总和知道很多，但只有B知道乘积。
Now let's look at the first part of B's statement. What if A's number was 5?
咱们来看看B陈述的第一部分。如果A的数字是5呢？
That could be from 1+4 or 2+3, in which case B would have either 4 or 6.
5有可能是1+4或2+3，这两种情况下，B的数字分别会是4或6。
4 would tell B what A had, like he said, because there's only one option to make the product: 4 times 1.
若是4，B就会知道A的数字，如他所言，因为要让乘积是4，只有一种可能：四乘以一。
6, on the other hand, could be broken down three ways, which sum like so.
另一方面，6则有3种拆解方式，其总和如图所示。
7 isn't on B's list of possible sums, but 5 and 6 both are.
B的可能总和清单中没有7，但有5和6。
Meaning that B wouldn't know whether A's number was 5 or 6, and we can eliminate this option because it contradicts his statement.
意思就是B不会知道A的数字是5或6，我们可以剔除这个选项，因为和B的陈述矛盾。
So this is great -- 5 and 4 could be the override code, but how do we know it's the only one?
这样很棒，5和4有可能是撤销密码，但我们怎么知道只有这个可能性？
Let's consider if A's number was 6 -- which would be 1+5, 2+4, or 1+2+3, giving B 5, 8, or 6, respectively.
咱们来看看A的数字若是6会如何--6可能是1+5、2+4、1+2+3则B分别会是5、8、6。
If B had 5, he'd know that A had 6. And if he had 8, the possibilities for A would be 2+4 and 1+2+4.
如果B的数字是5，他就会知道A的是6。如果B是8，A可能会是2+4或1+2+4。
Only 6 is on the list of possible sums, so B would again know that A had 6.
只有6出现在可能的总和清单上，所以同样的，B会知道A的数字是6。
To summarize, if A had 6, he still wouldn't know whether B had 5 or 8.
总结一下，如果A是6，他不会知道B是5还是8。
That contradicts the second half of what B said, and 5 and 4 must be the correct codes.
不符合B陈述的后半段，所以5和4就一定是正确的密码。
With seconds to spare you override the missile launch, shoot yourself out of the torpedo bay, and send the sub to the bottom of the ocean.
在只剩几秒的时候，你撤销了核弹发射，从鱼雷舱把你自己发射出去，送潜水艇到海底长眠。
