Blog post # 139:
Our Limbic Extremity
The Intellect Unattached
Weird Plagiarism
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Poll results:
Question: What?
5 votes total received.
The answers:
"What?", and "This answer. Pick me!" each received 2 votes.
"You're lying!" Received 1 vote.
"I told you so", "Vibrant spittle", "Transcend the spiral", "I think so", "Wink-wink", "How?", and "None of the above" each received 0 votes.
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The news is so often so depressing, as we all know.
From now on, I am calling the "Press"
the "Depress".
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Puzzle!
Take the 12 playing-cards, ace,2,3,4,...,10,jack,queen. It is the beginning of the day. Let's say you have to take 12 pills over the next few days.
You shuffle the "deck" and draw a card, which you then don't return to the deck. You then take your first pill on the hour noted by the card you drew. If it is a number, you take the pill at that hour. If it is an ace, you take your first pill at 1. If it is a jack, you take your first pill at 11. And if it is a queen, you take your first pill exactly 12 hours from now.
Okay, every time you draw a card, you wait that many hours (with ace=1, jack=11, queen=12, of course) from when you last took your pill to take your next pill. You then don't return the card to the deck.
After all 12 pills are taken, you notice a strange thing. The numerical values 1 through 12 for the clock's HOURS the pills were taken each occur exactly once (over the course of several days). Question: Is there a particular order of the cards that allows each hour (on possibly different days) to occur exactly once?
PS: I'm using a 12-hour clock.
Here's is an incorrect partial solution as an example: First, draw a 3. 3 o'clock. Then draw a 5. 3+5 = 8 o'clock. Then draw a 6. 8 + 6 = 2 o'clock. (Remember, the hours go to 12, then start over again at 1.) Then draw a 7. 2 + 7 = 9 o'clock. Then draw an 11. 9+11 = 8 o'clock. But you already took a pill at 8 o'clock, so this solution is wrong.
Clue: If there is a solution, the first card must be a 12. This is easy to see, since if any other card was the 12 (queen), then we would take our pill the next time at the same time, with PM instead of AM or vice versa. And so, the last time we took our pill (just before we drew the 12) would occur twice -- forbidden.
Good luck! I have not solved this myself. But...
Someone on a math/puzzles newsgroup calculated via computer (I hope they used a computer -- otherwise, I'm scared) that there are 3856 solutions to my puzzle. That seems like a lot. But there are 39916800 possible permutations of the 12 cards, where it is given that the first card is a 12 (queen). (So, we are really asking about the number of permutations of the 11 remaining cards, since the queen must be first.) That means that 1 in about every 10350 permutations -- permutations with it given that 12 is the first card -- is a solution. That makes it highly unlikely you will find a solution by chance.
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If that wasn't mentally-straining enough, here's the poem I wrote today! (Arrgg, you Leroy!)
Into Perpendicular Decisiveness
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These games are simplified
into inaccurate ambiguity.
They are scribbled oddly
into grids of intellect,
Into perpendicular decisiveness
certain of its lines,
But not of its intersections.
Oh, these gambits
Are simply guessed
but not obvious, despite
The very multiplicity of
possibilities drawn.
For, columns become rows;
yet horizontality is
Never diagonal within
this paper page. Yes,
Obscured are our minds from
our minds. Such
Is desired. However, it is problematic.
Such squares
Become x's, and only once
do they do so twice.
Oh, the games are
irrational and random.
They complicatedly are considered
by geniuses and
Are proposed as weirdnesses
uttered then alternated.
Ah, alternately we
win and lose this
One play upon a
notebook's scrawl. And
We are knowing regarding
the counterintuition
That is but countered.
For, we play just
Rectangular games
turned orthogonally. We
Are players of such
juxtaposition positioned
With subdivided directions
Each circumvented
by their erroneousness.
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PS: I plan to put up a new poll in a few days.
Leroy
Friday, December 10, 2010
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