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.