My take on this: The naive approach (which is what I did) is to assign each second to a binary number and work out how high the number goes - this works out at

My take on this: The naive approach (which is what I did) is to assign each second to a binary number and work out how high the number goes - this works out at 2^12. The clever way other people did - and which I've been thinking about all morning, is: The complete answer to the test is a set of true/false answers that we map to 23-bit vectors. We know that there 2^23 vectors, which is way more than the 90*60=5400 ways we have to identify each of them. Pretty sure we're all clear on that. The hard part comes when we start saying that we can let some of the bits be wrong. This lets us say that some of the 2^23 23-bit vectors are pretty much the same as some others. I think of it like grouping vectors together into groups that are all related to each other by flipping up to 3 bits. So let's take an example with 4-bit vectors instead because I find it more intuitive. This is a test with 4 true/false answers. Our naive method would make us think we need 2^4=16 seconds to correctly identify true/false answers. But if we look at what happens when we say one of those answers can be wrong, it gets interesting. Let's look at the first 4-bit vector: [0 0 0 0] Now we can let any bit be wrong: [0 0 0 0] [0 0 0 1] [0 0 1 0] [0 1 0 0] [1 0 0 0] ...and we have FIVE possible vectors identified by the SAME second. Just by looking at the first vector we've reduced our space from 2^16 to 2^16-5. And remember we can group all of our 4-bit vectors together like this. If you let 2 answers be wrong you have larger, fewer groups. If you let 3 answers be wrong you only need one bit of information! Your cheating friend gets up on second 1 or second 2. This is my intuitive way of looking at it anyway. For the actual maths and such look at other answers...