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Post by servant on Oct 5, 2009 16:51:04 GMT -5
Because I love riddles. Please try not to google. The moment someone arrives at the correct answer, I'll post a new riddle! This is just for fun. WARNING: Riddles may contain maths.RIDDLE #1: The Three SiblingsMeet Anna, Lebrant and Ronald. The three came from the same father and mother, and since their youth have been quite competitive of one-another. One day, Lebrant called together his two siblings and said, "I will become this country's next president!" Of course, the two siblings claimed they could become the president in his stead. The three agreed to a competition. Ultimately, each of the three became a candidate for president. In fact, they became the only candidates for president through some sheer luck. Now, knowing that there are 300,000 vote-eligible inhabitants of the country ( including the three siblings), and assuming each of the three to cast their vote for themselves, what is the minimum amount of votes one of the siblings needs to be sure of victory?Please provide a reasoning with your answer! Just an answer will not be considered the proper answer.
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Post by LEO LION on Oct 5, 2009 17:09:04 GMT -5
My reasoning would be 1 vote each, since they're all running only against each other and there's 300,000 possible votes (the three voting for themselves) and if none of the 299,997 vote. I'm probably wrong, but that's what my brain comes up with.
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Post by servant on Oct 5, 2009 17:21:32 GMT -5
I guess I should have clarified.
All eligible voters will vote. Their vote will always be either on Anna, Lebrant or Ronald. There is no such thing as abstaining to vote, and no votes are cast on multiple candidates or no candidates at all. In total, 300,000 votes will be cast, after which the one with the highest amount of votes will become president.[/color]
Now that that is clarified, please refer to the question in the first post. :-)
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Post by prythian on Oct 5, 2009 17:31:59 GMT -5
That's easy, then.
150000, counting their own vote.
Say all the remaining voters outside the siblings vote for only one of the other siblings? The remaining sibling will still cast their vote for themselves, leaving the other sibling with 149999. Thus, 150000.
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Post by servant on Oct 5, 2009 17:51:48 GMT -5
That is, indeed, the correct answer! I see that one was too simple for you. Let's see how you deal with this: RIDDLE #2: Trap's High HatExcuse the art. But still, this is an upside down tophat, as you may astutely observe. I've gotten myself nine slips of paper. Each of these slips is marked with a number, then folded. The numbers are 1, 2, 3, 4, 5, 6, 7, 8, 9. I grab a scoring board. You know, the 00-00 boards. Then, I decide to play a game. I will, at random, draw one slip from the hat, then add it to the first number in the sequence. I then put it aside, and grab a new slip from the hat. I add that slip's depicted number to the second number in the sequence. Then, I return both slips to the hat, shake it up good, and then draw two more slips for the last remaining numbers. So if I drew 2, 4, *shake it up* 4, 2, I'd put the scoreboard at 24-42. I then declare this a score. Here's the question. What is the total amount of different score combinations possible, adhering to the rules established in the paragraph above?
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Post by Rhea on Oct 5, 2009 19:48:01 GMT -5
This is all from Pre-Algebra so don’t shoot me if I’m way off.
For the first slip you can have 1, 2, 3, 4, 5, 6, 7, 8 or 9. That’s 9 possibilities. Since you’ve already removed one, for the second slot you can have 8 possible numbers. After returning those two slips, for the third slot you’ll have 9 possibilities again, and for the last slot you’ll have 8. To figure out the number of possible combinations all you have to do is multiply (or write it out, but that takes FOREVER). So it’s 9*8*9*8 to get 5184 possible combinations.
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Post by servant on Oct 6, 2009 15:49:26 GMT -5
Very good... Let's see how you deal with this. RIDDLE #3: Welcome to #LiarsYou are chatting with five people. You know that of these people, there are three who always tell lies, and two who always speak truth. You may ask one of them a question of great importance, so it's vital you ask the one who speaks truth. As they are great sports, each of the five gives you a hint about one of the others. Ultimately, you get these five hints: Chatter #1> "What Chatter #4 says is the truth." Chatter #2> "If anything, #1 is the one to trust." Chatter #3> "I guarantee you that Chatter #5 is telling the truth." Chatter #4> "You can't believe in Chatter #3." Chatter #5> "Chatter #2 is telling lies. Totally." These hints alone should allow you to discern who speaks the truth and who lies. So, who would you ask your question of great importance? How did you decide?
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Post by Rhea on Oct 8, 2009 16:32:02 GMT -5
To figure it out I just wrote out what the truth would be if certain people were lying or telling the truth. I found one that worked out:
#1 lies. #2 lies. #3 tells the truth. #4 lies. #5 tells the truth.
If #1 is lying then a true statement would be that #4 lies. According to my answers, that is correct. If #2 is lying then a true statement would be that #1 lies. I’ve already said that was the case. If #3 is telling the truth, then #5 is also telling the truth. That’s correct as well. Is #4 is lying, then the truth is that #3 is telling the truth, which he is. And finally, if #5 is true then #2 is lying. This is also correct.
Answer: I could ask ask chatter #3 or #5 and expect a truthful response. Preferably #3 because too many "totally"'s and my ears would bleed.
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Post by Akito on Oct 9, 2009 19:46:06 GMT -5
^ Same answer she had. 3 and 5 are the truth tellers.
Simply, you initially assume either 1 told the truth or 1 told the lie and logically follow the sequence and determine the one that has the right proportions. In this case, was if 1 told a lie.
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Post by servant on Oct 9, 2009 21:56:20 GMT -5
You're utterly right. >_>; I'll post a new one once I've finished drawing the arts for it.
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Post by Akito on Oct 10, 2009 1:43:41 GMT -5
Ho, let me give one. It can be either -really- easy or -really- hard. It actually had mathematicians and various other people several months to get, or so the internet says.
"How many of each animal did Moses take on his ark?"
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Post by Heroic Bilby on Oct 10, 2009 1:52:36 GMT -5
The Ark of the Covenant? I'd have to go look it up, I don't remember any animal decorations on it. I only remember(I think) that there were 2 angels on the top. Unless they count?
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Post by Dante on Oct 10, 2009 2:09:47 GMT -5
None. Zero, zip, zilch. Moses didn't take the animals on the ark, Noah did.
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Post by lysander on Oct 10, 2009 4:05:25 GMT -5
Oyo! I want to steal some of Trap's limelight too! >.> Well, I kinda just want to say a riddle, if that's okay? Sorry Trap! ^_^;
This is probably googled pretty easily, but it's a fun one: The man who made it doesn't want it; the man who wants it doesn't need it; the man who needs it doesn't know it. What is it?
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Post by servant on Oct 10, 2009 9:13:20 GMT -5
Offhand I'd guess a coffin. Think I've read that riddle before. That all aside... I'd have drawn a real person, but stickfigure will have to do because everyone be intruding on mah territory! RIDDLE #4: Pro golf is proMeet Patrick. Patrick is a pro golfer. He has the uncanny ability to putt a ball exactly 3, 5, 7 or 11 meters. Sadly he lacks the ability to putt them any other distance, but all is forgiven in the face of such imperfections. Above is a picture of Patrick at the world championships golf. He is 20m from the last goal, and knows he'll have to putt the ball in. Knowing that any meter in excess will roll that amount past the goal, what is the smallest number of putts Patrick can use to win?(So if there's 4 meters left and he putts 5, he is one meter in excess)
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