Ok, I'm just making this thread because I'm bored... And because I want to confuse the hell out of people 
There are not many rules...
No cheating.
No helping others.
No using the web to find an answer
No help from others
More might be added soon.
Only one guess per week.
I'm not going to tell you the answer, even if you ask.
The prize is 50 Frih$ - Donations are accepted!
Week ones challenge:
If all X's are Y's, all Y's are Z's and all Z's are A's, which of these statements are not true.
1. All X's are Y's
2. All Y's are Z's
3. All X's are A's
4. All A's are Y's
Week twos challenge:
http://www.frihost.com/forums/vt-62337.html#526887
Week threes challenge:
http://www.frihost.com/forums/vt-62337.html#533107
Last edited by Hogwarts on Tue Jan 16, 2007 3:16 pm; edited 6 times in total And the prize for the first week goes to.... hunnyhiteshseth
Congratulations, hunnyhiteshseth 50 frih$ will be awarded immediately.
The correct answer was 4, while most people thought this was 3
But, that is not true.
| Quote: |
| If all X's are Y's, all Y's are Z's and all Z's are A's, which of these statements are not true. |
If you look at that, all X's must by Y's.
After a bit of thinking, you'll realize that not all A's are X's.
Just because all X's are A's, does not mean that all A's are X's.
For example, look at squares and quadrilaterals. All squares are quadrilaterals, while all quadrilaterals are not squares. You see? Cool contest you got going here, so when is the next challenge?
| Star Wars Fanatic wrote: |
| Cool contest you got going here, so when is the next challenge? |
Read the title 
(Hogwarts' weekly logic challenge) (It's weekly because I need to get enough frih$ for the prize)
I hope for this to be running all 2007 I'll never be smart enough to take part in this but I have sent you 100FrihS to help toward prizes - hope it helps.
I love it when people use their time and effort (as well as Frih$) to contribute to the forum.
Okay, Ladies, Gentlemen and Its.
For round two, you do not have to PM me the answer.
Question 2:
A recent study involving broken bones while skiing showed that every broken bone somebody had while skiing happend the last time a person went down the slope.
Should a rule be implemented to stop people going down the slope the last time?
Would this work? Why?
Last edited by Hogwarts on Mon Jan 08, 2007 12:36 pm; edited 1 time in total
| Munkey_boy's answer wrote: |
| you cant make a rule to forbid them going down the slope for the last time because you dont know when the last time will be untill they break a bone and cant go down it anymore. |
I wrote that on MSN to hogwarts but he said I had to submit it on the forums.Even if you stop people going down the slope the last time, then the one before the last time would be the last time they went skiing. So, they would broke bones again. So, effectively, you have to go on stopping them skiing the last time till they stop skiing at all. If they stop skiing then there is no chance of broken bones.
Um... a bit confused?
If you are, let me explain my (stupid) answer -
John goes skiing 8 times, he always breaks his bones the 8th time, the last time. So, he stops going the last time, i.e., the 8th time. Then, the last time would be the 7th time. But he has the tendency to break bones then, as it is the last time now. So, the rule eliminates all of John's "last times" - 7th, then, 6th, 5th, 4th, 3rd, 2nd, 1st and lastly, none at all. So, there is no chance of John breaking bones, at the cost of giving up skiing.
So, this rule is not effective to the point that you can't stop a person skiing the last time.
| Hogwarts wrote: |
| Would this work? Why? |
No, because the rule could only ever be: a) violated; or b) result in an infinite life of skiing.
Here's my convoluted explanation. 
Last time can mean previous time but it can also mean final time . Since we can't travel back in time this question must be refering to the final time a person goes down the hill.
You can't have a last time (final time) without having a first time. So everyone has to go down the hill at least once for this rule to even be possible. So far, no problem.
But once you start skiing, the minute you stop (for any reason, let alone broken bones) you've violated the rule by making that time your last time.
The only way to be a contientous, rule-abiding skier is to never, ever stop skiing.
Or something like that... I skipped the last contest which could have been my last one. As it turns out this is the first one I've done and may not be the last. Can't wait until the next.
Take care,
Eyvind | Eyvind wrote: |
| Hogwarts wrote: | | Would this work? Why? |
Or something like that... I skipped the last contest which could have been my last one. As it turns out this is the first one I've done and may not be the last. Can't wait until the next.
Take care,
Eyvind |
ARGH! MY head just exploded. 
| munkey_boy wrote: |
| you cant make a rule to forbid them going down the slope for the last time because you dont know when the last time will be untill they break a bone and cant go down it anymore. |
Correct.
Not the most detalied answer, but it has the bare essentials.
| Eyvind wrote: |
No, because the rule could only ever be: a) violated; or b) result in an infinite life of skiing.
Here's my convoluted explanation.
Last time can mean previous time but it can also mean final time . Since we can't travel back in time this question must be refering to the final time a person goes down the hill.
You can't have a last time (final time) without having a first time. So everyone has to go down the hill at least once for this rule to even be possible. So far, no problem.
But once you start skiing, the minute you stop (for any reason, let alone broken bones) you've violated the rule by making that time your last time.
The only way to be a contientous, rule-abiding skier is to never, ever stop skiing.
Or something like that... I skipped the last contest which could have been my last one. As it turns out this is the first one I've done and may not be the last. Can't wait until the next.
Take care,
Eyvind |
Correct, but you didn't get it first. Still, you will be awarded 51 frih$ for putting the most effort in, and also for confusing Captain Fertile
| saratdear wrote: |
Even if you stop people going down the slope the last time, then the one before the last time would be the last time they went skiing. So, they would broke bones again. So, effectively, you have to go on stopping them skiing the last time till they stop skiing at all. If they stop skiing then there is no chance of broken bones.
Um... a bit confused?
If you are, let me explain my (stupid) answer -
John goes skiing 8 times, he always breaks his bones the 8th time, the last time. So, he stops going the last time, i.e., the 8th time. Then, the last time would be the 7th time. But he has the tendency to break bones then, as it is the last time now. So, the rule eliminates all of John's "last times" - 7th, then, 6th, 5th, 4th, 3rd, 2nd, 1st and lastly, none at all. So, there is no chance of John breaking bones, at the cost of giving up skiing.
So, this rule is not effective to the point that you can't stop a person skiing the last time. |
Nearly correct. Although, it does say that the rule does not work, and why it does not work, so you'll be getting frih$ anyway
My answer?
No, it's not correct. The rule wouldn't work. Why? Because people who do break their bones break them the last time they go down the slope. The reason it's the last time is that after a person breaks their bone, they stop skiing.
Stopping them going down the "last time" would therefore not work, as it is unknown as to when somebody is going to break their bone, and stopping them going down the "last time", as previously mentioned, would stop them skiing alltogether.Thanks for the frih$
I think I am a little late in answering the 2nd question, but I will try next week.
Thanks, Hogwarts. Much appreciated.
Captain Fertile, sorry about burstin' yer noodle there mate!
But you know, after I posted my answer I checked the dictionary and found that one of the more rare meanings of "last" is actually
| Quote: |
| a wooden or metal form in the shape of the human foot on which boots or shoes are shaped or repaired. |
(check it, it's true!)
And this got me thinking that "last time" could have actually meant "the time when ski boots are repaired". Anyone who is skiing when their boots are due to be mended would certainly be risking broken bones! And if this were the case, forbidding people from skiing on their "last time" would actually be a rather effective rule. So the answer could, actually, be "Yes"!
Take care,
Eyvind Question 3:
Sorry I'm so late.... Forgot 
Post in this topic this week.
A person has 6 leather and 6 jogging shoes.
How many shoes would the person have to get out of his room in the dark to make sure he has at least one pair (i.e. set, left and right. Not a pair as in 2 of the same thing)?
I won't give away as much this week ... unless you "bust Captain Fertiles noodle... heh." Mmmm, let me see, I can't quite get it. I have a doubt. Hope this wouldn't be too stupid...
| Hogwarts wrote: |
How many shoes would the person have to get out of his room in the dark to make sure he has at least one pair (i.e. set, left and right. Not a pair as in 2 of the same thing)?
|
So, the man needs a pair of either jogging shoes or leather shoes, or can he have, for instance, a left pair of leather and a right pair of jogging shoes?
Hoping it doesn't seem too stupid. Assuming "6 shoes" here means "3 pair" (ie. 6 pairs of shoes in total with half of them being leather and half of them being jogging), and assuming he can't identify the shape or material in the dark by feel, then:
He would have to get 7. Anything less than 7 and it is possible he could pick up all the shoes for one side (all left shoes for example). He would need one more to get a right. This seventh one would necessarily match one of the leather shoes or one of the jogging shoes and thus would also guarantee him a matched set.
Take care,
Eyvind
He could possibly get it with 2-6, but to be sure he needs 7.
*sigh*
Ok guys, none of you have got it right.
But to speed things up, I'm going to say this
Don't make assumptions He will need to take out just 4 shoes to get a pair.
I solved it by using Pigeon Hole Principle.
| hunnyhiteshseth wrote: |
He will need to take out just 4 shoes to get a pair.
I solved it by using Pigeon Hole Principle. |
No, that's not the answer. Sorry 
I'm going to make this even easier and say this.
| Quote: |
| Assuming "6 shoes" here means "3 pair" (ie. 6 pairs of shoes in total with half of them being leather and half of them being jogging), and assuming he can't identify the shape or material in the dark by feel, then: |
Plus this..
| Me wrote: |
*sigh*
Ok guys, none of you have got it right.
But to speed things up, I'm going to say this
Don't make assumptions |
= A large clue.Only one guess per week!? Aw, man.... I wanna guess again He would have to take all of them to make sure he has at least one right, and one left in there someplace. For all we know, he could have 11 shoes that are right, and only one that is left.
| Hogwarts wrote: |
How many shoes would the person have to get out of his room in the dark to make sure he has at least one pair (i.e. set, left and right. Not a pair as in 2 of the same thing)?[/size]
I won't give away as much this week ... unless you "bust Captain Fertiles noodle... heh." |
hmm, well given what you've said none of them have to match or make a "pair", so given only the information you've allowed us no matter how many he removes, we can't say that he has at least one pair. Even if he does remove all of the shoes he may not have a pair. Perhaps all of the shoes are right shoes, or all of them are left shoes... we don't know
As a side note, none of the shoes are guaranteed to actually be in the room...
By the way, it would be "Fertile's noodle", not "Fertiles noodle" | Montressor wrote: |
| Hogwarts wrote: | How many shoes would the person have to get out of his room in the dark to make sure he has at least one pair (i.e. set, left and right. Not a pair as in 2 of the same thing)?[/size]
I won't give away as much this week ... unless you "bust Captain Fertiles noodle... heh." |
hmm, well given what you've said none of them have to match or make a "pair", so given only the information you've allowed us no matter how many he removes, we can't say that he has at least one pair. Even if he does remove all of the shoes he may not have a pair. Perhaps all of the shoes are right shoes, or all of them are left shoes... we don't know
As a side note, none of the shoes are guaranteed to actually be in the room...
By the way, it would be "Fertile's noodle", not "Fertiles noodle" |
Out of all the guesses, this one is the only correct one
Congratulations, you will be awarded 50 frih$. Remember to come back.. Today! When I think of another question Good work with the competiton. I'll give you 20 frih$ to help you continue to fund it.
| Hogwarts wrote: |
| Montressor wrote: | | Hogwarts wrote: | How many shoes would the person have to get out of his room in the dark to make sure he has at least one pair (i.e. set, left and right. Not a pair as in 2 of the same thing)?[/size]
I won't give away as much this week ... unless you "bust Captain Fertiles noodle... heh." |
hmm, well given what you've said none of them have to match or make a "pair", so given only the information you've allowed us no matter how many he removes, we can't say that he has at least one pair. Even if he does remove all of the shoes he may not have a pair. Perhaps all of the shoes are right shoes, or all of them are left shoes... we don't know
As a side note, none of the shoes are guaranteed to actually be in the room...
By the way, it would be "Fertile's noodle", not "Fertiles noodle" |
Out of all the guesses, this one is the only correct one
Congratulations, you will be awarded 50 frih$. Remember to come back.. Today! When I think of another question |
Honestly speaking, even I had thought of tthat in first place but then I thought it was obvious that shoes were in pairs because otherwise there was no problem.
But I think there I committed mistake, I should not have taken obvious as obvious.ohh Hogwarts, from now on, please keep exact answers to all questions that u ask. It seems that we rack our brains on all such questions, and somone comes up with a usual, childish answer to the question.
Anyways, it is a nice challenge.
Warm Regards,
Abdul Basit
| toughtrio wrote: |
| ohh Hogwarts, from now on, please keep exact answers to all questions that u ask. It seems that we rack our brains on all such questions, and somone comes up with a usual, childish answer to the question. |
Hmm, perhaps the challenge that Hogwarts is choosing to present is the ability to think more simply, anyone can use math to (based on assumptions) find the minimum number of shoes needed to be removed from the room, but it takes logic (and a large quantity of hints) to ignore those assumptions and see that the conventional answer (while presumably mathematically correct, and if we work off of the assumptions, very logical and orderly), is not always bound to be correct.
On a side note, given that it is Hogwart's Weekly Logic Challenge, Hogwarts has the right to run it any way he/she desires. Hogwarts isn't limited to obeying arbitrarily set rules (based on assumptions), but is only restricted by the rules of the forums.
