The worlds first riddle!

Woah, you are answering the problem quicker than I can type. I think I will go for a walk!

In a perfectly circular arena, I walk from the edge directly to the center. I then turn directly to my left, & walk in a straight line to the edge of the arena. I then turn to the right & follow along the edge for a total of 500 meters until I arrive at the point that I started from.

What is the circumference of the inner edge of the arena

[size=7]ARENA
500 is 3/4 of the circumference; so the circumference is 666 2/3 meters.
Perhaps I misunderstood. If the total trip was 500m, then the circumference is ~468.03 meters[/size]

Mark:

FENCE
200x250 (with the 200 bordering Creepy Street)




PODS
1.8%



The answer is 9/500, or 1.8%. The problem can be set up algebraically. For each trip, there is a 100% chance that you will arrive in pod A, B, or C. Let X be the chance that you arrive in pod A, & since it is twice as likely that you will arrive in A than B, there is only half the chance that you will arrive in B than A. So B is 1/2 X. By the same reasoning, there is only 1/6 the chance you would arrive in C than A, so C is 1/6 X. The equation thus looks like this: 100 = X + 1/2X + 1/6X. X=60, so A = 60, B = 30, & C = 10. These numbers correspond to the % chances you would arrive in each pod.

The answer is found by multiplying these three percentages together (i.e. .3 x .1 x .6). There is only a 1.8% chance that you would arrive in these 3 pods in this particular order for your only 3 trips to Exray.



ARENA
500 is 3/4 of the circumference; so the circumference is 666 2/3 meters.


666 2/3 meters. Draw a diagram of the route I took, and then think of it in terms of a pie chart. I walked exactly 3/4 of the way around the arena, so simply multiply the 500 meters I walked by 4/3 to get the answer.

This was really for our Australian friends when they go fo their ?'walkabout'.


TURKEY SANDWICHES
There has been a great deal of interest in how Mark arrived at the total 21.

I can't speak for him, but the long way round is:


The first thing we need to do is figure out how many people were involved. Let's let N represent the number of people at dinner on Thursday. Then, the number of orange rolls eaten on Thursday is 2N, and the number eaten on Friday is N-4 (remember that 4 people left on Thursday night). Since there were 7 orange rolls left Friday night, a total of 2N + (N-4) + 7 = 3N + 3 orange rolls were baked.

There were N pie slices eaten Thursday and 3(N-4) pie slices eaten Friday. There were 4 slices (two-thirds of a pie) remaining on Friday night. So, the total number of pie slices baked must have been
N + 3(N-4) + 4 = 4N - 8.

We can now figure out how many people were at dinner! There were originally equal numbers of orange rolls and pie slices, so:
3N + 3 = 4N - 8 which leads easily to the value N = 11.

So, there were 11 people at the house on Thursday, 7 on Friday, and (since 2 more people leave Friday night) 5 on Saturday. Now we're getting somewhere! Let's let M be the number of turkey sandwiches eaten by each person on Friday. Then a total of 7M sandwiches are eaten Friday, and 5(M-1) are eaten on Saturday. Since Friday's total is 11 greater than Saturday's, we have the equation
7M = 5(M-1) + 11 which leads to the value M = 3.

Now we have the answer: 7 people ate 3 sandwiches each on Friday -- a total of 21 sandwiches!



Talking of 7 and 3…



If 7 web programmers can format 1001 puzzles in 429 minutes, how long does it take 3 web programmers to format those 1001 puzzles



Oh, not 7 and three again>>

If you have 3 dice that are shaped as a tetrahedron, a cube, & an icosahedron, & you rolled each of them, each die would display a number from 1-4, 1-6, & 1-20, respectively. Assume that the tetrahedron & icosahedron are regular. If there is an equal chance that a die will display any of its numbers each time it is rolled.

What percent chance is there that the numbers rolled will total 7 if all three dice are rolled once, & the numbers they display are added together

[size=7]WEB PROGRAMMERS
1 minute per puzzle

DICE
7/240 = 2.91666...%[/size]

I'm off to Chicago until Thursday...

Chicago. Trade shows, conventions and wind. Enjoy

Now that Mark has left for a few days, we can get down to some serious problem solving.

I have six ?'Lego' type eight stud bricks of the same color. How many different ways are there of combining them

That is serious. Too many to enumerate.

Mark: (From outta state replies)


WEB PROGRAMMERS
1 minute per puzzle


1001 minutes. The total amount of work is 7 times 429 programmer-minutes, so 3003 programmer-minutes. So 3 programmers will need 3003/3 = 1001 minutes.



DICE
7/240 = 2.91666...% (…666 now that's exact)

14/480, 7/240, or approximately 2.91%, as there are 14 situations where the dice will show a total of 7 out of 480 possible combinations.


Eight stud bricks.

Too many to enumerate. .....


....Would be the correct answer if, and only if, you thought more than 915 million was ?'too many'

Anyhoo, what are you doing? I note the only convention in Chicago this week is Ladies Fashion.
You're not into ladies wear are you?


Enough already of this banter, I'm hungry>


Andy, Bart, and Chris are eating at Smiley's All You Can Eat Pizza Buffet. Andy made 2.4 times as many trips to the buffet as Bart, and Bart made 6 fewer trips than Chris. What is the smallest possible total number of trips the 3 made to the buffet, assuming that each person made at least one trip


Do not attempt this after a meal.


In training for a competition, you find that swimming downstream (with the current) in a river, you can swim 2 miles in 40 minutes, & upstream (against the current), you can swim 2 miles in 60 minutes.

How long would it take you to swim a mile in still water

Yes - literally :wink:

That ought to bring paulaj out of retirement...

[size=7]PIZZA BUFFET
Bart: 5
Andy: 12
Chris: 11
Total: 28

SWIMMING
24 minutes[/size]

Mark wrote, "That ought to bring paulaj out of retirement..."

Trick or treat? That comment may well send me into retirement.

BTW who is running the west coast, if you are in the east?


Mark:


PIZZA BUFFET
Bart: 5
Andy: 12
Chris: 11
Total: 28




The puzzle can be expressed algebraically as A + B + C = X, where X is the total number of trips. A = 2.4B. Because A has to be a whole number (Andy couldn't make 2.4 trips, for example), you quickly find that 12 and 5 are the lowest possible whole numbers for A and B, which makes C 11.



SWIMMING
24 minutes


You are able to swim downstream at 3 miles an hour, & upstream at 2 miles an hour. There is a difference of 1 mile an hour, which is the river helping you in one direction, & slowing you in the other direction. Average the two rates, & you have the rate that you can swim in still water, which is 2.5 miles an hour. You can thus swim a mile in still water in 24 minutes.



Hot news

All of the delegates at a convention in Chicago are majoring in cross dressing, business, or both. 73% of the delegates are cross dressing majors, & 62% are business majors. If there are 200 delegates, how many of them are majoring in both cross dressing & business




Come on people, you can do this…

12 members were present at a board meeting. Each member shook hands with all of the other members before & after the meeting. How many handshakes were there

70 & 156

Rap

Rap, I have a lower number for the meeting. Could be I messed up due to a cranial impediment, but…


I'm set on history-

To move their armies, the Romans built over 50,000 miles of roads. Imagine driving all those miles! Now imagine driving those miles in the first gasoline-driven car that has only three wheels and could reach a top speed of about 10 miles per hour.

For safety's sake, let's bring along a spare tire. As you drive the 50,000 miles, you rotate the spare with the other tires so that all four tires get the same amount of wear.

Can you figure out how many miles of wear each tire accumulates

Could it be 132? 37500 miles

Rap

It sure could

146+124-200 = 70
C(12,2)*2=132
50000*3/4 = 37500

Rap:

70


If 73% of the delegates are cross dressing majors, we know that 27% are not cross dressing majors. By the same reasoning, 38% are not business majors, because 62% of the delegates do major in business. So:
27 + 38 = 65

65% of the delegates are not majoring in both cross dressing & business, so 35% are double majors, a total of 70 delegates.


132

Think of it this way: the first person shakes hands with 11 people, the second person also shakes hands with 11 people, but you only count 10, because the hand shake with the first person was already counted. Then add 9 for the third person, 8 for the fourth, & so on. 66 hand shakes took place before & after the meeting, for a total of 132.


37500 miles


Since the four wheels of the three-wheeled car share the journey equally, simply take three-fourths of the total distance (50,000 miles) and you'll get 37,500 miles for each tire.


Mark: (a.k.a. Tonto)

146+124-200 = 70

C(12,2)*2=132

50000*3/4 = 37500



Report from the conference:

A man (no name, but if you think ?'local' you will not be too wide of the mark.) boards an aeroplane and takes his seat. As he settles in, he glances up and sees a most beautiful woman boarding the plane. He soon realises she is heading straight towards his seat. A wave of nervous anticipation washes over him. Lo and behold, she takes the seat right beside his. Eager to strike up a conversation, he blurts out, "Business trip or vacation?".

"Nymphomaniac Convention in Chicago," she states.

Whoa!!! He swallows hard and is instantly crazed with excitement. Here's the most gorgeous woman he has ever seen, sitting RIGHT next to him and she's going to a meeting of nymphomaniacs! Struggling to maintain his outward cool, he calmly asks, "What's your business role at this convention?"

"Lecturer", she says. "I use my experiences to debunk some of the popular myths about sexuality."
"Really," he says, swallowing hard, "what myths are those?"

"Well," she explains, "one popular myth is that African American men are the most well-endowed when, in fact, it is the Native American Indian who is most likely to possess that trait. Another popular myth is that French men are the best lovers, when actually it is men of Greek descent."

Suddenly, the woman becomes very embarrassed and blushes. "I'm sorry" she says, "I shouldn't be discussing this with you, I don't even know your name!"

"Tonto," he says, as he extends his hand. "Tonto Papadopoulos."

Have a good flight!



A palindrome is a word, number, or phrase that reads the same forwards or backwards. For example, the phrases "Madam, I'm Adam" and "Poor Dan is in a droop" are palindromes. The numbers 743347 and 828 are also both palindromes.

This challenge (should you be wearing your own undergarments) is to count how many numbers between zero and ten million are palindromes

The first ten are 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9 (any one digit number reads the same forwards or backwards). But 10 is not a palindrome (even though you could write it as 010, we don't want to allow zero as a leading digit).

Try, you're psychic!!! That's exactly the way it happened. And she was quite the maniac. Neither of us made it to our conventions, but we did dress up in each other's clothes.

[size=7]PALINDROMES FROM 0 TO 9999999
Consider the seven-digit number, ABCDEFG, where leading zeros are allowed. All palindromes are determined by the values of A, B, C, and D (E=C, F=B, and G=A). All possible values for A, B, C, and D lead to palindromes (some with leading zeros). The palindromes with leading zeros can be converted to shorter palindromes by removing the leading (and trailing) zeros. This will generate all palindromes of odd length up to seven.

Even length palindromes can be generated the same way with the six-digit number, ABCDEF. Here, A, B, and C determine the palindrome.

There are 9,999 possible odd length palindromes and 999 even length palindromes, not counting zero. Therefore, there are 10,999 palindromes less than 10,000,000.[/size]

Mark:

PALINDROMES FROM 0 TO 9999999

Consider the seven-digit number, ABCDEFG, where leading zeros are allowed. All palindromes are determined by the values of A, B, C, and D (E=C, F=B, and G=A). All possible values for A, B, C, and D lead to palindromes (some with leading zeros). The palindromes with leading zeros can be converted to shorter palindromes by removing the leading (and trailing) zeros. This will generate all palindromes of odd length up to seven.

Even length palindromes can be generated the same way with the six-digit number, ABCDEF. Here, A, B, and C determine the palindrome.


Poor Mark, he must be suffering from jet lag. Although his logic cannot be questioned, what the hell is he talking about? A,b,c diddle d. I ask; can anyone translate? I did not spend three weeks learning English just to translate Chicago speak.

What he meant to say was:

There are 10,999 palindromes between zero and ten million. It's best to count them according to the number of digits they have.
• There are 10 one-digit palindromes.
• There are 9 two-digit palindromes (11, 22, 33, . . . , 99).
• There are 90 three-digit palindromes, since there are 9 choices for the outside digits (we can't use zero there!) and 10 choices for the inside digit.
• There are 90 four-digit palindromes, since again there are 9 choices for the outside digits, and 10 choices for the inside digits. Note that both of the two inside digits must be the same.
• There are 900 five-digit palindromes, since there are 9 choices for the outside digits, 10 choices for the next-to-outside digits, and 10 choices for the middle digit.
• There are 900 six-digit palindromes, since there are 9 choices for the outside digits, 10 choices for the next-to-outside digits, and 10 choices for the two middle digits (which must both be the same).
• There are 9000 seven-digit palindromes for reasons similar to those described above.
The only eight-digit number to consider is 10,000,000, which is not a palindrome. So, the number of palindromes between 0 and 10,000,000 is
10 + 9 + 90 + 90 + 900 + 900 + 9,000 = 10,999

What could be easier?


It was at this point that I found I had missed Mark's last line, which read,


"There are 9,999 possible odd length palindromes and 999 even length palindromes, not counting zero. Therefore, there are 10,999 palindromes less than 10,000,000."


On which we all agree.


You are correct, my Dr says I am a psycho. However, "And she was quite the maniac. Neither of us made it to our conventions, but we did dress up in each other's clothes."

Whoa! That's way too much information for me to handle. Although, I suspect you managed ok.


At this point I would like to thank Rap for standing in at very short notice, in an effort to keep this train wreck going.




Having taken care of the Ugly, the Good and the Bad are dividing a bounty of 2000 silver dollars as follows. First, the Good divides the coins into two piles with at least two coins in each pile. Then the Bad divides each of those two piles into two (non-empty) piles and takes the largest and the smallest of the four piles, leaving the other two piles for the Good.

What is the maximum share the Good can count on no matter how smart - or greedy - the Bad is



a) Fred Flintstone and Barney Rubble in turn take pebbles from a pile of 2000 pebbles. They can take 1, 7 or 13 pebbles at a time. The player who takes the last pebble wins.

Find a strategy that allows Fred or Barney to win regardless of how the other one may play
Oh, Fred goes first.

(b) What is the winning strategy if the players are allowed to take 1, 2, 7 or 13 pebbles at a time