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Digit Sum


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Do you mean let N = 8765 X 10^4321?

 

8765^4321 is ridiculously big and makes the question unworkable, whereas A and B are straightforward to work out if N = 8765 x 10^4321 and one understands scientific notation.

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Fair enough, my mistake - I approached this question as though it was demonstrating scientific notation to make a point to grade school students. If it's 8765^4321 then I haven't a clue on the methods to use.

 

Out of interest where is this question from?

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I would be interested. As far as I can see, with my basic knowledge of maths, the solution will only be found by crunching huge numbers (possibly using a computer program). Unless there is a technique for this sort of thing which then takes the whole thing over my head. :(

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That seems rather low considering the value of N or have I got the wrong end of the stick here?

 

If we say N = 8765^2 what value for the sum of the digits of B do you get?

 

a=37

b=10

"C"= 1

 

so if you think about it c=2 could mean a great many things. b could be 11 or 20, a could be 29 or twenty 1's followed by an infinite number of 0's.

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but to return a number between 2 and 9 there would have to be a LOT of 0's in the decimal notation of 8765^4321, A or B.

 

Consider that 8765^78 = 3.42491497 × 10^307. That's a whole lot of digits if it's written out in decimal notation and without rounding and that's only a fraction of N.

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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 :P

 

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

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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)

 

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

= 8^4321 mod 9

 

8^2 = 64 = 1 mod 9

 

Therefore

 

S(8765^4321) = (8^1)*(8^2)^2160 mod 9

= 8 * (1^2160) mod 9

= 8 mod 9

Thought this might help a bit. :)

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Finding "C" by calculating the number was never going to happen, I was just thinking around the whole thing to get some idea of the values we are looking at.

 

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.

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Well' date=' 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 :P

 

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

 

 

there can be zeros :P but it is unlikely that you would get a lower number than if it were all 1's

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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? :-(

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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

[][][]

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Following this...

if mod 9' date=' 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? :-([/quote']

 

We're working in the decimal system aka base 10.

 

E.g. 156 = 1*10^2 + 5*10^1 + 6 * 10^0

So if we look at S(156),

S(156) = 1+5+6 = 2 mod 9

Now

156 = 1*10^2 + 5*10^1 + 6 * 10^0 = 1*1^2 + 5*1^1 + 6 * 1^0 mod 9

= 1 + 5 + 6 mod 9

The reasoning here can be used to prove a more general theorem/lemma.

 

 

Concerning the number of digits of C:

The number of digits of 8765^4321 is approx Log10[8765^4321] ie 17036

The sum of those digits is at most 9*17036 = 153324 ....A

Now the sum of the digits of a 6 digit number is at most 6*9 = 54 ....B

 

This sum, B, along the same reasoning as before, has to be = 8 mod 9

i.e. B is in {8,17,26,35,44,53}

The sum of the digits of each of these is 8. Thus C = 8

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  • 2 weeks later...
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)

 

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

= 8^4321 mod 9

 

8^2 = 64 = 1 mod 9

 

Therefore

 

S(8765^4321) = (8^1)*(8^2)^2160 mod 9

= 8 * (1^2160) mod 9

= 8 mod 9

Thought this might help a bit. :)

 

 

This is very useful. Can you explain why 8765^4321 mod 9 = 8^4321 mod 9?

Although I know 8765 mod 9 = 8.

 

Similarily, why 8^2160 mod 9 = 8?

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