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DeGe

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Posts posted by DeGe

  1. I am adding this using my mobile and I don't see any spoiler button, so the mod may please put this in a spoiler.
    Here is a general method to solve for more possibilities

     

    Let ab = x
    Then cd = 99 -x
    Abcd = 100x + 99 -x = 99 (x+1)
    Now the condition is that N (x,99-x) should be divisible by 99 (x+1)

    All N are divisible by 9 and 11 - easy to prove.
    Now we just need to see which x+1 can divide N.
    Some numbers I can identify quickly are
    1,98 divisible by 2
    4,95 divisible by 5
    7,92 divisible by 8
    9,90 divisible by 10
    19,80 divisible by 20
    24,75 divisible by 25
    39,60 divisible by 40
    49,50 divisible by 50

    There can be more possibilities. For example I don't remember the divisibility rules for 7,13, 27, 17, 19, 23 etc.

    So can not check for multiples of 3 or above numbers

  2. 75


    Add column 1 and 3 and reverse digits to get column 2

     

    8


    Multiple numbers on left and subtract numbers on right

     

    92


    Multiply numbers on left then subtract product of numbers on right

     

    6 & 5


    2nd number is a multiple of x and third number is first + x
    x decreases in each line from 4/3/2

     

    Not sure about this one


    If i had to guess, i would say 5867
    This is the only number where 2nd digit is greater than the first

     

    31


    Add all numbers and reverse the digits

     

    28


    Add equal amount in digit on top and then to digit on right
    Amount decreases 6/5/4

     

    7


    Add first and 2nd row digits, then subtract third row digit, then add 1
  3. Lets say the area of the triangle is A

    The area of reflected triangle in also A

    Lets say area of overlap is O

    The area we are looking for is 2A - O

    Area O is a kite with diagonals 2 and 5 (haven't drawn it but only pictured it... so not sure if this is correct)

    Therefore area (d1d2/2) of O = 5

    The area we want is 2A - 5

    Now for the aha moment:

    2A - 5 = A + 1/4A = 5/4A

    3A/4 = 5

    A = 20/3

    So the area we want (2A - O) = 25/3

  4. For starters:

    the second and third words seem to suggest "google batman"... last word of first line could be "looks".

    similarly, the second line suggests to "google" something...

    can you send a screen shot of where and how this phrase is written. That might contain more clues...

    The idea here would be to google some keywords and find the answer to your puzzle.

  5. The youngest child among the oldest children in each column is NEVER younger than the oldest child among the youngest children in each row.

    He is either older or of the same age.

    The same age happens when the youngest in any row is the oldest in that column.

    Number of ways in which the ages could be same can be calculated as below:

    Choose any place (chosen cell) in the 5x10 matrix. This can be done in 50 ways

    The kid in this cell must be the youngest in the row and oldest in the column. All other places can be filled randomly.

    So, there are 9 (in row) and 4(in column) "special cells" where combinations must be made. For other cells, the rest can be placed in 36! Ways.

    The 9 chosen kids in rows and 4 chosen in column can be palced in 9!*4! Ways.

    Now the only thing that remains to do is to choose 1 kid for the chosen cell and 9 kids for the rows and 4 kids for the columns in the special cells.

    For any given kid lets say there are x kids younger than him/her and 49-x elder kids

    Now we need to choose 4 kids out of x younger kids and 9 kids out of 49-x elder kids.

    Note that x varies from 4 to 40 ONLY.

    4C4 * 45C9 + 5C4 * 44C9 + .... + 40C4 * 9C9

    Using excel for this, the total combinations comes out to 9.378E + 11

    Now, the total possible ways of placing the kids is 50!

    Total ways in which the age is same is = 50 * 36! * 4! * 9! * the above combinations

    Relying on excel getting the calculations right, there is 1 in 200 times* that the age is same.

    So, the probability we are looking for is 1/200 = 0,5%

    * Excel calulated it to be 1 in 200.2 times. I simplified it to make the answer look elegant :D

  6. The first three points are always within a semi-circular arc.

    From 4th point onwards, the probability of lying within a semi-circular arc is 1/2. For 4th the probability is 1/2. For 5th, it will be 4th lies within the arc and the 5th also; so 1/4... and so on

    The overall probability is then 1/2(n-3) for n>3. For n<3 it is 1.

  7. There is ambiguity to define "line length" when the line has non-zero width and is not all straight. If it curves there is both an inside and outside circumference. Similarly at corners different parts of the line have different lengths.

    I'll take a simple-minded approach and say first that it's clear the entire page can be covered without lifting the pen or retracing. Then I'll say area = length x width.

    Length = 11 x 8.5 / 1 = 93.5 inches.

    Agree! Another Aha puzzle.

  8. attachicon.giflotr ok.png

    Dig along the green lines

    1 + 1 + √(0,52 + 0,52) = 2,7071

    I think maybe Sam could dig even less, following the pink lines BMD + MC + AO (green), but I don't know, how to find point M.

    It would be minimum for M chosen at a distance of (3 -root(3)) / 6 = 0.211 from both sides

    This gives total length of digging as 2.64

  9. The positive result of the test is possible by 2 ways:
    Someone "X" has the STD and the checker gave the right result or X does not have STD and the checker gave the wrong result.

    Random probability that X has STD = 0.001
    Prob that the test gave correct result = 0.93

    So, in a sample population, number of correct positive results would be = 0.93 * 0.001 = 0.0093

    Random probability that X does not have STD = 0.999
    Prob that the test gave wrong result = 0.07 (showed positive for an actually negative person)

    In the sample population, number of false positive results would be = 0.999 * 0.07 = 0.06993

    So, for every one correct positive there are = 0.06993/0.0093 = 75.19 false positive results

    So, the probability that W has STD given that the result shows positive is = 1/(1+75.19) = 0.013

  10. There is either some info missing or I have misunderstood the question. I am getting 2 solutions that satisfy the conditions:

    Lets say the boxes are arranged as:

    c d

    a b

    g h

    e f

    Then;

    a+b = e+f

    c+d = g+h

    a+c = g+e = d+h

    d+b = h+f = a+e

    b+f = 6

    c+g = 12

    Subtract 1 and 3; b-c = f-g

    The, b+g = c+f

    b+g+c+f = 18

    Therefore, b+g = c+f = 9

    Also, substract 2 and 3

    then c-h = h-e

    c+e = 2h

    Now b+g = 9, there are 4 possibilities; 1/5; 5/1; 2/4; 4/2

    All the conditions are satisfied with the first 2 possibilities as below:

    b f c g e h a d 1 5 4 8 2 3 6 7

    b f g c e h a d 5 1 4 8 6 7 2 3

  11. post-54025-0-41888200-1382340758_thumb.j

    The digging should be done along the red lines as shown in the figure above
    AB² = x² + 0.25
    BC = 1 - 2x

    Total length of digging = L = 4AB + BC

    L = 4(root(x² + 0.25)) + 1 - 2x

    Now we need to find the minimum value of this f(x)
    dL/dx = 8x/(2 root(x²+0.25) - 2

    This dL/dx should be zero for the minima
    This gives: 4x/root(x² + 0.25) - 2 = 0
    4x² = (x² + 0.25)
    x = 1/2(root3)

    With this x , the total digging length is 2,732

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