The worlds first riddle!

Drop in the ocean
A stitch in time plus nine
Tip top

Shari wrote:

Drop in the ocean
A stitch in time plus nine (Saves nine, but I didn't know how to show ?'saves')
Tip top (Top tip) same difference.




HLOIVPEE


Custer's
STAND
STAND
STAND<



Cat'

Custers last stand
Love in Hipe???
Cat apostrophy?

[size=8]QUADRATIC EQUATION
6, 1

live in hope
catastrophe (bit of a stretch, don't ya think?)[/size]

Shari:

Custers last stand

Cat apostrophy (Catastrophe)


Mark:


QUADRATIC EQUATION

6, 1

Whoops, I have: The sum of the roots is 7 and their product is 6. Did Shari help you with that one?


live in hope
catastrophe
"(bit of a stretch, don't ya think?)"

Yes, yes it was, and I will forever carry the pain with me.
Are you having…




CRILI FESIS


SLIIVINGN





Shari has a cube of dimensions 10cm x 10cm x 10cm it is made up of smaller cubes of dimensions 1cm x 1cm x 1cm. If the outermost layer of smaller cubes are taken off, how many smaller cubes would be left

[size=8]mid-life crisis
living in sin[/size]

SHARI'S BOOBS, I MEAN CUBES
[size=8]8x8x8 = 512[/size]

Mark:


SHARI'S BOOBS, (what exactly were you measuring?) I MEAN CUBES
8x8x8 = 512



totalcubes=10*10*10=1000
innercubes=8*8*8=512
outercubes=totalcubes-innercubes=488


mid-life crisis
living in sin

Whichever of the above I don't care, just setup this wet T shirt competition why don't'ya.





An officer wishing to arrange his men in a solid square found by his first arrangement that he had 39 men left over. He then started increasing the number of men on a side by one, but found that 50 additional men would be needed to complete a new square.

How many men did the officer have



Costnothdiveing


SSEERLVINFG AREA

Done. You missed it (didn't know where to send the invitation). Shari won. It wasn't even close. I didn't get any pictures, but I've got memories to last a lifetime...

[size=8]SQUARE MEN
1975[/size]

MMMAAAWWWWWWWAHH!

Oh cruel fate, how you do mock me so. I feel I am but a pawn in a bigger game.


Two squares are chosen at random on a chessboard. What is the probability that they have a side in common

Yes, yes....I took home the twin trophies. They were bronzed! lol

"They were bronzed!"

Is that because you have been out in the sun?

No sun here sweet pea! I should go tanning though. I am thinking about getting a massage for my birthday.

"I am thinking about getting a massage for my birthday."

There is no mail delivery on a Sunday, so no messages, silly.

Cheese ball!

You are sooo sweet. In my neck of the woods, that is such a term of endearment, and hardly ever heard.

I got the message.

trlanoslastiton

"trlanoslastiton"

Re-arrange the word to form a well known phrase…

?'A LOST NOSTRIL TAN'. Well, I hope you find it soon.

Or, do you mean, 'SANTA I STROLL NOT' - walking is good for the heart.

[size=8]CHESSBOARD SQUARES
This turned out to be much easier than I first thought.

Let NxN be the dimensions of the chessboard (8 for this problem).
The number of ways to select two squares is C(N*N,2).
There are N*(N-1) ways to select pairs that share a vertical side and N*(N-1) ways to select pairs that share a horizontal side.

Therefore the probability that the selected squares have a side in common is:
2*N*(N-1) / C(N*N,2)

For N=8, the answer is 112/2016 = .0555...[/size]

Mark:

SQUARE MEN
1975

The officer did have 1975 men. When he formed a square measuring 44 by 44, he had 39 men over. When he tried to form a square 45 x 45, he was 50 men short.


CHESSBOARD SQUARES
This turned out to be much easier than I first thought.

?'Oh boy, as soon as I read that I knew there was going to be trouble.'

Let NxN be the dimensions of the chessboard (8 for this problem).
The number of ways to select two squares is C(N*N,2).
There are N*(N-1) ways to select pairs that share a vertical side and N*(N-1) ways to select pairs that share a horizontal side.

Therefore the probability that the selected squares have a side in common is:
2*N*(N-1) / C(N*N,2)

For N=8, the answer is 112/2016 = .0555...

?'It would appear that we differ by a margin of 50%'

The number of ways of choosing the first square is 64. The number of ways of choosing the second square is 63. There are a total of 64 * 63 = 4032 ways of choosing two squares.

If the first square happens to be any of the four corner ones, the second square can be chosen in 2 ways. If the first square happens to be any of the 24 squares on the side of the chess board, the second square can be chosen in 3 ways. If the first square happens to be any of the 36 remaining squares, the second square can be chosen in 4 ways. Hence the desired number of combinations = (4 * 2) + (24 * 3) + (36 * 4) = 224.

Therefore, the required probability = 224/4032 = 1/18. Check mate I think. No, I hope.



A square, whose side is 2 meters, has its corners cut away so as to form an octagon with all sides equal.

What then is the length of each side of the octagon, in meters



STAND?
Hard to


YOU. CLEAR UP