Can you solve the rogue submarine riddle? - Alex Rosenthal

Smuggling yourself aboard the rogue submarine was the easy part. Hacking into the nuclear missile launch override—a little harder. But now you’ve got a problem: you don’t have the override code. 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.

From your hiding spot, you’ve been able to learn the following: The big boss didn’t trust any minion with the full information to launch nuclear missiles on their own. So he gave one launch code to Minion A, the other to Minion B, and forbade them to share the numbers with each other. When the order came, each entered their own number and activated the countdown. That was 50 minutes ago, and there's only 10 minutes left before the missiles launch.

Suddenly, the boss says, “Funny story—your launch codes were actually related. 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.” After a moment of awkward silence, A says to B, “I don’t know whether you know my number.” B thinks this over, then responds, “I know your number, and now I know you know my number too.”

That’s all you’ve got. What numbers do you enter to override the launch? Pause now to figure it out for yourself.

Answer in 3
Answer in 2
Answer in 1

Ignorance-based puzzles like this are notoriously difficult to work through. 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. So let's start with A's first statement. It means that B could conceivably have something with the potential to reveal A’s number, but isn’t guaranteed to. That doesn’t sound very definitive, but it can lead us to a major insight.

The only scenarios where B could know A’s number are when there’s exactly one valid way to factor B’s number. Try factoring a few and you’ll find the pattern. 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. 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.

Because the boss’s numbers must be less than 7, A’s list of B’s possibilities only has these 4 numbers. Here’s where we can conclude a major clue. To think B could have these numbers, A’s number must be a sum of their factors—so 3, 4, 5, or 6. We can eliminate 3 and 4, because if the sum was either, 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.

5 and 6, however, are in play, because they can become sums in multiple ways. The need to consider this is one of the most difficult parts of this puzzle. The crucial thing to remember is that there’s no guarantee that B’s number is on A’s list—those are just the possibilities from A’s perspective that would allow B to deduce A’s number. That ambiguity forces us to go through unintuitive multi-step processes like: consider a product, see what sums can result from its factors, then break those apart and see what products can result.

We’ll soon have to do something similar going from sums to products and back to sums. But now we know—when A made his first statement, he must have been holding either 5 or 6. B has access to the same information we do, so he knows this too. Let’s review what’s in each brain at this point: everyone knows a lot about the sum, but only B knows the product.

Now let’s look at the first part of B’s statement. What if A’s number was 5? That could be from 1+4 or 2+3, in which case B would have either 4 or 6. 4 would tell B what A had, like he said, because there’s only one option to make the product: 4 times 1. 6, on the other hand, could be broken down three ways, which sum like so. 7 isn’t on B’s list of possible sums, but 5 and 6 both are. 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.

So this is great—5 and 4 could be the override code, but how do we know it's the only one? 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. 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. Only 6 is on the list of possible sums, so B would again know that A had 6.

To summarize, if A had 6, he still wouldn’t know whether B had 5 or 8. That contradicts the second half of what B said, and 5 and 4 must be the correct codes.

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.