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DeGe

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Posts posted by DeGe

  1. By brute force, 3/4 of the space is white

    Now, I need to get the creativity points :blink:

    The white area is 100%!

    I think this result is mainly because we are considering area (in black) covered by points; which by definition have an area of 0.

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  2. 2 exchanges needed.


    The balloon with bag number 3 is not involved in any exchange. The ballons with 5 and 1 lb bags exchange once and the ballons with 4 and 2 lbs bags exchange once.
    Total time taken is 1000/3 = 333.3 secs; Both 4/2 and5/1 exchange the bags at exactly 166.67 secs.
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  3. Effectively, you are adding the numbers on the left. Now, all the numbers on the left are doubles (even) except the first number that you start with. So, irrespective of the number you have on the right, the addition on the left will be odd if you start with an odd number on the left.

    Therefore, this algorithm is not correct.

  4. Four people is rich enough to be instructive, and still tractable by hand.

    In case of 4 people, the immediate neighbours of the starter each have a starting probability of 1/2 of not being the last.

    For the one opposite, it does not matter whether the first toss is heads or tails. After the first toss, he also has a probability of 1/2 of not being the last. So his effective starting probability is also 1x1/2 = 1/2

    All have equal probability of getting the coin before the last person.

  5. I think it should be 25.

    Here's why:
    Basic assumption: Results of all expressions should be integers

    Simplifying all the expressions, it is clear that the number should be a perfect square.
    Next, in 2 or 3 expression with the decimals, the single digit should divide 10, or double digit number should divide 100, a triple digit should divide 1000... and so on
    Finally, the last expression yields an integer if the number is 16 or 25.

    Using all the above combinations, 25 seems to be the correct answer

  6. Looks like it is fair. The important point is not whether someone get the coin on first few throws but that they should not be the last one to get it. In such a case, the "endgame" would be symmetrical for any number of persons and for any position around the table.

    Meaning that, after the first toss of coin for example, the situation is the same as in the beginning except that one person is sure to get a free coffee. Effectively, it is the same situation but with one person less. And the positions of some persons will have improved while that of others worsened.

    In the end, there would be any two random persons left who would both have equal probability of getting the coin first (because the coin can be closer to either one of them).

  7. that bellies the assumption that he must have apples to drive. which is not the case. he only eats apples while driving if they are available.

    ps. you are also answering a different question.

    If he eats apples only if they are available, it is much simpler then

    He takes 3 trips carrying 1000 apples on each trip to a distance of 333 1/3 miles

    Now he has 2000 apples left.

    Now he takes 2 trips of 500 KMs each; he has 1000 apples left and has covered 833 1/3 miles

    He drives the remaining 166 1/3 miles and delivers 833 1/3 apples to the store.

    Although a better way could be for the driver to drive an empty truck to the local store and explain the problem. Let the local store send a non-apple-eating driver with the truck who makes 3 trips delivering all the 3000 apples.

  8. (x+1)3 - x3 = c²

    3x(x+1) = c² - 1

    3x(x+1) = (c+1)(c-1)

    if c is sum of consecutive squares

    c = y² + (y+1)²

    c = 2y² + 2y + 1

    Then,

    3x(x+1) = (2y²+2y + 2)(2y²+2y)

    Take out 2 and 2y common from the 2 expressions

    3x(x+1) = 4y(y²+y + 1)(y+1)

    y² + y +1 = y(y+1) + 1, then

    3x(x+1) = 4y(y+1) [(y(y+1) + 1)]

    Let, y(y+1) = a

    Then

    3x(x+1) = 4a(a+1)

    Effectively then, it needs to be shown that for some "x", there exists an "a" such that the sum of first x digits is 4/3 times the sum of first a digits.

    Need help to prove this!

    You can go further will the above analysis that since a = y(y+1), it must be even; so let a = 2n

    Then, 3x(x+1) = 8n(2n+1)

    Therefore, either x or x+1 must be a multiple of 8

  9. If any displaced person takes the seat of the first person, all the following persons take their respective seats.

    Let us define any displaced person taking the first person's seat as "closing the loop".

    So, if the first person takes his own seat, he closes the loop and everyone gets their own assigned seats

    Now, lets define combinations to see how many persons other than the first person are needed to close the loop

    If one person is needed to close the loop, it means that the first person sits on his assigned seat

    If two are needed to close the loop, it means that the first person takes the seat of xth person in the line and the xth person takes the first person's seat; the rest sit as assigned

    If three are needed to close the loop, it means for example that the first person takes the seat of 5th person, the 5th person takes the seat of 8th person and the 8th person takes the seat of 1st person. The rest sit as assigned

    And so on

    (Note that two persons needed could also mean that first person takes seat of the last; all others sit as assigned and the last person must take the seat of the first)

    Therefore,

    Persons needed Combinations possible

    1 n-1C0 (only the first person closes the loop)

    2 n-1C1 (first person + any one other closes the loop (could be the last person also))

    3 n-1C2 ...

    n-1 n-1Cn-2 (the person before last closes the loop)

    n n-1Cn-1 (the last person closes the loop)

    So, the possible ways of seating arrangement are:

    n-1C0 + n-1C1 + n-1C2 + … n-1Cn-1

    n-1 everywhere because the first person is already chosen by default; we are looking at how many others are needed

    n-1Cn-1 would mean that the 1st person takes the seat of 2nd, he takes seat of 3rd, he takes seat of 4th…..last before one takes the seat of last, and the last person has to take the seat of the first person

    Sum of this combinations is = 2^(n-1)

    So, all the people can sit in 2^(n-1) ways

    Out of this, there are n-1 combinations where the last person's seat is taken before he enters

    (except in the case of n-1C0, there is always 1 case where the last person's seat is already taken)

    So, the probability of last person NOT getting his assigned seat is:

    n-1 / 2^(n-1)

    The probability of getting the assigned seat is then:

    1 - [n-1 / 2^(n-1)]

    Bonanova is right. In the above solution, the number of cases where the last person does not get his seta is much higher than n-1. It is actually 2^(n-2) considering all possibilities where the "loop is closed" with n-2 persons instead of n-1 so that the last person's seat is always available.

    The probability will then be as suggested by Bonanova.

  10. If any displaced person takes the seat of the first person, all the following persons take their respective seats.

    Let us define any displaced person taking the first person's seat as "closing the loop".

    So, if the first person takes his own seat, he closes the loop and everyone gets their own assigned seats

    Now, lets define combinations to see how many persons other than the first person are needed to close the loop

    If one person is needed to close the loop, it means that the first person sits on his assigned seat

    If two are needed to close the loop, it means that the first person takes the seat of xth person in the line and the xth person takes the first person's seat; the rest sit as assigned

    If three are needed to close the loop, it means for example that the first person takes the seat of 5th person, the 5th person takes the seat of 8th person and the 8th person takes the seat of 1st person. The rest sit as assigned

    And so on

    (Note that two persons needed could also mean that first person takes seat of the last; all others sit as assigned and the last person must take the seat of the first)

    Therefore,

    Persons needed Combinations possible

    1 n-1C0 (only the first person closes the loop)

    2 n-1C1 (first person + any one other closes the loop (could be the last person also))

    3 n-1C2 ...

    n-1 n-1Cn-2 (the person before last closes the loop)

    n n-1Cn-1 (the last person closes the loop)

    So, the possible ways of seating arrangement are:

    n-1C0 + n-1C1 + n-1C2 + … n-1Cn-1

    n-1 everywhere because the first person is already chosen by default; we are looking at how many others are needed

    n-1Cn-1 would mean that the 1st person takes the seat of 2nd, he takes seat of 3rd, he takes seat of 4th…..last before one takes the seat of last, and the last person has to take the seat of the first person

    Sum of this combinations is = 2^(n-1)

    So, all the people can sit in 2^(n-1) ways

    Out of this, there are n-1 combinations where the last person's seat is taken before he enters

    (except in the case of n-1C0, there is always 1 case where the last person's seat is already taken)

    So, the probability of last person NOT getting his assigned seat is:

    n-1 / 2^(n-1)

    The probability of getting the assigned seat is then:

    1 - [n-1 / 2^(n-1)]

  11. remember right angled is not considered acute in the game's directions

    I don't see why 90° should be an issue. There are no coins that make 90° -- 10 sides for 360°; angle between any 2 vertices is a multiple of 36° and will never be 90°.

    I agree with gavin on the solution.

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