December 28, 2009

Find the sum of digits ?



Find sum of digits of D.
Let A= 19991999
B = sum of digits of A, C = sum of digits of B, D = sum of digits of C.



8 comments:

  1. UnknownDecember 30, 2009 at 4:12 AM

    Wrong puzzle !
    Should be : 1999^1999
    and also hint is A = B = C = D (mod 9)

    ReplyDelete
  2. UnknownDecember 30, 2009 at 4:13 AM

    btw the answer is 1

    ReplyDelete
  3. Secret SquïrrelDecember 30, 2009 at 6:54 AM

    B=1+9+9+9+1+9+9+9=38
    C=3+8=11
    D=1+1=2

    ReplyDelete
  4. UnknownDecember 30, 2009 at 4:01 PM

    Hi all,

    the question is 1999^1999

    sorry for the mistake

    ReplyDelete
  5. Secret SquïrrelJanuary 3, 2010 at 5:40 PM

    Ok, that makes it a nicer altho' harder problem.

    The series 1999^1, 1999^2, 1999^3, 1999^4 all have a digital root of 1; eg 1999^4==1 (Mod 9). I'm guessing that 3 cycles of extracting the digital root of the thousands of digits of 1999^1999 is sufficient to reach the digital root; ie 1.

    ReplyDelete
  6. how to ollieFebruary 4, 2010 at 2:55 AM

    I like your writing style. Nice blog.

    ReplyDelete
  7. UnknownFebruary 8, 2010 at 11:27 PM

    Let E be the sum of digits of D.

    Following the hint above (A, B, C, D and E are all equivalent mod 9) and using 1999 = 222*9 + 1 we get that E is equivalent to 1^1999 = 1 mod 9.

    Let n(x) denote the number of digits of x. We have

    n(A) <= 1999*log_{10}(1999) + 1 <= 1999*4 + 1 <= 8000.

    Therefore, B <= 9*n(A) <= 72000 which implies n(B) <= 5.

    On the same way we get:

    C <= 9*n(B) <= 45,
    n(C) <= 2, and
    D <= 9*n(C) <= 18.

    Every natural number in [1, 18] has a sum of digits in [1, 9]. In particular, E belongs to [1, 9].

    Being in [1, 9] and equivalent to 1 mod 9, E must be 1.

    ReplyDelete
  8. Viagra OnlineSeptember 22, 2010 at 11:23 PM

    Heck!
    Where did I let my calculator!?

    ReplyDelete