# A puzzle and a solution

Nair posed this question about how one can ‘measure’ any number using a power series (his example was a series in 3 – 1,3,9,27…, and numbers upto 40). The comments on his post are fairly explanatory. The basic idea is that any number can be represented as an expansion in any power series (all whole numbers, mind), as long as one is allowed coefficients upto the number used in the power series… or as one of the commenters on Nair’s post put it, you need all the numbers in the ‘base’ of ‘N’.

(The long and short of it is that balances have two scales (aha!), which means a maximum possible of three  ‘coefficients’, so only power series in 2 or 3 will work. DIY. Also, a power series in 2 needs only two coefficients; which means you can measure any number putting your weights 1,2,4… on one side of the balance alone. Again, DIY).

It’s somewhat harder to imagine this with numbers other than 2 or 3, because we have no analogues of weighing scales with 3 or more places, but it can be done in abstraction. It actually is, if you think of it. Everyday, in fact. (I gave the bus driver three 10s, a 5 and a 2 for a Rs. 37 daily pass just yesterday).

OK. So much for the solution. Here’s a puzzle (you didn’t assume the puzzle and the solution were connected, did you?):

The Monty Hall problem is well known. Can you explain why the probabilities of victory for the two remaining boxes aren’t the same? An attempt at an explanation below the fold:

The important player in the puzzle is the host, who knows the answers beforehand. Because he has this information, and uses the information when he chooses to open one of the two remaining doors, he’s changing the probabilities in the game, and your chances of victory are no longer equal to the a priori probabilities of victory for each door.