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03242009  #1 
Join Date: Dec 2008
Posts: 249

How Many Ones?
How many times does the digit “1” appear in all the whole numbers between 0 and 999,999? For example “1” appears 3 times in the number 1,101 and 2 times in 101,058.

03252009  #2 
Join Date: Mar 2005
Posts: 10,609

This is an interesting problem. My first thought is looking at every place value and calculating how many times "1" can possibly appear. I'm sure there's a variety of strategies.
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03272009  #3 
Join Date: Mar 2009
Posts: 48

This kind of challenges make the people think........ I will try to find the answer. 
04202009  #4 
Join Date: Mar 2009
Posts: 4

Using Mathematica the answer is 600,000...
Here is the code: For[x = 1; y = 0, x <= 999999, x++; y = y + StringCount[ToString[x], "1"]]; Print[y] 
04202009  #5  
Join Date: Dec 2008
Posts: 249

Quote:
There are 1,000,000 numbers between 0 and 999,999. We can think of each of these numbers as having six digits whether 000,001 or 999,973. So if there are 1 million numbers with six digits each then there are six million total digits. Since we are going from 000,000 to 999,999 and there are ten digits (0,1,2,…,8,9), then each digit is used the same number of times. So there must be 6,000,000 / 10 = 600,000 of each digit used Therefore, there are 600,000 “1s” between 0 and 999,999. Also, there are: 1 "1" between 0 and 9 20 "1's" between 0 and 99 300 "1's" between 0 and 999 4,000 "1's" between 0 and 9,999 50,000 "1's" between 0 and 99,999 600,000 "1's" between 0 and 999,999 etc 

05012009  #6 
Join Date: Apr 2009
Posts: 5

That's cool, I couldn't figure it out.

08202010  #7  
Join Date: May 2010
Posts: 60

I like that
Quote:
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09032010  #8 
Join Date: Sep 2010
Posts: 3


02202010  #9 
Join Date: Feb 2010
Posts: 388

904 times idk?

05162015  #10 
Join Date: May 2015
Posts: 136

if you added just one 1 it would be 1,000,000 but I still don't know what it is its just a guess
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