if one write down all the numbers from 1 to 10^9, How many times was the zero digit written?

Answer 1

Okay, the guy before me is right. I counted wrong.

The original answer I have was:

number of the form

1xx xxx xxx to 9 xx xxx xxx:
9 * 10^8 such numbers, 8 digits for x's, and 1/10 of the digits are 0's.
720 000 000

1x xxx xxx to 9x xxx xxx
9 * 10^7 * 7 * 1/10 =
63 000 000

+ 5 400 000
+ 450 000
+ 36 000
+ 2 700
+ 180
+ 90
+ 9

+ 9 for the billion, itself.

Okay, I yield, I need dinner.

Answer 2

788,888,898
= 9*10^8 - 111,111,111 + 9

First, assume every number 0 - 999,999,999
IS written with leading zeroes so that every number has 9 digits. Then, by symmetry, every digit 0 - 9 occurs the same number of times. As there are 10^9 numbers each with 9 digits, there are 9*10^9 digits and each digit 0-9 occurs one-tenth as many or 9*10^8 times.

Then, subtract all those leading zeros. There are
10^8 leading zeroes in the first place,
10^7 leading zeroes in the second,
10^6 in the third,
etc., to
10 leading zeroes in the eighth place, &
1 leading zero in the ninth place (this occured in the number 0 itself, which was written with all nine zeroes.)
So, 10^8+10^7+10^6+ ... + 1 or 111,111,111 too many (leading) zeroes: we subtract these.

Finally, add the 9 zeroes in 10^9

Answer 3

think of roughly this for a 2d. Your variety is from a million to 999 999. enable us to visualize us counting making use of bases of 10. f*10^5 + e*10^4 + d*10^3 + c*10^2 + b*10^a million + a*10^0 the place f has integers from 0 - 9 e d c b have integers from 0 -9 a has integers from a million-9 i've got not got plenty time authentic now, i will examine this when I get homestead. I wrote a application in Java to envision.. i'm getting 8 looks 600000 cases between a million and 999 999

Answer 4

45 times

Answer 5

hmm.
109?
is that the number you meant to type?
typo!