Riker's Mailbox

Friday, February 04, 2005

ENSUING

Ahem,

The arrival of the heretofore unspecified time-to-conclusion of my little contest from post previous is officially announced. I must say, for all the time that's been allotted for reader response, in consideration of the at least four people who read this blog, I'm a little saddened by the fact that only half of you ventured a guess. But, the feeling passes. On to bigger and better things...

...such as announcing the winner of the contest!

Shit, saddened again. Of the two entrants, Rob and Nichelle, one posted a valid solution set in the least amount of time without outside council, but entirely lacked theoretical explanation, while the other posted a valid solution set in slightly more time with outside council, but noted a very critical point on the theoretical side. Alas, no clear-cut winner will be determined, as both entrants were mostly correct in damn near the same amount of time (which I will note, were on the order of six times faster than the time it took me to find my solution - to my credit, I did it without a computer program while both entrants' solutions were derived from program output), so I will announce Rob and Nichelle to be co-winners of the grand prize six pack! Since neither contestant flat-out defeated the other, one six pack prize will be awarded; it will be rationed out equally (two for Rob, two for Nichelle, and two for me because I can do that... it's my damn contest, so shut up already... dick).

Typically at this point, the full contest solution would be presented for all to see. Since I like doing things differently, however, and since you guys know that about me, I'm deliberately not not going to be typical, which is different than what I usually do, and I will thereby present the solution I was looking for:

The Solution
Considering the method by which the students chose to alter their lockers, one can infer a certain number of things:

(1) The second student will affect all lockers that are a multiple of two, the fiftieth student will affect all lockers that are a multiple of fifty, and so on.

(2) Conversely, a locker number that is, for example, a multiple of two and of fifty, such as locker #100, will be affected by both students two and fifty. If those two affect locker #100, then the students that represent all other numbers that multiply together to equal 100 will also stop on this locker.

(3) This reveals that a locker of a certain number will be affected exactly as many times as the number of integer multiples it has, or in short: the number of factors a locker number has, that is how many times its state will change.

(4) Since they all started closed, and the sequence begins with student one, at the end of the sequence a locker will be opened if it has an odd number of factors, and it will be closed if it has an even number of factors.

NOTE: At this point I went ahead and started going through the numbers and figuring it out manually. After a bit I realized that there was a pattern emerging. The first thing I noticed was that the number of closed lockers between each open locker increased by two each time, ie., there were two closed lockers between the first two open lockers, then there were four closed lockers between the next two open lockers, then there were six closed, etc. Upon further inspection, I realized what else was going on.

(5) Most numbers have an even number of factors, simply because it takes two numbers to multiply together in order to be considered multiples of the number in question. This is true for every single number except for a select few: perfect squares. Perfect squares are the only numbers out there which have an odd number of factors, simply because one of their factors is counted twice... rather, a perfect square has an even number of factors due to the definition of multiplication, but a perfect square will always have an odd number of unique factors. A practical example, in case I didn't explain it well enough in words:

35 - this number is hit by the following: 1 X 35 and 5 X 7. As such, it will end up being closed, due to the even number of unique factors - 1, 5, 7, and 35.

36 - this number is hit by the following: 1 X 36, 2 X 18, 3 X 12, 4 X 9, and 6 X 6. As such, it will be closed, because being a perfect square, in order for there to be an even number of multiples, 6 has to be used twice; therefore, there are only an odd number of unique factors (1, 2, 3, 4, 6, 9, 12, 18, and 36), and it will therefore be open.

So, in extremely short form, all lockers whose numbers are perfect squares, will be open.
End of The Solution

This took me about half an hour to stumble through, and I bounced a bit of theory upon partner-in-crime and answer-man Joe Farnsworth. But, that's neither here nor there. What is definitely here, is that the contest is over and the (co-)winners have been announced.

But incidentally, a contest of only winners really indicates that no one's beaten anyone, and therefore invalidates the very notion that defines their label, in this particular case. As such, I have decided that I will simply buy a six pack and drink it. I will then tell Rob and Nichelle how it was. My prediction: Delicious.

2 comments:

  1. I ventured a guess! I was just wrong...

    ReplyDelete
  2. Anonymous6:47 PM

    I don't recall what my login name was -- meanwhile

    Fame and Notoriety is reminded of a Mt Dew commercial -- I'll do a little modification but a few of you will get it. For those few the lines:
    We got hosed Nichelle, we got hosed.

    We'll still love you Riker!
    -rsp

    ReplyDelete