Here I present one of my favourites; a problem of rare invention. From the 1998 Tournament of the Towns:
(Two-person case) Two robbers stole a bag of coins from a merchant. Each coin is worth an integer number of pennies. It is known that if any single coin is removed from the bag, then the remaining coins can be divided fairly among the two robbers (that is, they both get coins with the same total value in pennies).
Prove that after one coin is removed, the number of coins remaining is even.
(General case) A gang of robbers stole a bag of coins from a merchant. Each coin is worth an integer number of pennies. It is known that if any single coin is removed from the bag, then the remaining coins can be divided fairly among the robbers (that is, they all get coins with the same total value in pennies).
Prove that after one coin is removed, the number of coins remaining is divisible by the number of robbers.
In reading the problem, it appears that something quite fundamental about numbers is being claimed. Despite all the apparent freedom we have to assign values to coins, the condition of fairness alone is enough to force the number of coins to obey a particularly strict law.
I solved this puzzle quite a while ago, but I can still recall (i) its difficulty and (ii) my awe at its creation. Though I cannot quite remember how I came to have what I believe is the crucial insight required to solve it, I suspect that it came from trying out small cases, deciding what I wanted to be true, and then trying to show that I could have what I wanted after all.
Suppose there are robbers and coins with values . Let be the total value of all the coins in the bag.
The value of each coin leaves the same remainder upon division by .
Take any two coins and . The fairness condition implies that divides both and , so must also divide . That is to say, must be a multiple of ; equivalently, and leave the same remainder upon division by .
Notice that there was nothing special about or , and the above argument holds for any two coins.
This means we can write each coin value as , where is the (common) remainder when each value is divided by .
If we remove any single coin , the new total is divisible by . It follows that divides
and hence that divides . (*)
The solution now seems to be very close: all we are aiming for is to prove that divides (**), i.e. that the number of coins in the bag (after a single coin is removed) is a multiple of the number of robbers. In fact if only we knew that and had no common factors, then we could move straight from (*) to (**) and complete the proof on the spot.
Now here comes the crucial insight: incredibly, we can insist that and have no factors in common, for if we divide every coin value by the highest common factor (hcf) of all the values, the fairness condition is preserved!
To see this, imagine that a single coin has been removed, and the remaining coins have been fairly distributed among the robbers. If we divide each coin value by the hcf of all the values, then each robber’s earnings are reduced by exactly the same factor, and so the distribution is still fair. Note also that, as desired, and now have no factor in common; if they did, say , then would divide the value of every coin , which is impossible since we have already divided every value by the hcf of all the values.
This means is that at the start of our solution, we may simply divide all values by their highest common factor, and run through the argument presented above, where this time we are allowed to make the leap from (*) to (**), thus completing the proof.
This problem really has to be seen to be believed. It possesses many of the hallmarks of an outstanding mathematics problem: an interesting context, a simple premise, a surprising result and a difficult-to-find but elegant solution. It is the sort of problem I aspire to create, and I can only hope it brings you as much joy as it brought me.