Number Theory

HINT:
 

1 through 9 is obviously 45.

So look from 10 to 99

the tens digit it 1 through 9 = 45

but the ones digit it is 1 through 9 occuring 9 times 45*9

So the digit sum of 1 to 100 is

45 for the single digit numbers

and 45 (units digits) and 45 (tens digit) for the double digit numbers

Do you see the pattern?

Now just keep going you might have to do some exponent work the answer will be big

Since 1 to 99 =[99+1] * 2 * 4.5 =900. [4.5 being the average value of all 10 digits 0 - 9 =45 / 10 = 4.5 ]
Since 1 to 999 =[999+1] * 3 * 4.5 =13,500
Since 1 to 9,999 =[9,999+1] * 4 * 4.5 = 180,000
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Since 1 to 999,999 =[999,999 +1] * 6 * 4.5 =27,000,000
Now, you can clearly see a pattern:
Therefore: 1 to 10^100 - 1 =[10^100 - 1 + 1] * 100 * 4.5 =[4.5 x 10^102] + 1. This last 1 being the leading digit of 10^100.

This short computer code confirms the above answer:
n=1E100;s=0;cycle:if(n<10, return s=s+n*(n+1)/2, 0);place=0;p=1;loop:place=place*10+45*p;p=p*10;if(p*10<=n,goto loop,0);msd=int(n/p);n=n%p;s=s+msd*place + (msd*(msd-1)/2)*p + msd*(1+n);goto cycle
Answer: 4.5000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 01 E+102

What is the sum total of all individual digits from 1 to 10^100? In other words:
1+2+3+4+5+6+7+8+9+[1+0]+[1+1]+[1+2]+[1+3].......and so on to 10^100?
Any help or hint would be greatly appreciated. thank you.

\(\text{Let $ \mathbf{b} = $ numeral system } \\ \text{Let $ \mathbf{s} = $ sum of all digits from $1$ to $b^n$ }\)

\(\begin{array}{|rcll|} \hline \mathbf{s} &=& \mathbf{1 + n\cdot b^n \cdot \left( \dfrac{b-1}{2} \right) } \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline b &=& 10 \quad (\text{decimalism}) \\ n &=& 100 \\ && \text{from $1$ to $10^{100}$ } : \\\\ \mathbf{s} &=& \mathbf{1 + n\cdot b^n \cdot \left( \dfrac{b-1}{2} \right) } \\ s &=& 1 + 100\cdot 10^{100} \cdot \left( \dfrac{10-1}{2} \right) \\ &=& 1 + 10^2\cdot 10^{100} \cdot \left( \dfrac{9}{2} \right) \\ &=& 1 + \left( \dfrac{9}{2} \right)\cdot 10^{102} \\ \mathbf{s} &=& \mathbf{ 1 + 4.5 \cdot 10^{102} } \\ \hline \end{array}\)

For problems like these, I like to see if we can discover a pattern

Note that  the sum of the digits from 1 -9  inclusive  = 45

And adding the "1" in the next integer, 10, gives us  46

And we can write this sum as

Sum of the individual digits from 1 - 10^1  inclusive   = 1* 45  + 1

Next....the sum of the digits of the integers from 1-99 inclusive  =  900    

We can see this because we  have the digits  1-9  in the ones place repreated 10 times  =  10(45)  = 450

And we have the sum of the digits in the tens place as  10  ( 1 + 2 +3 +....9)  = 10 (45)  = 450

And adding the "1"  in the next integer (100) gives us this sum   =  900 + 1 =    20*45 + 1  

Following this, the sum of the individual digits of the integers from 1 -999  =   13,500   [check this for yourself]

And adding the "1" in the next integer  (1000)   gives us this sum:

13500 + 1    =    300* 45  +  1

Note the pattern that seems to be emerging......the sum of the digits of the integers from 1 - 10^n  inclusive  =

(n) followed by the number of zeroes calculated as (n - 1) * 45    +  1

So....for instance .....the sum of  the digits of the integers from 1 -10^1 inclusive  =

(1)  followed by  ( 1 - 1) zeroes * 45 + 1   =

(1) followed by no zeroes * 45 + 1  =

1*45 + 1

And the sum of the digits of the integers from   1 - 10^2 inclusive  =

(2)followed by (2-1) zeroes * 45 + 1   =

(2)  followed by 1 zero  * 45 + 1  =

20*45 + 1

Note that the sum of the individual digits of the integers from 1 -100^3   =

300*45 + 1......which follows our pattern

This seems to imply that  the  sum of the individual digits in the integers from 1 - 10^100 inclusive should be :

(100) followed by (100-1) zeroes * 45 + 1  =

100 followed by 99 zeroes * 45 + 1  =

10^101 * 45 + 1  =

4.5 * 10^102  +  1       as found by the Guest and heureka  !!!!