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yeah, I know those.
and why didn’t it work?
~a~
you didn’t do it right – go to the HTML formatting thread
must be the way his brain works, from the bottom 🙂
I always fell for those.
it’s like the one the teachers give out on the first day of school that say:
read all the directions first, then go back and do them.
1) draw a circle in the upper left corner of your page.
2) clap your hand four times.
…all random things till you get to the bottom one that says…
25) sit quietly with your hands on your desk and do not go back.
I was so mistrusting of teachers, that whatever they did I was always trying to figure out what’s the catch? I NEVER fell for these.
~a~
Suppose you have a barrel of wine that appears to be about half full. It can’t be exactly half full because nothing is exactly precise. How can you tell if it is slightly more or slightly less than half full, without using any measuring implements or removing any wine from the barrel?
squeak: Put someting directly down the middle splitting the barrel in two sides, and put all the wine on one side. If it is takes up more than one side it is more than half full if it takes up less than one side it is less than half full.
aussieboy – that will not work – when you put something in the barrel, it will displace some of the liquid making it appear as if it is more than 1/2 full.
squeak, tilt the barrel.
~a~
squeak-
Tilt the barrel until it is almost about to spill. Now look into the barrel, if you can see the bottom it is less than half full.
Mod72: I didnt think of that, but technically you can account for the amount that should be displaced.
doc, for that to work you’d have to have the barrel completely horizontal.
moish01: No. it would need to be on a 45 degree tilt making 2 triangles. and if you can see the bottom it means that there is less than half of the barrel full.
oh i read what he said wrong. yeah now it makes sense. (i thought he said “if you could see half of the bottom”)
Which is the longest suspension bridge? (not sure if its the longest in the world or just the US so ill stick with longest in the US.)
just googled it because i didn’t know. it’s the longest over water, but there’s a longer bridge in Thailand that’s over land.
mosih: Oh cool, but the one im talking about is usually confused with another more famous bridge which is the highest, but not the longest suspension bridge.
i know. i told you i googled it 😉
aussie: the Golden Gate Bridge in CA? just a guess!
Jax: Yes. That is what people usually guess.
The longest in the U.S. is the Verrazano Narrows Bridge at 4260 feet (measured by length of main span or distance between towers). The Golden Gate Bridge has a main span of 4200 feet. The Verrazano Narrows Bridge lost the title of longest suspension bridge in the world in 1981 but remains the longest suspension bridge in the U.S.
I posted a riddle some time ago as to why the towers are further apart at the top than at the base. Anyone remember that?
aussieboy
I think you may find this interesting. (From your screen name I assume you are Australian.) When construction on the Sydney Harbor Bridge began it was supposed to be the longest arch bridge in the world. The Bayonne Bridge, on which construction was started later on and finished earlier is longer by about 2 feet. The same gold scissors was used for the ribbon cutting ceremonies on both bridges. After the ribbon cutting in Australia the scissors were taken apart, half was kept in Australia and the other half was sent to the U.S.
Dr. Pepper, do you memorize these random facts?
Dr. Pepper: Actually I am not from Australia, but it is still intresting.
Good job, anonymisss, with the correct answer. Dr. Pepper, you too, except that I feel like you are the teacher’s desk copy edition of this thread. 😉 I’m going to try to find one to stump you one day.
*kapusta*
When I was studying civil engineering one of the textbooks would throw in some interesting facts. I don’t recall the vast majority of those facts except for a few that pertain to bridges (and tunnels) in and around New York.
(By the way- I did not graduate as an engineer.)
hey doc, i was actually going to ask if you were an engineer. between the math and physics and now this trivia…
There is one correct solution for a Rubik’s Cube.
How many other possible ways are there to arrange a Rubik’s Cube. (Without taking it apart or removing stickers… Also I’m referring to a classic Rubik’s Cube- a different color on each face with no logo on any of the tiles.)
Dr. Pepper, as long as you’re not this factual in real life. 🙂
ames: Google away just please dont post it on here unless you come up with the answer yourself.
Dr. Pepper: 531441 or 362880 or 2177280. any of those the correct answer?
I doubt any of those are the correct answer but I got thise answers by:
For 531441: 9 to the 6th power
For 362880: 9*8*7…*3*2*1
For 2177280: (9*8*7…*3*2*1)*6
Just thought I would show how I got the wrong answers.
ames-
Consider this a take home test- you may use any resources you desire but your answer must contain a full explanation or you will receive no credit.
kapusta-
Not it my personal life, but if I wasn’t like this in my professional life my colleagues would have a severe advantage over me.
aussieboy-
Nice try but you’re trying to solve it from the wrong dimension.
Here are two hints:
1. The correct answer is over 10,000,000 and
2. The last three digits of the correct answer are the same.
Good Luck Everyone
i was just going through old riddles and by the one of orange kangaroos
did anyone besides me come up with red otter in the democratic republic of the congo
Dr. Pepper-
The Rubik’s cube puzzle will take time and smarts, neither of which I have right now.
Here are the facts I think need to be factored into getting the answer:
1) Corner pieces. There are eight corner pieces, each of which can be on any corner, independent of the other seven corner pieces. Each piece can have three different positions on each corner.
2) Center pieces. There are six center pieces. They are not completely independent of each other, since two opposing colors will always oppose each other.
3) Side pieces. There are twelve side pieces, each of which can be on any side, independent of the other eleven side pieces. Each piece can have two different positions on each side.
What has to be calculated is corner-piece-combos * center-piece-combos * side-piece-combos.
Calculating by colored squares won’t work since all colors are locked immovably to one or two other colors, except for the center pieces.
I can only try
You’re off to a good start but let me make some corrections.
The center piece is one piece and remains fixed. On a “classic Rubik’s Cube” the orientation of the center faces won’t make a difference. (Had there been any writing or a logo then the orientation would have to be taken into account since the writing (or drawing) must be face up.) Therefore the center piece can be ignored for this problem.
The other mistake is that you are not taking into account the possibility of losing a degree of freedom. If one were to disassemble a Rubik’s Cube and put the pieces together at random- chances are that the resulting cube will have no solution. When assembling the cube most pieces may be placed in randomly but there will be a few that are forced into a specific location or position- hence the loss of a degree of freedom. As a hint: there will be one piece forced into a certain location and two pieces forced in a specific orientation.
Solution to Rubik’s Cube Riddle.
There are 8 corner pieces each one can be in any corner => 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 possibilities.
The first 7 corner pieces can be oriented in three different directions but the last one is fixed => 3 * 3 * 3 * 3 * 3 * 3 * 3 * 1 possibilities.
There are 12 side pieces the first 10 can be anywhere but the last two are fixed => 12 * 11 * 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 1 * 1.
The first 11 sides can be oriented in any direction but the last one is fixed => 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 1.
The centers are all connected and direction does not matter since the faces are a solid color (recall that we are talking about a classic Rubik’s Cube with no logo on any face) => one possible way.
(8 * 7 * 6 * 5 * 4 * 3 * 2 * 1) *
(3 * 3 * 3 * 3 * 3 * 3 * 3 * 1) *
(12 * 11 * 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 1 * 1) *
(2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 1) *
1
=
8! * (3^7) * (12!/2) * (2^11) * 1
=
43,252,003,274,489,856,000 possible configurations, of which one is correct and
43,252,003,274,489,855,999 are incorrect.
As you can see the number is greater than 10,000,000 and the last three digits are the same.
(The number is 43 quintillion, 252 quadrillion, 3 billion, 274 million, 855 thousand, 999)
What does this sign “^” mean?
aussieboy-
raised to the power of
3^7 = 3 * 3 * 3 * 3 * 3 * 3 * 3 = 2,187
Oh. So I was on the right track I just didnt take in to account that certain pieces could only go in certain positions.
If A + B = C then A + C =?
A + C = 2C – B
Two numbers multiply to 1,000,000. Neither number contains the digit 0.
What are the two numbers?
You mean that the 0 before the decimal point doesn’t count?
Dr Pepper, i used a calculator, but can it be a decimal with a repeating bar?
moish01-
There is a solution using two integers.
Using the second part of this post http://www.theyeshivaworld.com/coffeeroom/topic/the-riddle-thread/page/8#post-16043 you should be able to do it in your head.
Good luck.
if that was the case, moish, there would be tons of answers. I tried it too.
Dr. Pepper: By the way i completly did that randomly because I was bored and I wanted to see if someone would come up with an answer…and you did.
As for your riddle is there more than one answer?
ha i was just kidding – i knew it couldn’t be. there’s no way i’m even reading that other post to even try and figure it out. (and nothing gets figured out in MY head, by the way.) i read words, not numbers.
aussieboy- Is that the answer you had in mind?
There is only one integer solution to my riddle.
Dr. Pepper-
It has to be an even number multiplied by a number ending in a 5.
That would be 15,625 * 64
Solved by dividing 4, 8, 16, 32 and 64 into 1,000,000
I didn’t know that was the only solution.
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