Question

What is the sum of all 10 digit integers?

Answer 1

wow, that is not a very relevant problem to anything

Answer 2

The sum of all integers from 1 to n is n(n + 1)/2

Answer 3

Interestingly, it is exactly the same as: \[\Huge \int\limits_{999999999}^{9999999999}x-\frac{1}{2}dx\]

Answer 4

10+11+12.........+99= ??
In general to find the sum of all the numbers from 1 to N:
1 + 2 + 3 + 4 + . . . . + N = (1 + N)*(N/2)
A good way to approach such problems is to consider smaller problems
and look for a patterm.

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:

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.

From the distributive law of multiplication, we know that

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

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

So, the sum of digits of the numbers 10-99 is:

9(0 + 1 +...+ 9) + 10(0 + 1 +...+ 9) = 19(0 + 1 + 2...9)

Since (0 + 1 +...+ 9) = 45, this equals:
19 x 45 = 855

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

Answer 5

This problem is very simple.
All you need to know is what I wrote above, that the sum of all whole numbers from 1 to n is n(n + 1)/2.
You are looking for the sum of all whole numbers from, 1,000,000,000 to 9,999,999,999.
First, using the formula, find the sum of all whole numbers from 1 to 9,999,999,999 (the largest 10-digit number). Then subtract from that result all the whole numbers less than 1,000,000,000. That is, subtract the sum of all whole numbers from 1 to 999,999,999 that you again find using the same formula.