BetonaBG

Sadly I don't understand the various proofs in this thread but I'm still not convinced how 2 <= C <= 9 when the number of digits in N is so great.

Yeah digits in N are a lot' date=' but C is the Sum of the Sum of the Sum from the digits in N).

Ex:

['][][]

N = 23523525236235386529783562789364928374 => Sum 4833 = A

A = 4833 => Sum = 18 = B

B = 18 => Sum = 9 = C

C = 9

[][][]

Let S(x) = Sum of the digits of x

 

Thm: S(x) = x mod 9 (more generally Sb(x) = x mod (b-1) if we are working in base b)

Following this...

S(8765^4321) = 8765^4321 mod 9

= 8^4321 mod 9

if mod 9, this mean that the base (b-1) = 9 =>> b = 10

I don't see base 10 anywhere, or is it something else?

And can you explain what do you mean by base?

Well, my idea about proving that its between 2 and 9 looks like this:

(Note) Totally wrong from the right method to start the problem:

8765^2 = 8 digit number

so in the worst case 8765^4321 = will give us 4*4321 (17284) digit number:

Consider all those digit being 9s (Max) and 1s(min) I discard the case one 1 and lots 0s, just because 8765^4321 doesn't seem like having many 0s.

17284*1 min <= Sum <= max = 17284*9

17284 <= Sum <= 155556 (but we get Max and Min digit sum from)

A: 20000 <= SUm <= 99999

B: 2 <= B <= 45 (digit sum) (39)

C: 2 <= C <= 12 (digit worst case) (9)

C 2<= C <= 9

I don't know if even half of this is correct (I doubt it), so pls find some new approach

And yeah, Dave is right, we don't look to calcutate the number but to get the sum

C

btw I found a way to prove that its between 2 and 9.

But that still no good

I found a formula that might help out but I don't understand even the example

http://www.cut-the-knot.org/Curriculum/Arithmetic/PowerOfDigits.shtml

Its given from two of the professors in San Jose State University.

There is a whole sequence going on for the last 2 years, I'll try to find

old problems and give you a link if you are interest.

No, I mean what I said,

N = 8765^4321

[15 times Editing] Yeah, question is pain for me last 2 days

Let N = 8765^4321 be writen in decimal notation.

If A is the sum of the digits of N and B is the sum

of the digits of A, then what is the sum of the

digits of B?

Have Fun.

I'll say simple - Thanks

but my gratitude pales in comparatively with my words.

First of all, I'd like to say Hello to everyone as this is my first post

Now the fun part, I spend 2-3h in painful strugle and I was unable

to prove the following question. If anyone knows how to solve it

it will be largely appreciated

Prove for any positive integer n that:

2196^n – 25^n – 180^n + 13^n is divisible by 2004

Have Fun