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Posers and Puzzles

In the den

Joined 09 Jun '07 Moves 45641 What percentage of integers contain the digit 9 (at least once)?

127.0.0.1

Joined 18 Dec '03 Moves 16018 Assuming you don't use leading zeroes, the chances that a given integer has a '9' in it is:
1 - (8/9)(9/10)^n
Where 'n' is the length of the integer in digits.
I guess you might take the limit of that as 'n' approaches infinity? But that would make it very close to "all of them" I think

127.0.0.1

Joined 18 Dec '03 Moves 16018 What about this?
https://www.wolframalpha.com/input/?i=summation+from+n%3D1+to+infinity+of+%28%281-%288%2F9%29%289%2F10%29%5En%29%2F%2810%5En%29%29

In the den

Joined 09 Jun '07 Moves 45641 Originally posted by forkedknight
Assuming you don't use leading zeroes, the chances that a given integer has a '9' in it is:
1 - (8/9)(9/10)^n
Where 'n' is the length of the integer in digits.
I guess you might take the limit of that as 'n' approaches infinity? But that would make it very close to "all of them" I think Yes.

100% of the integers have at least one occurrence of the digit "9".

ðŸ˜€

tinyurl.com/y9ls7wbl

Joined 23 Aug '04 Moves 24791 Originally posted by forkedknight
Assuming you don't use leading zeroes, the chances that a given integer has a '9' in it is:
1 - (8/9)(9/10)^n
Where 'n' is the length of the integer in digits.
I guess you might take the limit of that as 'n' approaches infinity? But that would make it very close to "all of them" I think If N = 1

1 - (8/9)(9/10) = 1 - (8/10) = 1/5

One fifth of all the one digit integers is not 9. Fail formula.

127.0.0.1

Joined 18 Dec '03 Moves 16018 Originally posted by AThousandYoung
If N = 1
1 - (8/9)(9/10) = 1 - (8/10) = 1/5
One fifth of all the one digit integers is not 9. Fail formula. Sorry, I guess it's an off-by-one error:

p = 1 - (8/9)(9/10)^(n-1)

Joined 11 Nov '05 Moves 43938 Originally posted by wolfgang59
Yes.
100% of the integers have at least one occurrence of the digit "9".
ðŸ˜€ Do you say that all integers has at least one nine? Then I can give you a counterexample - 8.

127.0.0.1

Joined 18 Dec '03 Moves 16018 All, except for an infinite number of them

Joined 11 Nov '05 Moves 43938 But - and this is not a joke:
Among all integers there are exactly as many numbers with a nine in them as they are numbers without a nine in them. No approximately but exactly! Not as n reach infinity, but when n *is* infinity!
This is provable. Cantor showed us how.

In the den

Joined 09 Jun '07 Moves 45641 Originally posted by FabianFnas
But - and this is not a joke:
Among all integers there are exactly as many numbers with a nine in them as they are numbers without a nine in them. No approximately but exactly! Not as n reach infinity, but when n *is* infinity!
This is provable. Cantor showed us how. Cantor cant count.

127.0.0.1

Joined 18 Dec '03 Moves 16018 Originally posted by FabianFnas
But - and this is not a joke:
Among all integers there are exactly as many numbers with a nine in them as they are numbers without a nine in them. No approximately but exactly! Not as n reach infinity, but when n *is* infinity!
This is provable. Cantor showed us how. Well the the answer should be 50%, and I don't think that's true...

Joined 11 Nov '05 Moves 43938 Originally posted by forkedknight
Well the the answer should be 50%, and I don't think that's true... The number of all integers is infinitely many.

How much is 50% of infinity?

In the den

Joined 09 Jun '07 Moves 45641 Originally posted by FabianFnas
How much is 50% of infinity? A half.