Help!

i think you would just count up the 1's then the 2's and so on, then multiply and add.

so there is 1, eleven 1's in 10-19, eight 1's in 20-99, thirteen 1's in 100-110,

1+11+8+13 = 33

so there are 2, 12, and 102,  then eleven 2's in 20-29, seven 2's in 30-101.

(3*2) + (11*2) + (7*2) = 42

i see some patterns, but test it out by doing the 3's and 4's?

(4*3) + (11*3) + (6*3) =63  based on patterns

would the 4's = 84? test and see

The method used by Guest #2 works ONLY for numbers all ending in 9s. Therefore, the following is the more accurate method:

0, 1, 2, 3..............96, 97, 98, 99. Will add first and last=0 +99 =18

Will add second to second last =1 + 98 =18........and so on.

Since we have: 100/2 =50 pairs, each equaling 18, we therefore have:

50 x 18 =900

The only numbers left 100 to 110 itself, and since they have:1+2+3+4+5+6+7+8+9+10+2=57

900 + 57 =957 - Sum of all digits from 1 to 110.

Here is an extremrly basic mockup of a number table from 0-99 for you to reference as I make the adding process easier.

I notice a few patterns here. The tens digit remains the same. Let's attempt to add the numbers 0-9.

$\underbrace{0+\underbrace{1+\underbrace{2+\underbrace{3+\underbrace{4+5}+6}+7}+8}+9}\\ $

The numbers enclosed by the underbraces sum to nine (9+0=9, 1+8=9, 2+7=9, 3+6=9. 4+5=9). There are 5 lots of this occurring with the numbers one to nine, so the sum of the tens digit in the first columns equals $9*5\text{ or }45$. This phenomenon occurs ten times or once per column, so let's multiply this by 10. $45*10=450$. 

Now, let's do the ones column. The ones column is different because each column increments the amount by 1. The sum of the ones column can be represented by the following sequence. $0*10+1*10+2*10+...+8*10+9*10$. We can use algebra to group the common factor, so it simplifies to $10(1+2+3+...+8+9)$. Of course, we already know that the sum of the numbers 1-9 sums to 45 by our first calculation. 10 times that amount yields $450$. 

Of course, we have only dealt with the sum of the digits from 0-99. However, if you think about it, the sum of 100-109 should be the same as adding the ones digit and the hundreds digits. Of course, this is a special case since the tens digits are all 0. We already know that $1+2+3+...+8+9=45$. We know that the hundreds digit of the numbers 100-109 sums to $1*10=10$. However, we have forgotten 110, which has a value of 2. $45+10+2=57$ for the sum of the digits from 100-110. Now, let's add everything together.

$450+450+57=957$ or the sum of the digits from 0-110

Summed up on Wolfram/Alpha and it gives:

57 + sum_(b=0)^9( sum_(a=0)^9(a + b)) =957

https://www.wolframalpha.com/input/?i=57+%2B+%E2%88%91%5Ba%2Bb%5D,+a%3D0+to+9,+b%3D0+to+9

11  "1s"  in the hundreds digits  =  11

10 * 45  +  1  =  451  =  sum of  tens digits   

11 * 45  = 495  =  sum of units digits

So

11 +  10*45 + 1  +  11*45   =    957

Here is a very simple method to add them all up.

Write each number as two digits as follows:

01, 02, 03..........to 97, 98, 98

Now, since we have 100 numbers each of which consists of 2 digits, we, therefore, have:

2 x 100 = 200 individual digits. The average value of each digit is: 1+2+3+4+5+6+7+8+9 =45/10 =4.5.

200 x 4.5 = 900 - this is the sum of all numbers from 1 to 100. And to this add the sum of all numbers from 100 to 110 inclusive =57.

So, the total is 900 + 57 = 957 - the sum of all numbers from 1 to 110.