How did you do?

Puzzle 1 Answer:

Removing one small item (which we know has a weight of at least 100g) would mean the remainder of the large package would weigh between 2 and 4kg and so could be sent for £10. The last item could then be sent as an individual parcel for an additional £5 – a total cost of £15.

Puzzle 2 Answer:

I can make exactly 150 as follows: “Science” (16) + “Fantasy” (27) + “Sport” (42) + “Politics” (65) = 150. So, unfortunately it looks like my maths books are being left behind!

There isn’t actually a specific method to find this solution beyond trial and error. The problem is part of a larger class of problems called NP problems, which are the subject of a million-dollar question which asks whether or not a quicker method of solution exists in general.

Puzzle 3 Answer:

First, note that a locker will be open only if it has had its state changed an odd number of times (since they all start closed). Due to the rules in which they are opened/ closed, this means the number of the locker must have an odd number of factors. Therefore, the number that remain open is just the number of integers from one to 100 that have an odd number of factors. So, the question is which ones are they?

Take a simple example of the number 6. It can be made by 6 x 1, and 2 x 3, so in total it has four different factors. What about a prime number such as 5? Well this is 5 x 1 and nothing else, so it has two factors.

Now, in both of these cases we have an even number of divisors, and in fact this will almost always be true because they come in pairs. If you have one number as a factor, then it must be multiplied by something else to give the starting value. So, if the factors of any number come in pairs, how can we end up with an odd number?

We can have a repeated factor. For example, 3 x 3 = 9. In this instance we have a further two factors from 9 x 1, but overall the number 9 has three factors – an odd number!

So, to answer our original question, we just need to know which numbers have a repeated factor (like 3 in the example above). If we call this repeated factor k, then we need all numbers that are equal to k2 for some number k – in other words the square numbers!

These are relatively straightforward to list for the first 100 numbers: 1, 4, 9, 16, 25, 36, 49, 64, 81, 100 – which means we have exactly 10 square numbers, and so 10 numbers with an odd number of factors, and therefore 10 lockers that will be open at the end.