Jump to content


Welcome to BrainDen.com - Brain Teasers Forum

Welcome to BrainDen.com - Brain Teasers Forum. Like most online communities you must register to post in our community, but don't worry this is a simple free process. To be a part of BrainDen Forums you may create a new account or sign in if you already have an account.
As a member you could start new topics, reply to others, subscribe to topics/forums to get automatic updates, get your own profile and make new friends.

Of course, you can also enjoy our collection of amazing optical illusions and cool math games.

If you like our site, you may support us by simply clicking Google "+1" or Facebook "Like" buttons at the top.
If you have a website, we would appreciate a little link to BrainDen.

Thanks and enjoy the Den :-)
Guest Message by DevFuse
 

Photo
- - - - -

Alex gets even with writersblock. Or does he?


  • Please log in to reply
35 replies to this topic

#21 Dr Benak

Dr Benak

    Newbie

  • Members
  • Pip
  • 1 posts

Posted 07 June 2008 - 06:24 AM

Not sure if this is relevant but just say, for example, 10% of infinity, ?/10, is still ? but a differnet level of infinity correct?

like

(?/10) / ?

it wouldnt be ?/? as in 1 as in 100%.... it would be .1 as in 10%

Am I just shootin in the dark here or do I have a slembance of a point? Help me out here lol


Oh, you have ventured into the Land of Indeterminate Forms. Consult our guide, Monsieur L'Hôpital.
  • 0

#22 bonanova

bonanova

    bonanova

  • Moderator
  • PipPipPipPip
  • 5564 posts
  • Gender:Male
  • Location:New York

Posted 07 June 2008 - 07:33 AM

Oh, you have ventured into the Land of Indeterminate Forms. Consult our guide, Monsieur L'Hôpital.

Why is it indeterminate? ;)
What does L'Hôpital say about this?
Is 10% of infinity not 10% just the same?
What percent of all integers are even? Is that indeterminate? ;)

Alex gave an understandable and meaningful set of choices ... which one is right?
  • 0
The greatest challenge to any thinker is stating the problem in a way that will allow a solution.
- Bertrand Russell

#23 johnnyleonclark

johnnyleonclark

    Newbie

  • Members
  • Pip
  • 2 posts

Posted 19 March 2009 - 02:31 PM

The chance of geting a number that contains a 3 is better the larger your number is becuse you have more numbers containing three this increase in percentage stops at 50/50 becuse there are an infinite numbers that contain a3 and an infinite number of numbers that do not there for they are even. The problem with infinite is you just keep going till you reach the solution you want. ;)

Edited by johnnyleonclark, 19 March 2009 - 02:32 PM.

  • 0

#24 Hell's Angel

Hell's Angel

    Newbie

  • Members
  • Pip
  • 4 posts

Posted 19 March 2009 - 03:55 PM

After being done out of a pint of O'Doule's by writersblock
last night at Morty's, Alex conjured up a question calculated
to get him even.

After WB had downed his cool one, Alex proposed a double or
nothing puzzle. To the nearest percentage point, he asked,
what percentage of all numbers contain at least one 3?

For example, 13, 31, 33 and 103 all contain the digit 3 at least once.
But 1, 2, 4, 5, 6, 7, 8, 9, 10, 11, 12, 14, 15, 16, 17, 18, ... well
you get the idea ... don't contain the digit 3 even once.

Now I know there's a lot of numbers to check, so I'll make it
easy for ya, Alex said. I'll give you five multiple choices.
That gives you a 20% chance even if you guess, and much
better odds if you yer' the genius ya' make yerself out to be.

The percent of all numbers containing at least a single "3" is:

[1] 10%
[2] 13%
[3] 33%
[4] 67%
[5] 100%

What was writersblock's choice, and did he win another pint?



The series this forms is

Term 1 = 10^0
Term 2 = 10^1 + (Term1 * 9)
Term 3 = 10^2 + (Term2 * 9)
Term 4 = 10^3 + (Term3 * 9)

so on and so forth... the percentage would go pretty close to 100% i think..... for any single digit.. not just 3
  • 0

#25 filigaitas

filigaitas

    Newbie

  • Members
  • Pip
  • 1 posts

Posted 19 March 2009 - 05:00 PM

It wouldn't be 200% it would still be 100% to the nearest percentage. That is what is cool about this whole concept. To the nearest percentage point, 100% of numbers contain any given digit except zero.

If you don't believe it, follow this.

In a series of 1 -10 there is 1 digit that contains a 3. 1 of 10 = 10%.
In a series of 1 - 100, each set of 10 contains 1 as above, except for the 30's series which contains 10. 10 for the 30's series plus 1 for all the other sets of 10 = 19. 19 of 100 = 19%. You can continue this sequence for every power of 10. At 1000, there are 271 3's. 19 for each series of 100 plus 100 for the 300's series. Extrapolate this far enough and you will see that at 1E66 (1 followed by 66 zeros), 99.9E65 numbers contain at least one digit 3. This means that 99.9 percent of the numbers less than 99.9E65 contain a digit 3. Therefore to the nearest percentage, 100% of numbers contain a digit 3. As you approach infinity this number will get closer and closer to 100%, but never reach it.

And yes, this is true for all digits except zero.



Except zero? I think this riddle could be made with any digit, including 0. Once you get to really big numbers it is statisticly very difficult to get numbers without any particular digit. We are dealing with infinity so you need to be able to think in big numbers. After all, lets say a 1000000000 digit number is still small when talking infinity.
To kind of prove the point I made a much smaller scenario. I made a simple quickbasic program (I never thought I was going to use QB again!) in which I would get a point for every 10000 random numbers from 0 to 9 in which no 0s were created. It ran all night without a point. It got plenty of points on the test runs with numbers that were 20 and 30 digits but once you get to higher numbers is statisticly impossible to get a number without a 3 even though like a poster says there are infinite numbers without any particular digit. Now, if it ran all night without any points, imagine the odds with 10000000000000000 digits or much higher. We are dealing with infinity so it get really really close to 100% fairly fast.
  • 0

#26 bigdamnjim

bigdamnjim

    Newbie

  • Members
  • Pip
  • 5 posts

Posted 22 April 2009 - 05:56 AM

Okay, maybe I am missing something really important because I think the answer is 20%. Or, of the answers given, 13%. From 1 to 10 it is 10%. From 1 to 100 it is 19%. From 1 to 1000, it is 19.9%. From 1 to 10000, it is 19.99%
  • 0

#27 vinays84

vinays84

    Junior Member

  • Members
  • PipPip
  • 78 posts

Posted 22 April 2009 - 10:18 PM

WB gave the method of obtaining the correct answer it in the first post.
The formula is:
f(n) = (f(n-1)*9 + 1)/10 where f(n) is the percentage of numbers out of 10^n that contain any non-zero digit.
The formula is simply a weighted average: 1/10 of 10^n will be have the digit in 100% of its numbers and the other 9/10 will have it in (f(n-1)*100)% of its numbers. (Visualize it the 10 blocks).
You can create a spreadsheet to quickly find the trend.
f(100) = 0.999973439...That means that almost 100% of all the numbers less than a googol contain at least one 3.

Note: If the question is "What percentage of ALL numbers contain at least a 3?", the answer isn't "close to but never 100%", its exactly 100%.
This is because the "ALL" makes the formula limn->∞f(n), which is 1.
While this might not make sense (because you can list out numbers that don't contain a 3), it still is the correct answer. The reason is you're trying to take a percentage of a percentage of infinity, which isn't a number.

Edited by vinays84, 22 April 2009 - 10:20 PM.

  • 0

#28 bigdamnjim

bigdamnjim

    Newbie

  • Members
  • Pip
  • 5 posts

Posted 23 April 2009 - 05:18 AM

For every number that you show me with a three, I will show you at least eight others that don't. If you look at it as the probability that one of the digits is not three, you must stop looking as soon as one of the digits is three, otherwise you are counting them more than once.
  • 0

#29 Neeth

Neeth

    Newbie

  • Members
  • Pip
  • 1 posts

Posted 31 December 2010 - 03:18 PM

Guys guys, I think this math stuff is going crazy. You guys are saying it is 100% to contain a number at least one digit 3.

Well, lets look it this way. What about a number to contain 3 OR 4. According to the argument it is going to be 200%. That is very funny ... because there are infinite numbers that contains neither 3 nor 4. (Hope no one will ask for an example).


No, the trick here is that the percentage is approx 100 percent and not exact 100 percent.. For eg: 99 percent put of 10000 leaves around 100 numbers without 3...
  • 0

#30 dark_magician_92

dark_magician_92

    Advanced Member

  • Members
  • PipPipPip
  • 382 posts

Posted 02 January 2011 - 08:38 AM

After being done out of a pint of O'Doule's by writersblock
last night at Morty's, Alex conjured up a question calculated
to get him even.

After WB had downed his cool one, Alex proposed a double or
nothing puzzle. To the nearest percentage point, he asked,
what percentage of all numbers contain at least one 3?

For example, 13, 31, 33 and 103 all contain the digit 3 at least once.
But 1, 2, 4, 5, 6, 7, 8, 9, 10, 11, 12, 14, 15, 16, 17, 18, ... well
you get the idea ... don't contain the digit 3 even once.

Now I know there's a lot of numbers to check, so I'll make it
easy for ya, Alex said. I'll give you five multiple choices.
That gives you a 20% chance even if you guess, and much
better odds if you yer' the genius ya' make yerself out to be.

The percent of all numbers containing at least a single "3" is:

[1] 10%
[2] 13%
[3] 33%
[4] 67%
[5] 100%

What was writersblock's choice, and did he win another pint?

Spoiler for i guess

  • 0




0 user(s) are reading this topic

0 members, 0 guests, 0 anonymous users