Hi Ozyhibby;

Don't you think you are better off doing this on your own? Even if you get it wrong it is much better than handing in someone else's work and getting it right.

What have you done to solve this problem?

Supposing it was the numbers from 1 to 100. How many would that be? Surely the thought has occurred to you of writing it out and actually counting them?

Hi;

Well, what did she do by herself? What grade is she in? Would she be content with a lot of casework? Without seeing her work I can not tell what she really needs.

Try to understand that math is a hands on activity. To understand any of it you have to do it.

I have answered the question in later posts but I sure wished I could have got some interaction from you. Also, I do not agree that this is a useless problem.

Hi Ozyhibby;

I was hoping to get you to do a little math. You would have found it to be fun. That should not stop me from having all the fun. Also, some other people might see this.

Break the problem up into segments. This is how many numbers are needed to write 1 to 9.

This corresponds to [1,2,3,4,5,6,7,8,9,0 ] meaning that there is one 5 needed and no zeros. To get the number needed from 10 to 99 we just multiply each number in vector A) by 10 and add 9 to each.

Meaning 9 zeroes are necessary and 19 of every other number.

To get the number needed from 100 to 999 we just multiply each number in B) by 10 and add 90 to each.

So for instance we need 180 zeros. We are done with the numbers from 1 to 999.

To get 1000 - 1999 is very simple. We know that there are 1000 ones used in th first digit and the other 3 digits are the same as above. We just have to add A,B and C.

Remember the 1000 ones.

We add the amount needed from 1 to 999 and from 1000 to 1999.

Now we are only left with the numbers 2000,2001,2002,2003, ... 2011. We count them up by hand and get the vector:

We add them to the subtotal:

So the answer is the vector:

We need 1604 ones and 513 zeros.