Best math riddle I've seen in a while. [Archive]

brassnuts

01-28-2005, 06:39 AM

This was posted in a thread in the WPT forum. If you saw it, don't spoil it for everyone else. I thought it was a really good problem. Couldn't find the original post, so if anyone else who read it initially spots an error, please correct me.

Three logicians are standing in a circle, facing eachother. Each logician has a positive, unique, integer written on his forehead. One of the numbers is the sum of the other two. And, none of them have any prior knowledge to what their number is.

A man comes up to Logician A and asks, "Do you know what number written on your head?" He answers, "No."

He then approaches Logician B and asks, "Do you know what number written on your head?" He answers, "No."

He then approaches Logician C and asks, "Do you know what number written on your head?" He answers, "Yes, it's 50."

What numbers do A and B have on their foreheads?

This is not a trick question and there is on play on words.

Edit: Thanks to Evan for pointing out that huge mistake. It's late, I went skiing today, I'm tired.

MarkL444

01-28-2005, 06:47 AM

it feels like somethings gotta be missing. either that or im an idiot.

Evan

01-28-2005, 06:47 AM

yea, you fu[/i]cked up the question big time.

"I have to agree that this is a stupid and pointless post.
But, while we're at it, I gave this to Chris Ferguson on a plane and he got it in six minutes....
three logicians are standing in a circle.
each has a positive, unique integer written on his head.
one of the numbers equals the sum of the other two.
a fourth person comes by and says, "logician #1, do you know what number you have on your head". he says no. then: "logician #2, do you know your number". "no". then: "logician #3, do you know your number?". "yes, it's 50." ? --- what are the other two numbers? -wins_pot"

Evan

01-28-2005, 06:53 AM

You're welcome. Spell my name right. /images/graemlins/tongue.gif

brassnuts

01-28-2005, 06:54 AM

Haha... you're not an idiot. At least not because you couldn't solve this problem. I left out a huge piece of information.

Scotch78

01-28-2005, 08:29 AM

[ QUOTE ]
somethings gotta be missing.

[/ QUOTE ]

Scott

brassnuts

01-28-2005, 03:56 PM

Bump. No, I think by the time you had viewed the page, I had it corrected.

Nick B.

01-28-2005, 04:27 PM

did you want us to answer?

30 an 20

bholdr

01-28-2005, 04:36 PM

if it were 30 and 20, then the guy answering could have had 50 or 10.

i thought about 99 and 49- then he could have 50 or 198
24 and 26 then he could have 50 or 2
49 and 1 could mean 50 or 48
etc

i'm still missing something critical here....
how to get rid of the "or"?

mmbt0ne

01-28-2005, 04:47 PM

<blockquote>En réponse à:<hr />
if it were 30 and 20, then the guy answering could have had 50 or 10.

i thought about 99 and 49- then he could have 50 or 198
24 and 26 then he could have 50 or 2
49 and 1 could mean 50 or 48
etc

i'm still missing something critical here....
how to get rid of the "or"?

[/ QUOTE ]

This is exactly where I am.

Jman28

01-28-2005, 04:53 PM

[ QUOTE ]
[ QUOTE ]
if it were 30 and 20, then the guy answering could have had 50 or 10.

i thought about 99 and 49- then he could have 50 or 198
24 and 26 then he could have 50 or 2
49 and 1 could mean 50 or 48
etc

i'm still missing something critical here....
how to get rid of the "or"?

[/ QUOTE ]

This is exactly where I am.

[/ QUOTE ]

It has to do with the others not knowing.

If Logician A saw a 10 and a 20, he would know that he must be 30, because you can't repeat a 10.

-Jman28

Jman28

01-28-2005, 04:59 PM

[ QUOTE ]
did you want us to answer?

 20 and 30 

[/ QUOTE ] He's right, right?

-Jman28

DMBFan23

01-28-2005, 05:02 PM

[ QUOTE ]
[ QUOTE ]
if it were 30 and 20, then the guy answering could have had 50 or 10.

i thought about 99 and 49- then he could have 50 or 198
24 and 26 then he could have 50 or 2
49 and 1 could mean 50 or 48
etc

i'm still missing something critical here....
how to get rid of the "or"?

[/ QUOTE ]

This is exactly where I am.

[/ QUOTE ]

I bet the answer to your problem lies in

[ QUOTE ]
A man comes up to Logician A and asks, "Do you know what number written on your head?" He answers, "No."

He then approaches Logician B and asks, "Do you know what number written on your head?" He answers, "No."

[/ QUOTE ]

here.

pshreck

01-28-2005, 05:03 PM

You win the obvious award.

mmbt0ne

01-28-2005, 05:05 PM

50/3 is not an integer. If it was 60 the numbers would be 20 and 40. That is why I'm still not sure.

No, it's not 20 and 30, because C could still be 10 or 50.

daryn

01-28-2005, 05:07 PM

<blockquote>In risposta di:<hr />
if it were 30 and 20, then the guy answering could have had 50 or 10.

i thought about 99 and 49- then he could have 50 or 198
24 and 26 then he could have 50 or 2
49 and 1 could mean 50 or 48
etc

i'm still missing something critical here....
how to get rid of the "or"?

[/ QUOTE ]

huh???

it's the sum right? not difference. and you can only use positive integers.

edit: oh i thought for some reason you meant negative 49 when you wrote "49-".. for some reason.

nevermind

Jman28

01-28-2005, 05:08 PM

[ QUOTE ]
50/3 is not an integer. If it was 60 the numbers would be 20 and 40. That is why I'm still not sure.

No, it's not 20 and 30, because C could still be 10 or 50.

[/ QUOTE ]

Explanation:

It can't be 10. If Logician A saw the 20 on B and the 10 on C, he would know his number. Since he didnt't know, C can't be 10. 

-Jman28

LALDAAS

01-28-2005, 05:08 PM

50 is the sum of the other two so I say 25 and 25

Otherwise I have no clue I give up

Jman28

01-28-2005, 05:09 PM

[ QUOTE ]

huh???

it's the sum right? not difference. and you can only use positive integers.

[/ QUOTE ]

One of the numbers is the sum of the other two. Doesn't have to be C.

-Jman28

mmbt0ne

01-28-2005, 05:10 PM

<~~~ Still one level of thinking behind.

DMBFan23

01-28-2005, 05:15 PM

thanks.

Stork

01-28-2005, 05:18 PM

You don't read so good.

DemonDeac

01-28-2005, 05:19 PM

[ QUOTE ]
50 is the sum of the other two so I say 25 and 25

Otherwise I have no clue I give up

[/ QUOTE ]

thats ur best guess. the basic premise of this problem is that number dont repeat.

brassnuts

01-28-2005, 05:29 PM

[ QUOTE ]
did you want us to answer?

[/ QUOTE ]

Yeah, I just didn't want people who had seen this a few weeks ago to answer right away. Instead of, "Don't spoilt it," I should have said, "Give them some time to answer."

Anyways, Jman has given the correct and answer and the explanation. Though, I think he should have answered in white. Good job, Jmang.

Jman28

01-28-2005, 05:31 PM

Firstly, Nick answered before me, and in white.

I replied to that agreeing with him without stating the answer.

I didn't mean to spoil it but I wanted to answer that one question. Sorry guys. Maybe I'll see if I can edit.

-Jman28

brassnuts

01-28-2005, 05:34 PM

Here I am saying to answer in white, and I didn't even see Nick do it. But, yeah, Nick was right, and you gave the explanation.

RocketManJames

01-28-2005, 06:21 PM

In Edit: I mucked something up... lemme think more.

-RMJ

brassnuts

01-28-2005, 06:31 PM

If person 2 were looking at 24 and 2, he could be either 26 or 22.

Keep it up though, you're on the right track.

RocketManJames

01-28-2005, 06:36 PM

Ok, I got it now...

Person A: 20
Person B: 30
Person C: 50

Person A sees 30/50, concludes he is 20 or 80. Cannot say for sure so he answers NO.

Person B sees 20/50, concludes he is 30 or 70. Cannot say for sure so he answers NO also.

Person C sees 20/30, concludes he is 50 or 10. But if he were 10, then Person B would have seen 20/10, and concluded he was 30 or 10, but knowing that he couldn't be 10 due to the distinct integer clause, he would have known he was 30. But, since he didn't figure this out, person C now knows this was not the case, and knows he is a 50.

-RMJ

sexypanda

03-09-2005, 03:44 PM

I came up with an actual logical solution to this after I told my friend the riddle and she wasn't satisfied with "its 20 and 30". It's much easier to rationalize the solution once you know it, but here's the logic/math (I may be wrong, but I think it all works out):

So, each guy can narrow down their number to a set of two numbers, so guy C knows he's either A+B, or |A-B|. Guy C was able to narrow down his search to 50 and x but was then able to cancel out x due to the fact that Guy A and B didn't know their number. So in this case, we can figure that if he had x Guy A would know his number and this only happens if B=2C or C=2B (Example if guy A saw 10 and 20, he'd know he was 30, 20 and 40 he'd know he was 60, etc.). So now we can conclude that B=2X or B=1/2X. We can also conclude that if it was x then A=B+C (or A=B+x), but since it isn't C=A+B. So we have:

C=50
C=A+B
A=B+x
B=2x or 1/2x => x=2B or 1/2B

so...

50=A+B
50=(B+x)+B since A=B+x
50 = 2B+x
50= 4B or 2.5B

B= 12.5 or 20, since B has to be an integer its 20
=> 50=A+20 => A= 30

TheJackal

03-09-2005, 04:45 PM

75 and 25, he can see the other two numbers and he knows the other two numbers subtracted from each other equal his number, so his number is 50.

jakethebake

03-09-2005, 04:59 PM

I'm kicking all these guys' asses and stealing their lunch money.

AviD

03-09-2005, 05:06 PM

Nix

pshreck

03-09-2005, 05:10 PM

[ QUOTE ]
I don't get this...the riddle can't be this simple as I solved it in under 10 seconds.

But all you are asking is 3 variables, all of which are unique positive integers. Two of them add together to equal the third, and we know one of the integers is 50. Solve for the other two?

I find it hard to believe Chris Ferguson took 6 minutes to figure this out...seriously if that is the question, it took me less than 10 seconds.

[/ QUOTE ]

Show us how you got it, and what you think the answer is?

AviD

03-09-2005, 05:11 PM

I Nixed as I now see the problem. /images/graemlins/wink.gif

Klepton

03-09-2005, 05:13 PM

TheJackal

03-09-2005, 05:14 PM

What is the answer it has not been posted w/ explaination.

AviD

03-09-2005, 05:20 PM

OK, so it took me a couple of minutes! /images/graemlins/smile.gif

AviD

03-09-2005, 05:22 PM

[ QUOTE ]
75 and 25, he can see the other two numbers and he knows the other two numbers subtracted from each other equal his number, so his number is 50.

[/ QUOTE ]

This is the "simple" approach I originally used...and its easy to solve quickly. But there are only two numbers that it can absolutely be.

That is, I can say the other two are 20 and 30 or 19 and 31 or 18 and 32...but there are only two that it really can be and you can be certain it is them.

For example if you see 20 and 30...how do you know you don't have 10 on your forehead? He is CERTAIN he has 50.

AviD

03-09-2005, 05:23 PM

If one guy has 20 and the other guy has 30...why do I have to be 50 but not 10?

sexypanda

03-09-2005, 05:27 PM

[ QUOTE ]
If one guy has 20 and the other guy has 30...why do I have to be 50 but not 10?

[/ QUOTE ]

If he had 10, the guy with 30 would see the other two have 10 and 20 and know he had 30... the fact that A and B didn't know allows him to rule that out.

AviD

03-09-2005, 05:37 PM

[ QUOTE ]
[ QUOTE ]
If one guy has 20 and the other guy has 30...why do I have to be 50 but not 10?

[/ QUOTE ]

If he had 10, the guy with 30 would see the other two have 10 and 20 and know he had 30... the fact that A and B didn't know allows him to rule that out.

[/ QUOTE ]

NIXED AGAIN /images/graemlins/smile.gif