Home » General Chat » Circular File » Puzzle Thread Dec. 15
 Puzzle Thread Dec. 15 |
Mon, 15 December 2003 23:15  |
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 Ron |
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Here are this week's puzzles. The usual 24hour timelimit applies. Send me your completed answers via Private Message.
There are six total puzzles this week, one is a bonus puzzle.
If you get ANY five of the below puzzles correct, you have one chance to become the Weekly Puzzle Master. If you get all six correct, then you get two chances.
1. A friend of yours announces that he has a number of coins (American) in his pocket that add up to precisely one dollar. When he tells you how many coins he has, you ask if any of them is a half-dollar, and he answers no. You quickly realize that you can't tell for sure what coins he has, because with the number of coins he says he has, there are six different combinations that produce precisely one dollar.
How many coins does your friend have in his pocket?
(100 cents = 1 dollar. American coins are the following (excluding the half-dollar): P=1 cent, N=5 cents, D=10 cents, Q=25 cents)
2. Note first that the number 150 is expressible as the sum of distinct squares, as shown below.
150 = 100 + 49 +1 = 10^2 + 7^2 + 1^2
We'll spare you the trouble, but you can take our word for it that every number greater than 150 is also expressible as the sum of distinct squares. But there are 37 numbers lower than 150 that cannot be expressed in this fashion. What is the largest one?
3. Melchior, Gaspar, Balthazar, and Roy, seated in that order, were playing a game of dominoes.
The game has 28 dominoes, one for each combination of the numbers 0 to 6. In the traditional game, players must place matching combinations next to each other.
Melchior began with domino 6:4 (6 pips on left side, 4 pips on the right side)
After placing the tenth piece on the table, the 3:6 next to the 0:3, Gaspar ended the game. No one could make another move.
The total number of pips of all of Melchior's dominoes (that he played) was 29; Gaspar's (that he played) were 23; Balthazar's (that he played) were 20; and Roy's (that he played) were 6.
Determine which
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[Updated on: Tue, 16 December 2003 20:48]
Ron Miller
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| Re: Puzzle Thread Dec. 15 |
Tue, 16 December 2003 23:33  |
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| Ron |
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Only HiltonL and LEit turned in at least 5 correct answers, and LEit has been chosen as this week's Puzzle Master.
Here are the answers:
1. Your friend must have 15 coins in his pocket. Here the six ways that you can create a dollar from 15 coins (no half-dollars allowed):
Q D N P
1) 3 1 1 10
2) 2 1 7 5
3) 1 1 13 0
4) 1 5 4 5
5) 0 9 1 5
6) 0 5 10 0
Believe it or not, 15 is the only number such that there are precisely six ways of creating a dollar from that number of coins (no half-dollars).
2. 128 is the largest integer that cannot be expressed as the sum of distinct squares. Note that 128-100=28, which is not expressible, and similarly 128-81=47 which is not expressible, etc.
3. For the game to be over, seven dominoes on which the same number appears had to have been played. One of them will be a double, and the other six will have six different numbers. Therefore, in order to be playable, the other three dominoes (we know 10 in all were played) had to have six different numbers.
Since the 6 was played twice (in the first and last moves) and we know the scores of each of the players, we can deduce that the number repeated was 6. Therefore, the seven dominoes from 0:6 to 6:6 had to be in the game.
The six numbers remaining range from 0 to 5, and one of the dominoes, as we are told, was 0:3.
The total of Roy's dominoes was six points, and this can only be achieved with 0:3 and 1:2.
Therefore, the third dominoe that has no 6 among its numbers is 4:5.
Gaspar had 3:6, and we were told his score was 23. We only need to account for the 14 remaining points. We can only arrive at this number with 6:0 and 2:6.
Knowing this, Balthazar had only one way of achieving 20 points: with 4:5 and 6:5.
Finally, we know that Melchior had 6:4 and his total score was 29 points. Therefore, his other two dominoes were 6:1 and 6:6.
With this, we know which dominoes were in play, and who had each one.
M plays 4-6
G plays 6-2 on the 4-6
B plays 4-5
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Ron Miller
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| Re: Puzzle Thread Dec. 15 |
Wed, 17 December 2003 00:59 |
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| BlueTurbit |
 | Lt. Commander
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The answer to the last puzzle is wrong.
This is not what happened. What really happened was...
The king put out this stupid decree about the blue eyes and after he was done the islanders all stood around confounded and looked at each other and said: "This is stupid. We can't figure this crap out."
"Were not rocket scientists. We're friggin' islanders who climb coconut trees and make canoes to pick up the tourists with."
"If we were that smart we would have a high paid job in the mainland, not be running our buns off on this big friggin' island trying to see everybody at least once everyday."
"I bet even intermediate Stars! players couldn't figure this crap out. No way Jose."
"And if the king wouldn't have hid all the water we could see our eyes in the reflection"
"Golly this friggin' island is fifteen miles from corner to corner and that's a lotta damn runnin' without water."
"To hell with this. Lets get drunk"
And they all got together and cracked some coconuts and popped open a few bottles of rum and got plastered at a pina colada bash singing songs like:
"You're so good to be true, I can't keep my eyes off of you...."
until they were barely able to crawl into to their huts and crash at about 4 AM in the morning.
In the morning after sunrise nobody showed up for their eye check, so the king asked his advisor what he should do. His advisor told him he should go from hut to hut to check their eyes.
So the king did just that, and all he found was that everyone had red eyes and none of them were in good enough shape to be leaving the island anytime soon.
After he checked the last one, he thought: "Oh bat poop, I really messed up this time. I must be the one with blue eyes!"
So immediately he went back to his advisor and told him off and blamed him for everything gone wrong.
He called all the islanders together and publicly chastized his advisor.
"You gave bad advice to your king." he said.
"A
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| Re: Puzzle Thread Dec. 15 |
Wed, 17 December 2003 06:18 |
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Well no...assuming they all think they way Ron presented the solution then yes they would all leave on the 10th day.
Myself I'd look at another blue eyed person and wonder when they'll realise they have blue eyes.
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| Re: Puzzle Thread Dec. 15 |
Wed, 17 December 2003 11:16 |
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| HiltonL |
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| mazda wrote on Wed, 17 December 2003 12:46 |
Well done LEit !
I also was bamboozled by number 5. 
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I certainly prefer a puzzle where I can make a plan of attack. I stared at number 5 and had no method to break it down or anything - very frustrating! To other people that's the kind of puzzle they love.
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Plus I answered 16 for number 1.
I have now found a 7th way of using 16 coins to give a dollar.
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I'm not sure that I actually want to admit this, but I sat down and worked out the number of combinations from 5 coins up to 14 by hand. When I realized I wasn't getting anywhere, I wrote up an excel spreadsheet to show me the number of combinations. When I finally realized I had fallen 1 short on my by-hand method... 
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Don't you need to show the case k=2 for the induction, as k=1 is strictly different ?
(i.e. for k=1 you know there are 0 or 1 and you know there are 1 becasue the king says so, but for k=2 you know there are 1 or 2 and it can't be 1 becasue that person didn't leave on day 1).
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No, I don't think so.
Applying the inductive step for k=2 translates into English like this:
You see 1 other person with blue eyes. You know, because P(1) is true, that if he is the only person with blue eyes he will leave immediately in the morning of the first day. When you meet him in the morning and he HASN'T left, you realize that there is not only 1 blue-eyed person on the island - there must be 2. And the second must be (shock and horror) YOU! You promptly leave on the second day.
So because P(1) is true, and given the reasoning above, which is valid for ANY k, not just 1 and 2, you and your blue eyed friend leave on the second day. P(1) is true, P(2) is true. And so on, that is the way of induction  . P(1) is a special case, but then everything follow FROM that special case.
Cheers
Hilton
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| Re: Puzzle Thread Dec. 15 |
Wed, 17 December 2003 14:44 |
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| Ron |
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| LEit wrote on Wed, 17 December 2003 12:18 |
Oh, and Ron, what's the secret to getting columns to line up on these forums?
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Use CODE tags and COURIER font. The CODE tag prevents wrapping and other strange things, while the COURIER font is a non-proportional font, so each character is the same horizontal amount of space.
Ron Miller
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Finally, dude!!!
