Hard to believe but I am familiar with manjyome's stuff. Anyway why should we consider his/her answer. After all it is only a probable answer( 3 out of 5). Berry's idea, underneath manjyome's post is right on the money. I don't have any way at present to firm up manjyome's work. If you do I would like to see it.

]]>one more thing

the number of digits in 'n' is (characteristic of log n) + 1

]]>

btw, For your information ...

D = 10

]]>Found this page:

http://nz.answers.yahoo.com/question/index?qid=20080620225029AAeW6Uh

with this proof:

We know that Iterative Sum of Digits = n mod 9( also called the digital root)

D = (1999^1999) mod 9 = (1999 mod 9)^1999

We know:

1) 1999 mod 9 = 1

2) 1^f = 1, for any value of f.

D = 1^1999 = 1

This is what I have been saying all along.

]]>TRY TO PROVE THAT THE SUM OF DIGITS OF 'D' WILL BE A SINGLE DIGIT NUMBER

in this question you don't even have to find out the values of b,c,d

]]>This process:

A=1999^1999

B=29656

C=28

D=10

E=?

is the definition of the digital root. That is computed by 1999 is congruent to 1 mod 9 so 1^1999 =1. This saves all the computing. E=1. How was that?

]]>you found out D .... but in the ques you need to find out the sum of digits of D ... which turns out to be 1+0 = 1

i ll give u the last and most important hint:

let E = sum of digits of D

find a way to prove that E would be a single digit number

then u can easily say that E would be 1

reason:

if the digital root of a number is 1 and the number is single digit number ... then the number is got to be 1

You are absolutely right. I was confusing the digital root with the digital sum.

I got the following from:

http://en.wikipedia.org/wiki/Digit_sum

Digital root of 199 =1 (199 is congruent to 1 mod 9)

Digital sum of 199 =19 but still not 10 (No easy way to compute. Just sum the digits)

Digital root of 84001 = 4 (84001 is congruent to 4 mod 9)

Digital sum of 84001 = 13 (No easy way to compute. Just sum the digits)

Based on what I think you require, I am stuck on following the rules for digital sums so:

A=1999^1999

B=29656

C=28

D=10

Thanks a lot for listening, please respond.

]]>The sum of the digits of 199 = 19 = 1 because 199 is congruent to 1 mod 9

if A = 199

den b = sum of digits of A = 10 (not 1)

bobbym wrote:

(by expanding this number, I assumed you didn't want to sum it again else B=1 C=1 D=1)

u r absolutely right here ... i didn't want it to sum it again.

bobbym wrote:

This is as far as I follow the problem please provide more guidance.

u r absolutely correct here ... u followed it absolutely correctly...

**HINT: Prove that sum of digits of D will be a single digit number**

**if u find that sum of digits of D will be a two digits number ... then the ans may be one of... 19,28,37,46,55,64,73,82,91coz all of the above two digit numbers are congruent to 1 under MOD 9 ....**

**but if the sum of digits of D is a single digit number then the answer will be 1 .... i hope u get wot i say**

Thanks for responding.

The sum of the digits of 199 = 19 = 1 because 199 is congruent to 1 mod 9, The sum of the digits of B = 1999^1999 = 29656 (by expanding this number, I assumed you didn't want to sum it again else B=1 C=1 D=1) and C = Sum of the digits of B = 28. D = The sum of the digits of C =10. I assumed you would reduce this 10 down to 1. This is as far as I follow the problem please provide more guidance.

]]>HINT : according to ur proof

sum of digits of 199 = sum of digits of 28

or 19 = 10

Here is my answer to #1, I hope it is what you require.

As you know the way to sum all the digits of any number is to take that number mod 9.

1999 is congruent to 1 mod 9 so 1999^1999 is also going to be congruent to 1 mod 9.

Therefore D = sum of the digits of 1999^1999 = 1

Also for fun:

B=29656

C=28

bobbym

]]>