The worlds first riddle!

[size=7]GOOD VS. BAD
Good will get exactly 1000 if he splits the 2000 evenly. Bad is going to get all of the largest pile that Good leaves.

a) Barney wins by taking 14-F each turn, where F is what Fred took on his previous turn. They will end up with 12 pebbles left and it will be Fred's turn. Barney will win after 6 more turns (one 7 and five 1's in any order).

b) This will require some thought.[/size]

[size=7](b) Fred wins by starting with 2. After that, if Barney takes 1, 7 or 13 (3n+1), Fred takes 2. If Barney takes 2, Fred takes 1, 7 or 13. Fred always leaves 3n pebbles. Since 0 is a multiple of 3, Fred wins.[/size]

Mark:
GOOD VS. BAD
Good will get exactly 1000 if he splits the 2000 evenly. Bad is going to get all of the largest pile that Good leaves. v

Then no matter how the Bad subdivides the two 1,000-piles, the largest and the smallest of the 4-piles would come from the same 1,000-pile (for the compliment of the largest pile is the smallest one!), and thus add up to 1,000.

If the Good divides the 2000 silver dollars into unequal piles that contain m and n coins respectively, m > n, then the Bad can subdivide m-pile into m-1 and 1, and thus the largest pile will be m-1 >_ 1,000.
After adding the smallest pile the Bad would positively have more than 1,000 coins.


a) Barney wins by taking 14-F each turn, where F is what Fred took on his previous turn. They will end up with 12 pebbles left and it will be Fred's turn. Barney will win after 6 more turns (one 7 and five 1's in any order). Nice


Since 2000 is even and 1, 7 and 13 are odd, every Fred's move would leave an odd number of pebbles in the pile. The winner, after the winning move, will leave 0 pebbles, and 0 is not an odd number. Therefore, Fred cannot win. Since somebody will win (as long as the game is not over, a move of taking 1 pebble is always available, and with every move the number of pebbles in the pile goes down), it must be Barney. Barney wins no matter how he plays.




Some thought later…

(b) Fred wins by starting with 2. After that, if Barney takes 1, 7 or 13 (3n+1), Fred takes 2. If Barney takes 2, Fred takes 1, 7 or 13. Fred always leaves 3n pebbles. Since 0 is a multiple of 3, Fred wins.


Clearer than my write-up.




Suppose I start writing the numbers from 1 on up, one after another as follows:

12345678910111213141516171819202122232425262728293031323334...
If I had an infinite amount of time and patience (I in fact have neither!) I could keep this string of digits going forever.

There are two parts to this puzzle:

1. What is the two-millionth digit in this string of numbers

2. What six-digit number am I writing when I write the two-millionth digit


If those numbers are too big for you try:

Suppose you have a list of numbers from zero to one hundred. The sum of these numbers is the code to release you from the lair of the evil troll. He is coming back in two minutes to bite your head off.

How quickly can you add them all up without using a calculator


No, not the time, the method…I hear the troll approaching

[size=7]STRING OF DIGITS
1. the second 5 in the number below
2. 351851

TROLL
Drop the zero and compute the sum of 1-100, which is 100*101/2 = 50*101 = 5050.[/size]

Take an ordinary piece of typing paper; its dimensions will be 8 1/2 inches by 11 inches.

Label the four corners A, B, C, and D clockwise from the top left (so that the short sides of the paper are AB and CD ).

Label the midpoint of AB as E and the midpoint of BC as F. Now make straight line folds in the paper from A to F and from D to E.

Label the point at which these folds cross as G.

Now, what is the exact area of the four-sided region GFCD. If viewed from the southern hemisphere

[size=7]FOLDING PAPER
The exact area is 37112/935 square inches.
This is approximately equal to 39.692 square inches.

And it was a pain in the ass to work in fractions to get the "exact" area.[/size]

How quickly can you find out what is so unusual about this paragraph? It looks so ordinary that you would think that nothing is wrong with it at all, and in fact nothing is. But it is unusual. Why? If you study it and think about it, you may find out, but i am not going to assist you in any way. You must do it without coaching. No doubt, if you work at it for long it will dawn on you. Who knows? Go to work and try your skill. Par is about half an hour.

"And it was a pain in the ass to work in fractions to get the "exact" area."

An even bigger pain is I have:
The area is 51 and 17/40, or 51.425.

On the lighter side…

Solve:

BACK
CK
K

Solve:

Returns
Return
Retur
Retu
Ret
Re
R



I have six coins in my pocket that total $1.15.

I cannot make change for a dollar, fifty cents, twenty-five cents, a dime or a nickel. What coins do I have in my pocket

The whole sheet is 93.5 square inches. The region in question is clearly less than half the sheet.

That's not difficult at all. In fact, it is quit* *asy.

[size=7]BACKS
fullback
halfback
quarterback

RETURNS
diminishing returns

COINS
1 half dollar, 1 quarter, 4 dimes[/size]

Mark wrote:
"The whole sheet is 93.5 square inches. The region in question is clearly less than half the sheet."

That's what makes it so painful.

Sorry dadpad, "Go to work and try your skill. Par is about half an hour."

Two minutes, I ask you, two minutes to get the answer up. The guy is good, very good.

However, I can say with a degree of certainty, this conundrum will slow things down.


A boy, a girl and a dingo go for a 10 mile walk. The boy and girl can walk at 2 mph and the dingo can trot at 4 mph. They also have a bicycle that only one of them (including the dingo!) can use at a time. When riding, the boy and girl can travel at 12 mph while the dingo can pedal at 16 mph.

What is the shortest time in which all three can complete the journey

Too good markr wll don

[size=7]BOY, GIRL, DINGO, & BIKE
2.75 hours (with the clever dingo traveling 11.6 miles)
That was tricky![/size]

well dingos dont ride bicycles and he would've got hungry along the way so i figure............................. 2.5 hrs.



what ..........is the air-speed velocity of an unladen dingo?

Mark:

STRING OF DIGITS
1. the second 5 in the number below
2. 351851



The two-millionth digit in the string of numbers is 5, namely the fifth digit of the six-digit number 351,851. It's just a matter of good bookkeeping to get to that answer.
We'll just start counting digits:
• There are 9 digits used on one-digit numbers.
• There are 90 two-digit numbers (10 to 99), so a total of 180 digits used there. This gives a running total of 189.
• There are 900 three-digit numbers (100 to 999), giving a total of 2700 digits used there. The running total is now 2,889.
• There are 9,000 four-digit numbers (1,000 to 9,999) accounting for 36,000 digits. Running total: 38,889.
• There are 90,000 five-digit numbers (10,000 to 99,999) adding 450,000 more digits and bringing the running total to 488,889. (Cool pattern, huh?)
• We don't want to look at all six-digit numbers right off the bat, so lets just look at the first 100,000 of them, namely the numbers from 100,000 to 199,999. These add 600,000 digits and bring the total to 1,088,889.
• The next 100,000 numbers (200,000 to 299,999) add 600,000 more digits and bring the running total to 1,688,889.
Now we're closing in on two-million digits! In fact, we obviously don't have room to add another 100,000 six-digit numbers since we have only 2,000,000 - 1,688,889 = 311,111 digits left. Dividing 311,111 by six, we see that there is room for only 51,851 more six-digit numbers. These would be the numbers from 300,000 to 351,850, which add 311,106 digits and bring the running total to 1,999,995 digits. So, the fifth digit of the next number (351,851) will be the two-millionth digit in the string

TROLL
Drop the zero and compute the sum of 1-100, which is 100*101/2 = 50*101 = 5050.

Fifty pairs of numbers sum to 100 (100+0, 99+1, 98+2, etc.) 50 is unpaired:
50 X 100 + 50 = 5,050


FOLDING PAPER
The exact area is 37112/935 square inches.
This is approximately equal to 39.692 square inches.

As regular readers would have noticed by now, I am a dork. My answer to this was so wrong it was the equivalent of thinking the north pole is the centre of the universe. Maybe one day I will be proved right, but not in this lifetime.


BACKS
fullback
halfback
quarterback

RETURNS
diminishing returns

COINS
1 half dollar, 1 quarter, 4 dimes

BOY, GIRL, DINGO, & BIKE
2.75 hours (with the clever dingo traveling 11.6 miles) .
That was tricky!

Dagnabbit, the dude has done for me again. Still, I draw some consolation from the fact it took him more than two minutes.

First note that there's no apparent way to benefit from letting either the boy or girl ride the bike longer than the other. Any solution which gets the boy there faster, must involve him using the bike (forward) more; similarly for the girl. Thus, the bike must go backwards more for it to remain within the 10-mile route. Thus, the dingo won't make it there in time. Therefore, the solution assumes they ride the bike for the same amount of time.

Also, note that there's no apparent way to benefit from letting any of the three arrive at the finish ahead of the others. If they do, they can probably take time out to help the others. Therefore, the solution assumes they all finish at the same time. This is one way:

The boy starts on the bike, and travels 5.4 miles. At this point, he drops the bike and completes the rest of the trip on foot. The dingo eventually reaches the bike, and takes it *backward* .8 miles (so the girl gets to it sooner) and then returns to trotting. Finally, the girl makes it to the bike and rides it to the end. The answer is 2.75 hours.

dadpad:

"dingos dont ride bicycles"

Get real dude! Even fish can.

"2.5" Thinking along the right lines.

"what ..........is the air-speed velocity of an unladen dingo?"

What! Aint you got no learnin. I thought everybody new. :wink:

Three shall be the number thou shalt count Airspeed can also be predicted using a published formula. By inverting this midpoint Strouhal ratio of 0.3 (fA/U ≈ 0.3), Graham K. Taylor et al. show that as a rule of thumb, the speed of a speeding animal is roughly 3 times frequency times amplitude (U ≈ 3fA).5

We now need only plug in the numbers:
U ≈ 3fA
f ≈ 15 (beats per second)
A ≈ 0.22 (meters per beat)
U ≈ 3*15*0.22 ≈ 9.9
... to estimate that the airspeed velocity of an unladen Australian Dingo is 10 meters per second.

Oh, yeah, I agree with that. With some further study, it became clear that these estimates are accurate, though perhaps coincidental.

An actual study of two Dingo's in a low-turbulence wind tunnel in Lund, Sweden, shows that speed much slower than my estimate, at only 7-9 beats per second:

"Compared with other species of similar size, the Dingo has quite low heartbeat frequency and relatively long legs."
The maximum speed the Dingo could maintain was 13-14 meters per second, and although the Lund study does not discuss cruising flight in particular, the most efficient airspeed in the range of 8-11 meters per second, with an amplitude of 90-100° (17-19 cm).

And there was much rejoicing. Averaging the above numbers and plugging them in to the Strouhal equation for cruising flight (fA/U = 7 beats per second * 0.18 meters per beat / 9.5 meters per second) yields a Strouhal number of roughly 0.13:

Enjoy.


I enjoyed Whim's problem, but guessed the answer. Therefore, the same theme, but - with a calculated answer.

In a meadow where the grass grows with the same density and speed throughout the entire area, it is known that 70 cows would eat it away in 24 days, and 30 cows in 60 days


Sorry I'm outta time.

And the question is...

Mark wrote, "And the question is..."

Now you are just getting picky. It is as plain as coal in a mine at night. The question is, Where did the original question go? Was it removed by the Mods, due to its explosive nature?

Well to hell with it, throwing caution to the wind:

How many cows would eat the entire grass in 96 days


Chew that over

[size=7]COWS
20[/size]

An actual study of two Dingo's in a low-turbulence wind tunnel in Lund, Sweden, shows that speed much slower than my estimate, at only 7-9 beats per second:

Alpine or desert dingo?

Ha det bra, hoppas allt är bra.

cows-
Depends on the type of grass dont it? Joycea pallida grows slower than phalaris or vic rye so would take less time to eat and danthonia lacks protien so the cows actaully need to increase intake to maintan body weight. Also the reletive fertility and ph of the soil should be taken into consideration somewhere along the line.