Question

Hello, I need to solve this:

I need to find a sum of digits for all natural numbers:
1) from 1 to 2000 (1 and 2000 includes)

Answer 1

@ash2326

Answer 2

There's this story of Karl Friedrich Gauss and his teacher, who was so lazy, he'd just ask his students to add up sums like this (MANUALLY!) which would, understandably take the whole period, while the teacher just slacks off... until Gauss found a formula for solving such sums quickly :D

Formula for the sum of all numbers from 1 to n is
\[\huge \frac{n(n+1)}{2}\]

Good hunting :)

Answer 3

So, i need to find n:

\[a _{n}=a _{1}+(n-1)d\]
\[2000=1+(n-1)1\]
n=2000
\[\frac{ 2000(2000+1) }{ 2 }=2001000\]?

Answer 4

Yeah, that's it :D
n is the last number you add

Answer 5

but I have another problem:
find a sum of digits:
a) 1 to 10^n
I get n=10^n
\[\frac{ 10^{n}(10^{n}+1) }{ 2 }\] is this an answer?

Answer 6

sum of digits?
or sum of number?

Answer 7

Digits

Answer 8

like 321=3+2+1

Answer 9

and for all numbers from 1 to 2000

Answer 10

well, is it an wrong formula?

Answer 11

Hmmm... I may have to rethink this... stand by :D

Answer 12

So, you take the sum from 1 to 2000, and add the digits?

Answer 13

well, you need to find a sum of a digits from 1 to 2000.. for exmple: 1+2+3+4+5+6+7+8+9+1+0+1+1+...+2+0+0+0

Answer 14

any ideas?

Answer 15

?

Answer 16

That seems beyond my knowledge o.O
I'll look into it.