Hard Addition Question

[MathStudent1999]

The sum of all of the digits of the integers from 1 to 2008 is?

Can someone tell me how to calculate a problem like this?

 

[mmm4444bot]

I erased my comments.

 

[lookagain]

The sum of all of the digits of the integers from 1 to 2008 is?

mmm4444bot,

the proposed question is not about the sum of the integers
from 1 to 2008.

It is about the sum of all of the > > digits < < of those integers.


Your link concerns only the sum of the integers themselves.

 

[TchrWill]

The sum of all of the digits of the integers from 1 to 2008 is?

Can someone tell me how to calculate a problem like this?

Try adding them up as you see them.

The numbers 1 through 9 account for 9 digits.
The numbers 10 through account for [(99-10) +1]2 = 180 digits
The numbers 100 through 999 account for [(999-100) +1]3 = 2700 digits

l'll let you carry it to its finale.

 

[lookagain]

The sum of all of the digits of the integers from 1 to 2008 is?

Can someone tell me how to calculate a problem like this?

Tchrwill,

the original problem means the total sum of the digits in the all of the numbers,
not the total number of digits in those numbers.



MathStudent1999, is that what you meant?


Or, MathStudent1999, did you mean the total number of digits (as done by TchrWill)?



Examples:


The total number of digits in the numbers from 1 to 10, inclusive, is 11.


versus


The sum of all of the digits of the numbers 1 through 10 is 46.

(1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 1 + 0 = 46.)

 

[guitarguy]

This is a tough arithmetic problem.

To sum up the digits from one to nine add 1+2+3+4+5+6+7+8+9 which will equal 45. This sequence will occur 200 times in the integers 1 to 2000.

Use the same reasoning for the ten's and hundreds digits.

Keep at it and let me know if you need more help.