Results 1 to 5 of 5

Math Help - Calculating 10^30-90

  1. #1
    Senior Member Mukilab's Avatar
    Joined
    Nov 2009
    Posts
    468

    Calculating 10^30-90

    Sum of the digits in the sum: (10^30)-90

    Am I correct in saying it's 253?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    -1
    e^(i*pi)'s Avatar
    Joined
    Feb 2009
    From
    West Midlands, England
    Posts
    3,053
    Thanks
    1
    Quote Originally Posted by Mukilab View Post
    Sum of the digits in the sum: (10^30)-90

    Am I correct in saying it's 253?
    10^{30} >> 90 so the 90 can be ignored.

    10^{30} - 90 \approx 10^{30}


    I don't get the question though
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member Quacky's Avatar
    Joined
    Nov 2009
    From
    Windsor, South-East England
    Posts
    901
    I'm unsure of the correct notation here so please bear with me.
    Provided that x is a positive integer:
    10^x will produce a number which is 1 followed by x 0's. What I mean is that:

    10^3 = 1000, which is 1 followed by 3 0's
    10^5 = 100000 which is 1 followed by 5 0's

    10^{30} will be a 1 followed by 30 zero digits.

    If x is greater than one, as it is in your case, and an integer, it will end in 10 and be preceeded by (x-2) nines when 90 is subtracted.

    Eg:
    10^6 - 90
    = 1,000,000 - 90 = 999910
    x = 6
    x-2 = 4
    The result is four 9's followed by 10.

    10^9-90 = 999999910
    x=9
    x-2=7
    There are 7 nines followed by 10.

    So in your example:

    10^30-90
    x=30
    You will have 28 nine's followed by a 10.
    The sum of the digits will be:
    28\times9+1+0
    which is, as you say, 253
    Last edited by Quacky; March 2nd 2010 at 01:20 PM. Reason: Got my unknowns a bit confused.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member Mukilab's Avatar
    Joined
    Nov 2009
    Posts
    468
    Quote Originally Posted by Quacky View Post
    I'm unsure of the correct notation here so please bear with me.
    Provided that x is a positive integer:
    10^x will produce a number which is 1 followed by n-1 0's. What I mean is that:

    10^2 = 10, which is 1 followed by (2-1) 0's
    10^5 = 10000 which is 1 followed by (5-1) 0's

    10^30 will be a 1 followed by 29 zero digits.

    If x is greater than one, as it is in your case, and an integer, it will end in 10 and be preceeded by (x-2) nines.

    Eg:
    10^6 - 90
    = 1,000,000 - 90 = 999910
    x = 6
    x-2 = 4
    The result is four 9's followed by 10.

    10^9-90 = 999999910
    x=9
    x-2=7
    There are 7 nines followed by 10.

    So in your example:

    10^30-90
    x=30
    You will have 28 nine's followed by a 10.
    The sum of the digits will be:
    28\times9+1+0
    which is, as you say, 253
    thank you, perfect answer. I'm quite shamed I didn't know the method!

    EDIT: of course I didn't do it manually but it took me 7 minutes which is 3 minutes above my limited time per question :/
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member Quacky's Avatar
    Joined
    Nov 2009
    From
    Windsor, South-East England
    Posts
    901
    Actually, I'd made a mistake at the beginning. I've rectified it now.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Calculating mean
    Posted in the Statistics Forum
    Replies: 1
    Last Post: April 16th 2011, 05:27 AM
  2. Calculating APR
    Posted in the Business Math Forum
    Replies: 2
    Last Post: January 19th 2011, 07:09 PM
  3. Calculating y'
    Posted in the Calculus Forum
    Replies: 5
    Last Post: January 3rd 2010, 01:34 PM
  4. calculating P(X>=Y)
    Posted in the Statistics Forum
    Replies: 2
    Last Post: February 4th 2009, 12:30 PM
  5. calculating p-value
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: July 20th 2008, 09:06 PM

Search Tags


/mathhelpforum @mathhelpforum