| Author | Maths Problems |
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Great questions till now, keep them coming :)
Player banned by moderator Lord STB until 2013-11-12 13:27:08 // FR#17//Add. characters aren't allowed to post in main forums. |
| Sorry for post with multi. |
for SunnyDaisy:
Bow down to our queen SAMA!!! Queen sama always be the best from brow to heels!!! |
try my problem if you are good at probability theory. for anyone solved it analytically i give 3 promises (if not offended with game rules)
Let a=number of trinkets of first type, b=no. of trinkets of second type, c=no. of trinkets of third type, d=no. of trinkets of fourth type.
(So a+b+c+d = 28)
The chances that you get it right from the first time is:
(a!b!c!d!) / (a+b+c+d)! * (1/4)^28
This means the expected number of fights needed to complete the vault is:
(a+b+c+d)! / (a!b!c!d!) * 4^28.
Right? |
for ParaLeul:
i'm sorry it is far from right. |
| ^^Hmm, I gave it a try and it doesn't look to be right. But I'm not sure where I did a mistake. I used binomial probability.. |
Oh, sorry, I did a small mistake. It may still not be right, but my chances formula is indeed
(a+b+c+d)! / (a!b!c!d!) * (1/4)^28
So the expected number of fights would be
(a!b!c!d!) / (a+b+c+d)! * 4^28
It still is higher than it should be I think, but at least it gives a decent number.
Is this closer to the solution? |
for RevolutionRebel:
Getting 5 trinkets ensures atleast 2 of a kind.
We need 28 total trinkets, and if the probability of getting a kind of a trinket is 1/4, then the number of trinkets required should be 5*1/4*28 = 35
Is it right? |
for ParaLeul:
no it is not that easy. |
| + |
for Majblomma:
nobody leaves in the first day, and everybody leave in the next day.
very good problem, confused me in the first look. |
nobody leaves in the first day, and everybody leave in the next day.
it should be: nobody leaves in the first 49 days and everybody leaves in the 50th day |
| RevolutionRebel: It is correct that everybody will leave on the same day, but is not the next day (day two). |
| liuker: That is correct - well done! |
for liuker:
you must have been hinted by queen sama! little lk o baka baka! |
for RevolutionRebel:
this is a classical logic problem. I've seen it years ago :PPPPP |
| The best problem I've seen, which took weeks for me and a friend to solve is a chess problem. A game starts with white moving his king's pawn two steps. The game ends in the fifth move with knight takes rook mate. There is a story (http://www.chessbase.com/post/the-infamous-1999-chessbase-christmas-puzzle-160813) that even Kasparov didn't solve this (he might not have spent so much time on it though), but you really mustn't be a chess player to enjoy it. If you know chess rules you can start working on it. |
As an enchanter I get both large and small orders to enchant arts. At some occasions I have a huge queue and have loads of arts waiting for enchantments. At one day when I had 11 SoM:s waiting for enchantment, a customer came to my smithy. He talked nervously and glanced at the rack with the swords. Beside each sword, there was a label attached to the rack so I could keep track of them. When we finally agreed on the terms, the customer rushed over to the rack to place his sword among the others. Just a step from the rack he stumbled, fell into the rack and turned it over. All the swords lay shattered on the floor in a great mess.
“No worries” he said. “They are all alike so we can just place them back in any order.” That got me thinking. The huge production of SoM:s had made them almost perfectly identical. But I had heard that there were some fake alloys that looked just like high quality steel, but had inferior durability. Maybe this customer came with a fake SoM and wanted someone else to have it while getting away with the real thing himself. All alloys have slightly different density so a fake SoM should be either lighter or heavier than the original. I needed a balance scale to compare them.
I went over to my neighbor, the alchemist, and asked if I could borrow his balance. That greedy bloodsucker wanted me to pay for each use, and he wanted the payment in advance! He didn't know how to make gold, but he sure know how to get his hands on it. It was simply out of the question to use the balance 12 times to compare all swords. I started thinking about some more intelligent way to use the balance that would limit the number of uses. How many uses are the least needed and still know for sure which SoM is fake and if it is heavier or lighter?
Please, don’t just post a number, but also some explanation about how to do it. |
So you wanna try tough questions? xD
Here try this...
Arrange :
1) 22222
2) 2222^2
3) 22^22^2 (twenty two raised to twenty two raised to two)
4) 2^2^2^22
5) 2^2^222
Dare not use a calculator!
HINT : There is a method to solve it without calculators. :P |
For Lord harddude, 39.
Is the question just to arrange them in a different way? It is indeed very easy:
1) 22222 = 11111*2
2) 2222^2 = 2222*2222
3) 22^22^2 = 22^22*22^22
4) 2^2^2^22 = 2^2^2^21*2^2^2
5) 2^2^222 = 2^2^221*2^2
I didn't use a calculator. (sun)
If you want an exact number for each, I dare to guess that few will solve it even with a calculator. For example is
22^22^2 = 1.165729954*10^59 (approximately)
That is a total of 60 digits. Good luck finding a calculator that can handle it. |
