Marvin (for the solution): [spoiler] You have one part wrong, which affects the solution. The answer is not Friday.[/spoiler]
Marvin (for the question): [spoiler]I tried some calculations with rounding down and everything, but then it hit me. The answer must be 169.[/spoiler]
about Puce’s riddle
I even bolded it that the answer was Wednesday, not Friday xD I hope that is correct?
About Puce’s solution to Marvin’s riddle
Yes, took Magnus only a few seconds to figure out. 169 is correct. But I’m not sure if he knew it before or not^^
On Puce’s riddle:[spoiler]Working from the other way: If today is friday, then tomorrow is saturday. Yesterday = ‘tomorrow’ so yesterday is saturday… Today must be Sunday[/spoiler]Am I right?
Marvin: [spoiler] Stupid me. But no, it isn’t Wedesday either. [/spoiler]
Blenderman: [spoiler] Yes, you are correct.[/spoiler]
Still on Puce’s riddle
[spoiler]I disagree. The way I read the riddle, it cannot be Sunday. Let’s assume today is Sunday. Then “tomorrow” would be Monday. And if yesterday := Monday then today would be Tuesday. Not Friday. Or where is my error?
The way Puce and Blenderman intepreted it, “tomorrow” is viewed in the hypothetical time that today would be Friday. For me, “tomorrow” is viewed as if it was in the current real timeline.
Now if today is really Wednesday, then “tomorrow” would be Thursday. If we assign yesterday Thursday, then “today would be Friday”, exactly as it is in the riddle…[/spoiler]
[spoiler]Well, the way I translated it, I tried to mean “tomorrow” in the hypothetical sense while “yesterday” exists as normal. To the extend of my knowledge, the if-clause conveys that meaning.
Now the way I understand your argument, you solve it this way: Today is Friday. Yesterday is then Thursday. Since we said “yesterday was tomorrow”, then the day we’re looking for is yesterday for Thursday, which is Wednesday.
Looking at it the other way around, I see this: The man claims that if yesterday was tomorrow, then today would be Friday. So let’s assume that it in fact is Wednesday. Then yesterday (which is Tuesday) should be tomorrow. Then that day would be Monday. Which is not Friday.
The solution I and Blenderman have goes like this: Today is Friday. We had made yesterday tomorrow, and tomorrow is then Saturday. Since it is also yesterday from the actual day, today is Sunday.
To verify this: Let’s assume today is Sunday. We had said “if yesterday was tomorrow”, so it means “if Saturday was tomorrow”, which would point today as Friday.
I think the part we disagree on is this: Yesterday is not always tomorrow and tomorrow is not always yesterday.[/spoiler]
[spoiler]I think the main problem is that we both thought that the solution would be unique. But due to the ambiguity of the riddle itself: Is tomorrow or yesterday in the hypothetical time (and the other one in the current time). Because as you pointed it out, depending on HOW you interpret the actual riddle the solution would either be Sunday or Wednesday. I wouldn’t say either is wrong.
And speculating which one (between yesterday and tomorrow) was meant in a hypothetical way and which in the real one, I guess this is more a matter of taste and personal interpretation that based on a linguistic rule. At least as far as my English goes…[/spoiler]
[spoiler]Well, I’m sorry to say Marvin, I really do think you are wrong. The key word being ‘if’. It’s not yesterday = tomorrow and therefore you can turn it around, even though I did (quite wrongly) use the = sign in my explanation.
The riddle says: ‘If yesterday was tomorrow’. That isn’t the same as ‘If tomorrow was yesterday’. There is truth and there is imagining. The true values are ‘today’ and therefore ‘yesterday’, the other words are made up… that’s why the word ‘if’ is used. If you work from the other side, from the ‘if’ side… so taking ‘friday’ and than trying to get the answer like I did, the statement would be ‘If tomorrow was yesterday’.
Let’s reason backwards from our answers. It is sunday. If yesterday (saturday) was tomorrow, then tomorrow would be saturday. Then it would be friday. That is correct.
It’s wednesday. If yesterday (tuesday) was tomorrow, then tomorrow would be tuesday. Then it would be monday.
You simply have to keep track of what is before the ‘if’ and what comes after it.[/spoiler]
^ Blenderman put what I actually tried to mean in better words. 
Paul is 20 years old in 1980 but only 15 years old in 1985. How is this possible?
and i have one more 
he who makes it, sells it
he who buys it, doesnt use it,
he who uses it, doesnt know it
define “it”
[spoiler]For your first, the only solution I can come up with is that Paul is a character in a movie. In 1980, the character is 20, and in 1985 they make a prequel where he is 15.
Your second, however, is a coffin. He who makes the coffin, sells it. He who buys it is purchasing it for a recently passed loved one. And if you are using it… well, you are dead, so you couldn’t possibly know thet you are.[/spoiler]
Uh, this is the only riddle I can think of:
What walks on four legs in the morning, walks on two legs tn the afternoon, and three legs in the evening?
I think Paul might be living BC… so as Paul get’s older the years get smaller… towards 0
It took me a few days to get it, and an hour or so to write it out, but here’s my solution:
1st WEIGH: Separate the balls into 3 groups with 4 balls each. Weigh two of the groups, group A against group B. Group C is not weighed.
Case 1: They weigh the same. In this case, you now know the eight balls from groups A and B are normal, and the odd ball is one of the 4 balls in group C.
2nd WEIGH: Separate group C into 3 sub-groups: 2-1-1. Weigh the group of two balls (group a) against a third ball and one of the normal balls (group b). Make sure you remember which is the normal ball. The remaining ball from the group C would not be weighed.
Case 1.1: They weigh the same. All balls on the scale are normal. [Color=449721]The remaining ball from group C is the odd ball.[/color] No need to use the scale again, unless you want to find out if it’s heavier or lighter than the normal balls.
Case 1.2: They weigh differently. In this case, note which group was heavier. [Color=449721]The odd ball is either one of the three balls.[/color]
3rd WEIGH: Weigh the two balls from group a against each other.
Case 1.2.1: They weigh the same. [Color=449721]The odd ball is the one from group b.[/color]
Case 1.2.2: They weigh differently. You now know the odd ball was from group a. [Color=449721]If group a was heavier, then the odd ball is the heavier one. Otherwise, the odd ball is the lighter one.[/color]
Case 2: They weigh differently. You now know that the 4 balls in group C are all normal. The odd ball is either in group A or B. You also know which group is heavier.
2nd WEIGH: Split group A into two groups of two. Add a normal ball to the first group, and a ball from group B to the second group; thus weigh two balls from group A + one normal ball (group a) against the remaining two balls from group A + one ball from group B (group b). Three balls from group B will be left over.
Case 2.1: They weigh the same. The odd ball is one of the three remaining balls from group B. Since you know whether group B was heavier or lighter, you now know if the odd ball is heavier or lighter.
3rd WEIGH: Weigh two of the three balls in the leftover group.
Case 2.1.1: They weigh the same. [Color=449721]The odd ball is the third ball.[/color]
Case 2.1.2: They weigh differently. [Color=449721]If group B was heavier, then the odd ball is the heavier one. Otherwise the odd ball is the lighter one.[/color]
Case 2.2: They weigh differently.
Case 2.2.1: Group A was heavier and group a is now lighter (or vice-versa). You now know that the odd ball was one of the two balls from group A which were in group b, and that the odd ball is heavier (or lighter).
3rd WEIGH: Weigh those two balls against each other. [Color=449721]The heavier (or lighter) ball is the odd ball.[/color]
Case 2.2.2: Group A was heavier and group a is still heavier (or lighter). There are two possibilities, either one of the two balls in group A that were also in group a is the odd ball and is heavier than normal, or the ball from group B which was added to group b is the odd ball and it is lighter.
3rd WEIGH: Weigh the two balls from group A that were also in group a against each other.
Case 2.2.2.1: They weigh the same. [Color=449721]The odd ball is the ball from group B which was added to group b.[/color]
Case 2.2.2.2: They weigh differently. [Color=449721]The heavier (or lighter) ball is the odd ball.[/color]
Haha. I had to search my own posts with ‘balls’ as a keyword to find my old post again. Wow. That was from five years ago. Time sure does fly. I did a quick check in that part to see if my riddle was solved. That doesn’t seem to be the case.
I had forgotten the solution to the riddle myself so I decided to work it out again. Happy to say I managed to solve it so I can compare my solution with yours. Reading up on it, all I can say is… Your solution is complicated :rolleyes: At least, it is definitely different from mine. I’ll post my solution later and you can check that too.
After spending a longer than anticipated time understanding your solution with a pen and paper, I can confidently say that you are correct Thanks for solving a 5 year old post.
Here’s my solution to my riddle.
[spoiler][code]Group the balls into three groups of four. We’ll call these groups A, B, C.
Additonal notes: A[1] would stand for the first ball in group A. A[1,2] would stand for balls 1 and 2 of group A. ‘=’ would stand for an equal reading while ‘!=’ stands for not equal.
1st Weigh: A and B
IF =
We know that the odd ball is in C
2nd Weigh: C[1] and C[2]
IF =
We know that the odd ball is in C[3,4]
3rd Weigh: C[1] and C[3]
IF =
We know that C[4] is the odd ball
IF !=
We know that C[3] is the odd ball
IF !=
We know that the odd ball is in C[1,2]
3rd Weigh: C[1] and C[3]
IF =
We know that C[2] is the odd ball
IF !=
We know that C[1] is the odd ball
IF !=
We know that the odd ball is in A or B
For simplicity’s sake, assume that the lighter group is A
2nd Weigh: A[1]+B[1,2] and A[2]+B[3,4]
IF =
We know that the odd ball is in A[3,4]
We know that the odd ball is lighter than the others
(Remember that A was the lighter group)
3rd Weigh: A[3] and A[4]
We know that the lighter ball is the odd ball
IF !=
Take note of which group is lighter and which group is heavier
We know that the odd ball is either one of the B balls in the heavier side or the A ball in the lighter side
(Remember that the A group was lighter group in the 1st Weigh)
IF A[1]+B[1,2] is lighter
We know that the odd ball is in B[3,4] or is A[1]
3rd Weigh: B[3] and B[4]
IF =
We know that A[1] is the odd ball
IF !=
We know that the heavier ball is the odd ball
(Remember that A[2]+B[3,4] weighed heavier in the 2nd Weigh)
IF A[2]+B[3,4] is lighter
We know that the odd ball is in B[1,2] or is A[2]
3rd Weigh: B[1] and B[2]
IF =
We know that A[2] is the odd ball
IF !=
We know that the heavier ball is the odd ball
(Remember that A[1]+B[1,2] weighed heavier in the 2nd Weigh)
[/code][/spoiler]
^^ very nice solution Alvin!
Very elegant
There are 25 race horses that each run at different speeds. You can only race 5 horses at a time, but you have no stopwatch or any way to time them. However, you know that all the horses always run at constant speeds. (i.e. The fastest horse will always be the fastest in every race.)
What is the minimum number of races you need to determine the fastest 3 horses?
Sample way of determining the fastest 3 horses (but not necessarily the optimal way):
Race 1: race 5 horses
Race 2: race the fastest three horses from Race 1 with 2 new horses
Race 3: race the fastest three horses from Race 2 with 2 new horses
Continue until all 25 horses have been raced. The fastest 3 in the last race are the fastest 3 of all.
Total races needed: 11 <-- correct answer is less than or equal to 11
I’m tentatively thinking 7.
Divide horses into 5 groups.
I think I need to rethink this. Oh 8 races should be enough. 
Please pm me Alot before I go crazy…
I think 9 races.
The races are conducted in 3 rounds. The first round, the horses are divided up into 5 groups of 5 horses. Each group runs a race, making 5 races in this round.
The second round has 3 races. All of the first-place horses from each of the first-round races run against each other, all of the second-place horses run against each other, and all of the third-place horses run against each other. The fourth- and fifth-place horses are eliminated.
The horse that comes in first out of all the first-place horse is the fastest overall, and need not run in another race.
Out of the horses that came in first place in the first round, the ones that came in second and third in the next round go on to the final round. The fourth- and fith-place horses are again eliminated.
Out of the horses that came in second in the first round, the ones that came in first and second in the next round go on to the finals. The third place is eliminated because that horse had been beaten by 3 other horses already: one in the first round and 2 other horses in the second. Fourth- and fifth-place are eliminated, as before.
Finally, out of the horses that came third in the first round, only the horse that came first in the second round goes on to the finals. The other horses were already outrun by 2 horses in the first round, and 1 or 2 horses in the second round. Fourth- and fifth-place are eliminated, as before.
Thus, the horses that run in the final race are the second and third place out of the first-placers, the first and second out of the second-placers, and the first out of the third-placers. The winner and runner-up from this round are the second and third ranked overall, joining the winner of the first-placers’ race in the top 3.
So, there are 5 races in the first round, 3 race in the second, and one final race, making 9 in total.
Considering this further: No, I was wrong, and moogle’s solution was closer than mine. It is 7 races.
As before, in the first round, each group of horses runs a race, making 5 races in the second round.
Then, in the next round, there is just one race, with the first-place winner from each group from the first round. As before, the first-place winner from this race is first-place overall, and need not run another race.
In the final round, both the second-place and third-place winners from the first-round group that the overall first place ran in need to run, because all of the top 3 might have ended up in that group.
The second place horse from the second round, and the runner up to that horse from the second round also need to compete in the final round. The lower places don’t need to run, because the third-place horse is slower than the second-place horse from the first round, which is slower than the first-place horse from that round, which is slower than the first-place horse from the second round. Thus, we know there are 3 horses that are faster than those horses.
Likewise, the the third-place horse from the second round is, at best, third-place overall, so none of the runner-ups which that horse beat in the second round need to compete in the final round.
Therefore, in the final round, the 5 horses are the second- and third-place horses from the second round, the second- and third-place horses that the overall first-place horse outran, and the runner-up from the first-round to the horse that came in second-place in the second round. The top 2 horses from the final round are second- and third-place overall.
So, there are 5 races in the first round, one in the second round, and one in the final round, for 7 races in all.