Question

.What is the sum of all the digits that appear in the numbers from 1 to 100 (including 1 and 100)? (Notice that the question is asking for sum of the digits, not the sum of the numbers)

Answer 1

i don't know , that's why i asked..

Answer 2

so the digits 1to 10= 46
11 to 20 = 56 = 10 + 46
21 to 30 = 66 = 20 +46

develop the pattern

Answer 3

Digits in the units place from one to ten sum to 45. We need these digits ten times. That would be 450. The digits in the tens place also go from one to nine, again ten times for another 450 that would be 900 so far. The only digit in 100 that adds to the total is the one in the hundreds place. I think the total is nine hundred one.

Answer 4

Let's first ask, what is the sum of the digits 0-9? Clearly, this is
0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9, which equals 45.

Next question: What is the sum of the digits of the numbers 10-19?
We know that these numbers all have a "1" in the tens place, and a
number 0-9 in the units place. There are ten of these numbers, so
there are ten 1's and the digits 0, 1, 2, 3, 4...9. So, the sum of
these numbers must be 10x1 + (0 + 1 +...+ 9).

What about the sum of the digits of the numbers 20-29? Again, we need
to add ten 2's and the digits 0-9, which is 10x2 + (0 + 1 +...+ 9).

Do you see the pattern? So, when we add up the digits of all numbers
0-99, the expression is:

(0 + 1 +...+ 9) + The numbers 0-9
10x1 + (0 + 1 +...+ 9) + The numbers 10-19
10x2 + (0 + 1 +...+ 9) + The numbers 20-29
10x3 + (0 + 1 +...+ 9) + The numbers 30-39
10x4 + (0 + 1 +...+ 9) + The numbers 40-49
10x5 + (0 + 1 +...+ 9) + The numbers 50-59
10x6 + (0 + 1 +...+ 9) + The numbers 60-69
10x7 + (0 + 1 +...+ 9) + The numbers 70-79
10x8 + (0 + 1 +...+ 9) + The numbers 80-89
10x9 + (0 + 1 +...+ 9) And, finally, the numbers 90-99.

Thank goodness we can simplify this! From the distributive law of
multiplication, we know that

(10x1 + 10x2 +...+ 10x9) = 10(1 + 2 ...9).

We also know that there are ten additions of (0 + 1 +...+ 9),
which is another way of saying that there are ten times
(0 + 1 +...+ 9), 10(0 + 1 +...+ 9).

So, the sum of digits of the numbers 1-99 is:

10(0 + 1 +...+ 9) + 10(0 + 1 +...+ 9) = 20(0 + 1 + 2...9)

Since (0 + 1 +...+ 9) = 45, this equals:
20 x 45 = 900

We add the digit 1 from "100" because we want the sum of digits 1-100,
and the final answer is 901.

-Doctor Chuck, The Math Forum

http://mathforum.org/library/drmath/view/57919.html