The Total is 14, Part III
Logic puzzles require you to think. You will have to be logical in your reasoning.
Mr. Simkin, the new math teacher at school, was impressed by his students' ability to solve logic puzzles. He pulled aside three more students, and handed them each a sealed envelope with a number written inside. He told them that they each have a positive integer, and the sum of their numbers was 14.
Manny, Moe, and Jack each opened their envelopes. Mr. Simkin asks Manny if he knows anything about the numbers the other two are holding, and Manny says, "I know that Moe and Jack are holding different numbers."
Moe answers, "IN THAT CASE, I know that all three of our numbers are different."
Jack thinks for a bit, and then says, "Now I know all of our numbers."
Mr. Simkin turns to the class and asks if anyone in the class knows the numbers. Gretchen's hand shoots up into the air, and after waiting for a while to see if anyone else will get the answer, Mr. Simkin calls on Gretchen.
What numbers does she say they each are holding?
Answer
Manny has a 3, Moe has a 2, and Jack has a 9.
From Manny's statement, we can deduce that his number is odd. Since Moe did not know that they were all different until Manny said that, we know that Moe is not holding a 7, 9, or 11. (Otherwise, he would have already known they were all different.) If he were holding a 1, 3, or 5, he would not be able to be sure his number was different than Manny's. If he were holding a 4, 8, or 12, he could not know that Manny and Jack didn't have the same number, since there are odd pairs that would bring the total to 14. Therefore, Moe must be holding a 2, 6, or 10.
There are 12 triples for (Manny, Moe, Jack) that satisfy all three statements. They are:
(1, 2, 11)
(1, 6, 7)
(1, 10, 3)
(3, 2, 9)
(3, 6, 5)
(3, 10, 1)
(5, 2, 7)
(5, 6, 3)
(7, 2, 5)
(7, 6, 1)
(9, 2, 3)
(11, 2, 1)
Since Jack can reason flawlessly, he knows these are the possibilities. In order to make his statement, his number has to be a unique solution. Therefore, he must be holding a 9 or an 11. If he were holding 11, though, he would have known from Manny's statement that Manny had a 1, Moe had a 2, and he had an 11. Since he didn't know them all until after Moe spoke, he must have a 9, leaving Manny a 3 and Moe a 2.
Mr. Simkin suggests to Gretchen that she may have a career in law enforcement if she can further hone these impressive deductive reasoning skills.
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Comments
NomadShadow   
Apr 26, 2006
| You are incorrect, because if Jack has been able to reason correctly he would've known that Manny cannot have either 7, 9, or 11. And thus there are only 8 possible triples, deleting the last two you have(the ones containg a 7 for Manny).
We know Jack doesn't hold 11 because it wouldn't need him to think. But after deleting the last two triples, there are 3 possible unique answers that Jack might have (Manny, Moe, Jack):
(3,10,1)
(3,6,5)
(3,9,2)
From this we know that Manny must have a 3, but only Jack and Moe could know the right answer.
In short, this teasor cannot be solved given just the information above. |
mercenary007 
Apr 26, 2006
| I dont' know what you are talking about. I think you need to read the answer again. I find the reasoning to be sound. Also think logically... if the teaser did not have an answer it would not be approved so therefore your thinking must be erroneous.  |
tsimkin 
Apr 27, 2006
| Nomad -- Manny could have had a 7, 9, or 11. Mr. Simkin asked him what he could say about the other two, and he said they were different. This is indeed something he could say if he held a 7, 9, or 11. |
NomadShadow
Apr 28, 2006
| Ok, this is a word play. Manny did know that they all had different numbers, yet she didn't say that, she only said the other two are different. Hmmm didn't think of that. But in that case aren't you missing two other possible triples. The full list should be (assuming that Manny can have 7, 9, or 11):
(Manny, Moe, Jack)
(1, 2, 11)
(1, 6, 7)
(1, 10, 3)
(3, 2, 9)
(3, 6, 5)
(3, 10, 1)
(5, 2, 7)
(5, 6, 3)
(7, 2, 5)
(7, 6, 1)
(9, 2, 3)
(11, 2, 1)
Still in this case the answer would be 9. |
tsimkin
Apr 28, 2006
| You are right -- fixing it now |
vanessaruck1989
May 06, 2006
| I got that one right away.  It was pretty neat though.  |
stormtrooper
May 08, 2006
| Good stuff.  |
mathwhiz1224
May 09, 2006
| This logic problem was easy but I think it is because I am in 7th grade that I already learned this stuff but it was pretty fun to read light but fun I needed the mind explore of where that was hiding in my mind so well good job and I liked it |
ava199345
Aug 14, 2007
| I'm confused. I read the entire teaser about 10 times and i still cannot figure out how you can deduce that Manny's number is odd from what he said. |
tsimkin
Aug 15, 2007
| ava199345 -- you know that Manny's number is odd because the only way he would know that the other two had different numbers (which he said) is if they COULDN'T be the same. If he had an 8, for example, the other two could have different numbers (like 4 and 2) or the same number (like 3 and 3). If he had a 7, though, there is no combination of numbers that the others could have had where they both would be the same. If this still doesn't make sense, PM me. Thanks! |
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