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Circle's brain teaser challenge

  1. Dec 11, 2006 #1
    I have selected a few brain teasers at random for the most part that I found at least mildly amusing and that I could not immediately find in this sub-forum.
    I wouldn't say that any of them are very hard, but some require sideways thinking and I think that makes them fun to do.

    It is almost guaranteed that there will be a couple of problems that have been posted before, so just don't do those ones if you recognize them, and if you link to the original post containing that puzzle I will remove it from my thread so that we can make sure that these are all new.

    1. How many cigars can a homeless man make from twenty five cigar butts if he needs five butts to make one cigar.

    2.A Middle Eastern man died, leaving 17 of his camels. His will specified that they be divided among his three sons as follows:
    -1/2 to the oldest son
    -1/3 to the second son
    -1/9 to the youngest son
    The three sons were puzzling over how this could be done when a wise man happened to ride by on a camel. How did the wise man solve their provlem?

    3.Supply a digit for each letter so that the equation is correct. A given letter always represents the same digit(The answer is not that all letters=0):

    ABCDE
    .....x4
    ------
    EDCBA

    4.A woman goes into a hardware store to buy something for her house. She asks the clerk the price, and the clerk replies, "The price of one is twelve cents, the price of thirty is 24 cents, and the price of a hundred and forty-four is 36 cents." What does the woman want to buy?

    5.What is the next letter in this sequence? O T T F F S S

    6. If a plane crashes directly on the line between two states, where are the survivors buried (Had to post this one, I remember it from grade school)

    Have fun!

    Make sure to put answers in a "" tag.
     
    Last edited: Dec 11, 2006
  2. jcsd
  3. Dec 11, 2006 #2
    Answer (highlight to see)

    1. 6 (that is 5 cigars initially and after these are smoked down to butts, the 6th)
    2. This doesn't have a solution without cutting up the camels. Nor do the fractions add up to 1, so there will be some meat left over. The following procedure is often provided as if it were a solution:
    The wise man lent one of his own camels to the sons. Of the 18 camels, he gave 9 (1/2 of 18) to the eldest, 6 (1/3 of 18) to the middlest, 2 (1/9 of 18) to the youngest, and took back his own camel at the end.
    3. I haven't solved this one. However I note that A = B = C = D = E = 0 would work.

    eom
     
    Last edited: Dec 11, 2006
  4. Dec 11, 2006 #3
    I have changed question 3, the answer is not 00000
     
  5. Dec 11, 2006 #4
    Answers (highlight to view)

    4. House numbers.

    5. E (the first letter of the word eight)

    6. In cemeteries (after a considerable delay as they survive).

    eom
     
    Last edited: Dec 11, 2006
  6. Dec 11, 2006 #5
    highlight to see my response

    correct on all you have answered so far, very nice. I admit that the solution to the camel problem is not very elegant. I remember kids would say "You don't bury survivors." in grade school in response to question 6, but I think your logic is better in this case since eventually they do get burried. Do question 3 :). I think that one has the most rewarding solution.
     
    Last edited: Dec 11, 2006
  7. Dec 11, 2006 #6
    Answer (highlight to see)

    3. 87912 = 4 * 21978

    eom
     
  8. Dec 11, 2006 #7
    Did you use "brute force" for the third one? ;)
     
  9. Dec 11, 2006 #8
    I really liked the logical solution to #3. Good job on all of them Jimmysnyder, I hope lots of people have fun doing these. :smile:
     
  10. Dec 11, 2006 #9
    Yes.

    Text added to satisfy a curious criterion.
     
  11. Dec 23, 2006 #10
    #2

    If we look at the second and third son, they both share the same denominator. They combine sum in fraction is a little less than half. This implies the oldest son much have a little more than half fraction. The oldest son must have 9 camels.


    This leave the remaining 8 camels to be subdivided by the second and third son. The second son has 3/9, and the youngest has 1/9. which is given. This implies the seond son must inherent 3/4 of the remaining 8 camels( which is 6), and the youngest must inherent 1/4 of the remaining 8 camels( which is 2). to test : 9+ 6+2= 17 !

    I am not sure how to make my words white
     
    Last edited: Dec 23, 2006
  12. Nov 12, 2008 #11
    Dear Jimmy,

    Your very clever to have solved this problem. In fact it's very popular and I have heared it many times, admiring the wisdom of that gentleman all the time.

    Other than the famous solution you have also said that it doesn't have a solution. Quite obviously the fractions add up to 17/18 which is not equal to 1. Hence thare always will be some meat left over.
    Observing that it is a real world problem (I have heared the names of acutual Chracters, though I donot remember them). Even if the fractions did add up to 1 but you had to cut the camels, it would be no solution. Owners must have the full camels, cutting it into pieces would be a horrible thing for an Arab when it could be used as a "desert plane". The beauty of the solution is that no camel is to be cut and every stakeholder gets about 5.9% more than his acutual share if you were going to cut down these camels. And that is well within famous (10%) limit followed by engineers in general. Every body should have been quite happy and satisfied with his share.

    I hope you follow my point and agree that the presented solution is the best possible and most intelligent one.
     
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