Ive heard that one before and this is the answer…
The surgeon is his mother 
A man lives in a high building and when he takes the elevator to work he goes all the way down, but when going back home he only takes the elevator to few floors below his and takes stairs rest of the way. Why does he do this?
Because he’s too short to reach up on the highest floors’ buttons. But of course it’s no problem when going to ground floor, because those buttons are the lowest.
Don’t know any off the top of my hat and don’t want to check all topics if they were posted already^^ But maybe I’ll post something later.
I remember this one from a few days ago: A man said: “If yesterday was tomorrow, then today would be Friday.” What day is it then?
Quite a braintwister I admit, but still fairly easily solvable. In order for it to "today would be Friday ", yesterday had to be a thursday. So if yesterday was tomorrow, then it has to be Wednesday. If it’s wednesday, then tomorrow is Thursday. And if yesterday was Thursday, then today it would be Friday.
Gah I still can’t think of a good one.
But try this one. Next number in line:
4 , 9, 25, 49, 121, ?
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