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.

Guest Sep 9, 2019

#1**0 **

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

Guest Sep 9, 2019

#2**+1 **

that was me

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sorry, the sum of the digits from 1 - 100 is actually

45 for the single digit numbers +

45 (tens digit) and 45 * 9 (ones digit) this is double digits

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So the sum of the digits from 1 - 100 is actually 450

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The sum of the digits from 1 - 1000

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Is 450 +

45 + 45 * 9 + 45 * 81

So 1 - 1000 is 4545

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So try to get 1 - 10^100

CalculatorUser
Sep 9, 2019

edited by
CalculatorUser
Sep 9, 2019

edited by CalculatorUser Sep 9, 2019

edited by CalculatorUser Sep 9, 2019

edited by CalculatorUser Sep 9, 2019

edited by CalculatorUser Sep 9, 2019

edited by CalculatorUser Sep 9, 2019

edited by CalculatorUser Sep 9, 2019

edited by CalculatorUser Sep 9, 2019

edited by CalculatorUser Sep 9, 2019

#3**+1 **

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

.

.

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**

Guest Sep 10, 2019

#4**+2 **

**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.**

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

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

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

heureka Sep 10, 2019

#5**+1 **

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 !!!!

CPhill Sep 10, 2019