DeGe

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Going further on gmgm's thinking, all words except SLIM use a "U". SLIM should in fact have been SLUM. So i would pick "I" as the answer

75
Add column 1 and 3 and reverse digits to get column 28
Multiple numbers on left and subtract numbers on right92
Multiply numbers on left then subtract product of numbers on right6 & 5
2nd number is a multiple of x and third number is first + x
x decreases in each line from 4/3/2Not 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 first31
Add all numbers and reverse the digits28
Add equal amount in digit on top and then to digit on right
Amount decreases 6/5/47
Add first and 2nd row digits, then subtract third row digit, then add 1 
Let S1 and S2 be the first 2 terms
Then S3 = S1 + S2
S4 = S1 + 2S2
S5 = 2S1 + 3S2
S6 = 3S1 + 5S2
... and so on
Given any Sn and Sm, you just need to solve 2 equations having 2 variables S1 and S2
You can then find out all the S 
Iron Man
Kneel
Maxwell

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

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.

"b" ... based on my interpretation of the last sentence "what is not there"... the word "sudtle" should be "subtle"
For the second, is there a picture you wanted to upload that did not get uploaded or is "vwQRZ1Y.png" the clue?
 1

Thanx everyone for an interesting debate.
I wish real life were also as clear as pcia wants the op to be. I won't check it again, perhaps.
 1
 1

From experience, the 4 legged stool will be unstable. If you have a table one of whose legs is even slightly shorter (read surface is uneven), it will rock. A 3 legged stool on the other hand will be stable.

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 49x elder kids
Now we need to choose 4 kids out of x younger kids and 9 kids out of 49x elder kids.
Note that x varies from 4 to 40 ONLY.
^{4}C_{4} * ^{45}C_{9} + ^{5}C_{4} * ^{44}C_{9} + .... + ^{40}C_{4} * ^{9}C_{9}
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

(3+3) * square root (8+8)

Place two 345 triangles A1B1C1 and A2B2C such that B1 and B2 are offset by 4 and 3 units on x and y axis respectively

Considering the point raised by Rob, the answer stands corrected as 1/2
^{(n2)} 
I would say that the expected number of balls left in the urn is 1.
The number of balls would tend to get even. For b>w for example, the probability of black being drawn also increases... 
The first three points are always within a semicircular arc.
From 4th point onwards, the probability of lying within a semicircular 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^{(n3)} for n>3. For n<3 it is 1.

Let's take a global average and say that in average, in a family there are N boys and M girls. Then, each of the N boys have M sisters and each of the M girls have M1 sisters.
Therefore, in average men have more sisters than women.

You assume a constant rate of growth of grass over the entire duration.
I'm not sure about this and pls correct me if I am wrong: If some grass is eaten on any given day, shouldn't the rate of growth of grass be slower the next day?

There are 101 different possibilities from 0 to 100 up sides.
Of these, 50 are cases where there are more up sides (from 51 to100).
The probability then is 50/101 = 0.495 
There is ambiguity to define "line length" when the line has nonzero 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 simpleminded 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.

Dig along the green lines
1 + 1 + √(0,5^{2} + 0,5^{2}) = 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

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 = 0.93So, 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

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; bc = fg
The, b+g = c+f
b+g+c+f = 18
Therefore, b+g = c+f = 9
Also, substract 2 and 3
then ch = he
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:

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 
3 dark stars are enough to block all the light as shown below:
Will that work in three dimensions?
Good point.
Add 2 more dark stars then.
8th Graders olimpyad problem
in New Logic/Math Puzzles
Posted · Edited by bonanova
Added spoiler
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