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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:

boban said...

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

boban said...

btw the answer is 1

Secret Squïrrel said...

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

Eldhose said...

Hi all,

the question is 1999^1999

sorry for the mistake

Secret Squïrrel said...

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.

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Cassio said...

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.

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