Please turn on javascript in your browser to play chess.
Posers and Puzzles

Posers and Puzzles

  1. 05 Mar '08 22:51
    Prove:

    0/2 + 0/4 + 0/8 + 1/16 + 1/32 + 0/64 + 1/128 + ... = 1/9

    Where the denominators are the powers of 2, and each numerator equals the number of odd digits in the denominator.
  2. 06 Mar '08 15:57
    i take your word for it
  3. Standard member adam warlock
    Baby Gauss
    06 Mar '08 16:33
    Originally posted by David113
    Prove:

    0/2 + 0/4 + 0/8 + 1/16 + 1/32 + 0/64 + 1/128 + ... = 1/9

    Where the denominators are the powers of 2, and each numerator equals the number of odd digits in the denominator.
    I really want to see how this one is done. I tried for a little bit but couldn't get much done and for the moment I'm not seeing how it can be done either.
  4. 06 Mar '08 18:22 / 1 edit
    I think it has something to do with number divisibility with 9. But I can't remember what properties were required for the number to be divisible with 9. Eh, that was just a wild guess.

    P.s. And, yeah, I'd also like to see this done.
  5. 06 Mar '08 18:59 / 2 edits
    Originally posted by David113
    Prove:

    0/2 + 0/4 + 0/8 + 1/16 + 1/32 + 0/64 + 1/128 + ... = 1/9

    Where the denominators are the powers of 2, and each numerator equals the number of odd digits in the denominator.
    1/9, you say it is 0/2 + 0/4 + 0/8 + 1/16 + 1/32 + 0/64 + 1/128 + ... = 1/9
    it is actually not so.
    The correct is 0/2 + 0/4 + 0/8 + 1/16 + 1/32 + 1/64 + 0/128 + ... = 1/9
    or if we show only the indices = 000 111 000 111 ...

    Go to the binary system, 1/9 (10) = 1/1001 (2), do the division 1/1001 and you get the result .000111000111... (a pattern of 3 zeroes followed by 3 ones and again and again).

    Just simple arithemtics.

    If you want a proof that .000111000111... is really 1/1001 binary = 1/9 decimal, here it comes, again in binary arithemtics:

    say that x = .000111000111...
    then 1000000x = 111.000111...
    and 1000000x-x = 111 exactly.

    Now we safely can go back to decimal arithemtics again.
    64x-x = 63x = 7
    x = 7/63 = 1/9

    And voilà we've shown that 1/9 is really .000111000111...
  6. Standard member adam warlock
    Baby Gauss
    06 Mar '08 19:03
    Originally posted by FabianFnas
    1/9, you say it is 0/2 + 0/4 + 0/8 + 1/16 + 1/32 + 0/64 + 1/128 + ... = 1/9
    it is actually not so.
    The correct is 0/2 + 0/4 + 0/8 + 1/16 + 1/32 + 1/64 + 0/128 + ... = 1/9
    or if we show only the indices = 000 111 000 111 ...

    Go to the binary system, 1/9 (10) = 1/1001 (2), do the division 1/1001 and you get the result .000111000111... (a pattern of 3 ...[text shortened]... arithemtics again.
    64x-x = 63x = 7
    x = 7/63 = 1/9

    And voilà We've shown that 1/9 is realy
  7. Subscriber joe shmo On Vacation
    Strange Egg
    06 Mar '08 19:11
    Originally posted by FabianFnas
    1/9, you say it is 0/2 + 0/4 + 0/8 + 1/16 + 1/32 + 0/64 + 1/128 + ... = 1/9
    it is actually not so.
    The correct is 0/2 + 0/4 + 0/8 + 1/16 + 1/32 + 1/64 + 0/128 + ... = 1/9
    or if we show only the indices = 000 111 000 111 ...

    Go to the binary system, 1/9 (10) = 1/1001 (2), do the division 1/1001 and you get the result .000111000111... (a pattern of 3 ...[text shortened]... arithemtics again.
    64x-x = 63x = 7
    x = 7/63 = 1/9

    And voilà We've shown that 1/9 is realy
    The pattern that he is asked for is different than the one you have shown, the series for the numerators according to his stipulations are

    0,0,0,1,1,0,1,1,2,1,0,1,2,2,2,3,4,1,1,3,4,3,1,5,5,2,5,2 for the first 28

    that doesnt look like 0,0,0,1,1,1,0,0,0,1,1,1..........
  8. 06 Mar '08 19:23 / 1 edit
    Originally posted by FabianFnas
    1/9, you say it is 0/2 + 0/4 + 0/8 + 1/16 + 1/32 + 0/64 + 1/128 + ... = 1/9
    it is actually not so.
    The correct is 0/2 + 0/4 + 0/8 + 1/16 + 1/32 + 1/64 + 0/128 + ... = 1/9
    or if we show only the indices = 000 111 000 111 ...

    Go to the binary system, 1/9 (10) = 1/1001 (2), do the division 1/1001 and you get the result .000111000111... (a pattern of 3 ...[text shortened]... .
    64x-x = 63x = 7
    x = 7/63 = 1/9

    And voilà we've shown that 1/9 is really .000111000111...
    The fact that there is another series whose sum is 1/9 does not mean that the sum of "my" series can't also be 1/9.

    Your series is the only one in which the denominators are 2,4,8,... and whose sum is 1/9 IF EACH NUMERATOR IS 0 OR 1.

    This doesn't mean that if the numerators are not restricted to being 0 or 1 there isn't another series whose sum is 1/9.
  9. Standard member adam warlock
    Baby Gauss
    06 Mar '08 19:29
    Originally posted by David113
    The fact that there is another series whose sum is 1/9 does not mean that the sum of "my" series can't also be 1/9.

    Your series is the only one in which the denominators are 2,4,8,... and whose sum is 1/9 IF EACH NUMERATOR IS 0 OR 1.

    This doesn't mean that if the numerators are not restricted to being 0 or 1 there isn't another series whose sum is 1/9.
    A hint on how it's done. Puhlease. Nothing too giving just a little hint! I love solving this type of things.
  10. 06 Mar '08 19:41
    Originally posted by adam warlock
    A hint on how it's done. Puhlease. Nothing too giving just a little hint! I love solving this type of things.
    I'll wait a few days before giving a hint... maybe someone will solve it?
  11. Standard member adam warlock
    Baby Gauss
    06 Mar '08 19:44
    Originally posted by David113
    I'll wait a few days before giving a hint... maybe someone will solve it?
    how much number theory notion is needed? Can't you just tell what branch of math is more used?
  12. 07 Mar '08 01:53 / 2 edits
    Originally posted by David113
    Prove:

    0/2 + 0/4 + 0/8 + 1/16 + 1/32 + 0/64 + 1/128 + ... = 1/9

    Where the denominators are the powers of 2, and each numerator equals the number of odd digits in the denominator.
    wait. I'm confused. If you add anything positive to 1/2, it goes up. Since 1/9 is less than 1/2, it's mathematically impossible, isn't it?
  13. Subscriber joe shmo On Vacation
    Strange Egg
    07 Mar '08 04:36 / 1 edit
    Originally posted by kayakninja
    wait. I'm confused. If you add anything positive to 1/2, it goes up. Since 1/9 is less than 1/2, it's mathematically impossible, isn't it?
    where do you see 1/2 in that series? what you should be seeing is 0/2 which = 0 ,
  14. Subscriber joe shmo On Vacation
    Strange Egg
    07 Mar '08 04:40
    Are there any 0's in the numerator after 2^11?
  15. 07 Mar '08 05:44
    I don't know if this is a good way to go about it, but I'd like to present a different viewing angle to the problem.

    In any doubling series:

    A digit is even if the digit immediately after it in the previous number is 0, 1, 2, 3, or 4. [Since the digit is doubled, and doesn't carry a one, the digit must become even].

    A digit is odd if the digit immediately after it in the previous number is 5, 6, 7, 8, or 9. [The digit is doubled, and it carries a one, so it must become odd].

    Therefore the number of odd digits in a power of 2 is equal to the number of digits greater than 4 in the previous number. Maybe that can help ('?'