[PSUSMU]

In a tavern not far from the place where Ariosto and Fibonacci left Anchuria, three knights were arguing who among them was the bravest and strongest. The locals all agreed that each knight was brave and strong but declined to decide who was more so. Finally, the pretty barmaid said that she would set them a task to prove their worthiness. After thinking a little she wrote a number on each knight’s forehead. The numbers, of course, were magical: a) each knight could see what number was written on the foreheads of the other knights; b) each knight was not able to see the number on his own forehead; and c) the barmaid (who was well-known to be exceptionally truthful) informed them all that one of the numbers was a sum of the other two. “Now, try to guess the number on your forehead. Let the best one win!” said the barmaid and the tournament started.
- The first knight said that he could not guess his number;
- Then the second knight said that he also was unable to guess his number;
- Then the third knight said, “I cannot guess my number, either”;
- Then the first knight said, “In that case, my number is 50.”

He was right, but the others argued that the test was not fair because he guessed twice. In the argument the numbers on their foreheads were erased. That is when Ariosto arrived and everyone appealed to him to sort things out. After hearing the story (exactly as it is recorded above), he thought a little and then said that he knows all three numbers as well as which number was written on which knight’s forehead. On hearing of such a feat of strength and bravery everybody agreed that Ariosto was by far the best knight present.

Question: What were the numbers and how Ariosto knew them?

[Bimmer84]

15, 35 and 50?

[SS32]

Originally Posted by Bimmer84

15, 35 and 50?

Those work, but I'm trying to get sets of equations to prove it.
X + Y = 50, assuming the sum of 2 of them is 50 and not X or Y
...I need one more unique equation to solve for one of the variables.

Edit:

X + Y = Z

Assuming Z = 50

X + Y = 50

X & Y could be any combination of numbers whose sum = 50 unless we know more about one of the variables.

[Chewy734]

You have 3 unknowns... X, Y, Z, and only two independent equations:
X+Y=Z
Z=50

You can't solve it for a unique solution. All you know is X+Y=50, and there are an infinite number of solutions for X and Y, since the riddle doesn't say the number has to be >= 0.

Unless of course I missed something, and I'm wrong

[AU335i]

If the numbers can not be the same and have to be greater than 0, then I know what the numbers are lol

[AU335i]

Originally Posted by Chewy734

Unless of course I missed something, and I'm wrong

pretty sure the dumb bitch of a barmaid left some key info out lulz

[MetsFan]

http://www.youtube.com/watch?feature...=jqpQPxWj8gE#!

Sorry, can't embed...

[SS32]

Cliffs? I don't have sound at work

[MetsFan]

Originally Posted by SS32

Cliffs? I don't have sound at work

Doh... I didn't have time to watch the whole thing so I just posted it.

[yakev724]

OP, you forgot 2 rules.

1) All 3 numbers are different
2) All numbers are > 0

[SS32]

That still doesn't give us much more to go on. X and Y can still be any combination of numbers whose sum is 50.

[yakev724]

No, X and Y have to be 20 and 30. Here's why:


Random Scenario 1

Say X is 10 and Y is 40 (Z is 50).

First, Z sees 10 and 40. He knows he can be 50 or 30, so he is unsure.

Next, X sees 40 and 50. He knows he can be 90 or 10, so he is unsure.

Next, Y sees 10 and 50. He knows he can be 40 or 60, so he is unsure.

Now Z assumes he was 30. In that case, X would see 40 and 30 and would be unsure of whether he was 70 or 10 (which checks out since he was unsure). Y would see 10 and 30 and would be unsure of whether he was 20 or 40 (which also checks out).

Now Z assumes he was 50. In that case, X would see 40 and 50 and would be unsure (10/90), and Y would see 10 and 50 and would also be unsure (40/60), which also checks out.

So Z doesn't have a way to know whether he's 30 or 50.


Proper Scenario

X is 20, Y is 30, Z is 50

First, Z sees 20 and 30. He doesn't know whether he's 50 or 10.

Next, X sees 30 and 50. He doesn't know whether he's 80 or 20.

Next, Y sees 20 and 50. He doesn't know whether he's 30 or 70.

Now Z assumes he's 10. In that case, X would see 10 and 30. He would be unsure of whether he was 20 or 40. Y would see 20 and 10. He could be 10 or 30, but if he was 10, he would violate the rules since both Z and Y would be 10. The only number he could be is 30, and he would have been sure of it first time through.

From this, Z knows he's 50 since if he was 10, Y would be sure of his number (30) on the first pass.




As far as I can tell, 20, 30 and 50 are the only numbers for which this works. Multiples would work too (50, 75, 125), but then the order would be screwed up.

[CamelBiscuit]

Do any of you clean bathrooms at MIT?

[SS32]

Originally Posted by yakev724

No, X and Y have to be 20 and 30. Here's why:


Random Scenario 1

Say X is 10 and Y is 40 (Z is 50).

First, Z sees 10 and 40. He knows he can be 50 or 30, so he is unsure.

Next, X sees 40 and 50. He knows he can be 90 or 10, so he is unsure.

Next, Y sees 10 and 50. He knows he can be 40 or 60, so he is unsure.

Now Z assumes he was 30. In that case, X would see 40 and 30 and would be unsure of whether he was 70 or 10 (which checks out since he was unsure). Y would see 10 and 30 and would be unsure of whether he was 20 or 40 (which also checks out).

Now Z assumes he was 50. In that case, X would see 40 and 50 and would be unsure (10/90), and Y would see 10 and 50 and would also be unsure (40/60), which also checks out.

So Z doesn't have a way to know whether he's 30 or 50.


Proper Scenario

X is 20, Y is 30, Z is 50

First, Z sees 20 and 30. He doesn't know whether he's 50 or 10.

Next, X sees 30 and 50. He doesn't know whether he's 80 or 20.

Next, Y sees 20 and 50. He doesn't know whether he's 30 or 70.

Now Z assumes he's 10. In that case, X would see 10 and 30. He would be unsure of whether he was 20 or 40. Y would see 20 and 10. He could be 10 or 30, but if he was 10, he would violate the rules since both Z and Y would be 10. The only number he could be is 30, and he would have been sure of it first time through.

From this, Z knows he's 50 since if he was 10, Y would be sure of his number (30) on the first pass.




As far as I can tell, 20, 30 and 50 are the only numbers for which this works. Multiples would work too (50, 75, 125), but then the order would be screwed up.

What if X saw 25 and 25?

[yakev724]

Originally Posted by SS32

What if X saw 25 and 25?

Can't happen; all 3 numbers must be different.

I added the 2 excluded rules in post 10.

[SS32]

You added those, but they weren't in the original riddle.

[yakev724]

Originally Posted by SS32

You added those, but they weren't in the original riddle.

The solution I posted works because a case of 2 numbers being identical (the only such case) can be excluded. Without that, there would be no solution (as far as I can tell).

I think OP found a bad source for the riddle or forgot the extra rules when typing it up. It's a pretty famous one and I doubt this is a different version of it.

[PSUSMU]

My manager sent it out (exactly as I posted) for our fun, weekly team puzzle. I came up with 20, 30, 50 as well but not sure if it's the correct solution.

[yakev724]

Originally Posted by PSUSMU

My manager sent it out (exactly as I posted) for our fun, weekly team puzzle. I came up with 20, 30, 50 as well but not sure if it's the correct solution.

It is; but again, it's missing the 2 rules I posted.

See the video link above for the best explanation. My post is just a synopsis.

[SS32]

If we say the numbers must be unique, then 20,30,50 work. If not, 25, 25, 50 is a viable answer as well. The video tells you two of the variables and sets more conditions, not the same problem IMO.

[BMWinNorthdakota]

What the fuck is shit this is wtf?

MATH???!?!?!?!?!!?!?!?!?!?!?!?!?!?!??! get out

[yakev724]

Originally Posted by SS32

If we say the numbers must be unique, then 20,30,50 work. If not, 25, 25, 50 is a viable answer as well. The video tells you two of the variables and sets more conditions, not the same problem IMO.

25 25 50 only works if you force each number to be >= 0. This isn't stated in the first post, so it isn't a solution if you go by first post rules only.

I think that actually the rules I posted might not even be necessary. The only solution is 20 30 50 regardless of the possibility for identical and/or negative numbers.