1. #1
    Deleted

    Simple math riddle.

    3 guys walk into a bar.

    They order a giant pitcher of beer, for the cost of 30 bucks, and each put down 10.

    The bartender takes the money and they guys go sit at their table.

    The bartender shortly after realizes that he made a mistake as the price for the pitcher of beer was actually 25, not 30.
    He ponders what to do, as he only has 1 dollar bills, and he can't share 5 bucks between 3 people equally.
    He comes up with the idea to tell the guys that the beer was 27, and not 25, as it really was, as this way, he can just give em back a buck each, and bank the remaining 2. He gives the guys each their bill, and they all praise him for his honesty.

    So, the guys originally paid 10 dollars each. They each recieved 1 dollar back which puts them at an expense of 3x9=27 dollars between them.
    The bartender who gave the money back kept the 2 bucks for himself.
    But if they guys each paid 9, and the bartender has 2, that equals 3x9 + 2 = 29. The price was originally 30 bucks.. where is the last dollar?

    29=30? Or are you missing something?

  2. #2
    Deleted
    25 + 2 + 3 = 30.
    3 * 9 + 3 = 30
    3 * 9 = 27

  3. #3
    Lets recap

    Cost of Booze: 25$
    Money Returned: $3
    Bartender Kept: $2

    25+3+2=30

    The 9$ each man spent includes the cost of the 2$ the bartender kept. And the bartender returns 3$. The final paragraph omits the 3$ returned to the men and duplicates the 2$ the bartender kept

    The math ends up looking like this

    Cost of Beer: $8⅓ each: $8⅓ x 3 = $25

    Cost of Money Bartender Kept: $⅔ each: $⅔ x 3 = $2

    Pocket Money Left Over: $1 each: $1 x 3 = $3

    Total: $30
    #1 Resto Tree Slap DPS

  4. #4
    I've heard this same problem with hotel rooms or something.

    Want a real math riddle? Prove that a set contains the elements it contains.

  5. #5
    Quote Originally Posted by manbeartruck View Post
    But if they guys each paid 9, and the bartender has 2, that equals 3x9 + 2 = 29. The price was originally 30 bucks.. where is the last dollar?

    29=30? Or are you missing something?
    -Each guy pays 9.
    3*9: 27
    -The bar keeper took 2.
    27-2: 25

    Why do you say that the keeper adds (+) 2 to the money that the guys spend? That's included in the 27 bucks otherwise they would not have payed 9 each, but 8 and 1/3.

  6. #6
    Deleted
    Quote Originally Posted by Tangster View Post
    Lets recap

    Cost of Booze: 25$
    Money Returned: $3
    Bartender Kept: $2

    25+3+2=30

    The 9$ each man spent includes the cost of the 2$ the bartender kept. And the bartender returns 3$. The final paragraph omits the 3$ returned to the men and duplicates the 2$ the bartender kept

    The math ends up looking like this

    Cost of Beer: $8⅓ each: $8⅓ x 3 = $25

    Cost of Money Bartender Kept: $⅔ each: $⅔ x 3 = $2

    Pocket Money Left Over: $1 each: $1 x 3 = $3

    Total: $30
    ^. is the anwser to the riffle in detail imo :P
    Had that figured out also ^^ but sombody already done the math for me :P

  7. #7
    Quote Originally Posted by Verdris View Post
    I've heard this same problem with hotel rooms or something.

    Want a real math riddle? Prove that a set contains the elements it contains.
    Thats not a riddle, thats barely even a problem since it is awnsered by an Axiom (a foundation of math that people take as fact without a specfic proof to prove it.)The Axiom of Regularity basically says that no set is an element of itself. http://en.wikipedia.org/wiki/Axiom_of_regularity
    However if that doesnt convince you, try the easy way out

    Problem: A set A contains the elements (x,y,z), prove A contains x,y,z
    Proof: Given A contains x,y,z, we know x is contained in A because it is assumed for the problem. y,z follow in this manner, thus A contains x,y,z

    That's kinda cheating tho, i wont get into why. The Axiom is the real answer :P
    #1 Resto Tree Slap DPS

  8. #8
    Deleted
    Well the way I see it you don't really count right.

    So they payed 30 then bartender saw it was 25 so gave them 3 back and kept too.

    And yes, 9x3=27. So the guys had already payed 30 out of which the bartender gave them 3 back. 30-3=27. I don't see any problem.
    You're counting it as the price considering how much they recived back, but in fact what matters is how much the bartender kept. He kept 27 and gave them 3. There's no dollar missing.

  9. #9
    no matter how many times people explain it to me I can't find the logical twist that makes the math work, it reasserts itself as a contradiction every time I read it.
    Quote Originally Posted by Aucald View Post
    Having the authority to do a thing doesn't make it just, moral, or even correct.

  10. #10
    Quote Originally Posted by Powerogue View Post
    no matter how many times people explain it to me I can't find the logical twist that makes the math work, it reasserts itself as a contradiction every time I read it.
    The money the barkeeper keeps in the end = 2 dolars
    The money those 3 pay (= 25 dollars)
    The other 3 dollars are given back (In 1 dollars bills, 1 for each person)

    So they basically pay 10 dollars each, minus the 1 dollar they get returned from the bar keeper. which makes it 9 dollars. But he also kept 2 dollars which they never get back.

    Recap

    Guys = 27 dollars paid (9x3 = 27)
    Register gains = 2 dollars ( instead of 25, they pay 27 so + 2 for the register)
    And in the end they all get 1 dollar back (3 dollars)

    27+ 3 = 30...The 2 dollars in the register is the 2 dollar he overpriced them, they aren't getting it back so you shouldn't involve it in your calculations since it's already there 27 = (25 + 2 dollars he over charged them)

  11. #11
    Quote Originally Posted by Tangster View Post
    Thats not a riddle, thats barely even a problem since it is awnsered by an Axiom (a foundation of math that people take as fact without a specfic proof to prove it.)The Axiom of Regularity basically says that no set is an element of itself. http://en.wikipedia.org/wiki/Axiom_of_regularity
    However if that doesnt convince you, try the easy way out

    Problem: A set A contains the elements (x,y,z), prove A contains x,y,z
    Proof: Given A contains x,y,z, we know x is contained in A because it is assumed for the problem. y,z follow in this manner, thus A contains x,y,z

    That's kinda cheating tho, i wont get into why. The Axiom is the real answer :P
    Trying hard to not derail the thread, but oh well.

    The way I'm reading that axiom is that it satisfies the old "Does the set of all sets contain itself?" paradox, but I'm not reading anything that satisfies that a set contains the elements it contains. Not that this is a paradox, just something really, really, really difficult to prove, when you don't resort to tricks like "by construction".

  12. #12
    Hi. What concept does this use? Can you please elaborate on the concept? Thank you

    - - - Updated - - -

    What is the concept behind this? Thank yoU!

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