Rob_Gandy

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Posts posted by Rob_Gandy


a=9 b=0 c=9 d= infinity and any similar variation.
I also get a=1 b=9 c=9 d=10.47368.....
with the help of JavaI get
a = 9
b = 1
c = 9
100*a + 10*b +1*c = M = 881
M/(a+b+c) = 881
a b and c are digits not integers, good idea though. If they were integers there would still be smaller answer then what you have found here. can you find that answer too?

the vendor paid $1.00 each for the hammers and so he makes $.50 profit in either way.

Oh this is how it starts, lightning strikes the heart. It goes off like a gun, brighter than the sun. Oh we could beat the starts, faliing from the sky. Shining how we want, brighter than the sun.  Brighter Than the Sun.

They were placed in the shape of a tetrahedron.
 1

You should open up the bottom of the box because the balls with a diameter of 4in would fall through any gap if the balls are stacked directly on top of each other and you should open the top if the balls are stacked in alternating slotted rows.

For the max number of slices
2n  2 n = 2 or 3
2n  2 + SUM_{}(I) + SUM_{}(I) + ... + SUM_{}(I) n > 3
Examples:
4 points gives a max of 2(4)  2 + SUM_{}(I) = 6 + 2 = 8
5 points gives a max of 2(5)  2 + SUM_{}(I) + SUM_{}(I) = 8 + 5 + 3 = 16.
6 points gives a max of 2(6)  2 + SUM_{}(I) + SUM_{}(I) + SUM_{}(I) = 10 + 9 + 7 + 4 = 30
Maybe?

15 degrees?

the answer is approx (if not exactly) 75 degrees.

28655 bees. After the 5th generation I realized what was going on.

you made the detector a square along the diagonal of the cube?
That is 1 x 1 m only because the side is 1 m, there is a larger square that will fit in the cube.
The detector would be sqrt(2) x sqrt(2)
Oops ignore that...

you made the detector a square along the diagonal of the cube?
That is 1 x 1 m only because the side is 1 m, there is a larger square that will fit in the cube.
The detector would be sqrt(2) x sqrt(2)

you made the detector a square along the diagonal of the cube?

I'm on the edge of glory and I'm hanging on a moment of truth. Out on the edge of glory and I'm hanging on a moment with you. I'm on the edge, the edge, the edge, the edge, the edge, the edge, the edge, I'm on the edge of glory and I'm hanging on a moment with you. I'm on the edge with you. Lady Gaga (The Edge of Glory [14])

I'm not sure how to prove it, but I think whenever possible she should choose the coin with the largest value of all that exist. If that isn't possible then she should look at the last two coins on each end and pick her coin so that Bob's choices are equal to or less than the coin she picks. Or pick such that Bob's choice are minimized.

My method is similar to James's. Flip the coin once for each person and assign them heads or tails. All who match the next coin flip (after everyone is assigned) stay in for another round. Continue until one person is left. If everyone is assigned the same thing then reflip and reassign everyone. Or have each person choose heads or tails then flip the coin and everyone who chose what the coin lands on gets to stay. Repeat until finished.

My first thought when I read this was about areas. Given a triangle on a unit circle the probablility of a 4th point (dart) being in the triangle is (the area of the triangle)/Pi. The average area of a triangle in a unit circle is 35/(48Pi). So wouldn't that make the probablility of the 4th dart being in the triangle 35/(48Pi^{2}) or approx. .07388? This is how I see this problem working in my mind.
If we know the average area of a randomly drawn triangle inside a unit circle, then the problem is solved. How do you find the average area of triangle? Where does 35/48Pi come from?

My first thought when I read this was about areas. Given a triangle on a unit circle the probablility of a 4th point (dart) being in the triangle is (the area of the triangle)/Pi. The average area of a triangle in a unit circle is 35/(48Pi). So wouldn't that make the probablility of the 4th dart being in the triangle 35/(48Pi^{2}) or approx. .07388? This is how I see this problem working in my mind.

R ~ 254.84 or sqrt{2([58Pi]
^{2}27^{2})} 
Assuming that the info from the Cheshire cat is true, we will know which cake is in the middle. So I would ask Dee "How would Dum answer the question 'Is the name of the queen's favorite cake alphabetically before the queen's least favorite cake?'" The opposite would be true. Then depending on the info I get from the cat I can decide on the correct order to eat the cake.

six nine fourty
 nine ten fifty

s
ixninefourtysx i our=six letters?

He built a fence around himself and said he was outside the fence.

The inside of your eyelid?

I'm not quite sure how to start on this one, but from drawing I think that the third road is approx. 4.4 miles long.

2.5 inches.
Drawing the square root of 3
in New Logic/Math Puzzles
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