It takes 867 digits to number the pages of a book consecutively starting with 1.?

Counting number of pages:

1-9 pages 9 digits.

10-99 pages 180 digits.

Remaining digits (867-189)=678 digits.

226*3 = 678. So there are (226+99) = 325 pages.

Counting number of 3's:

Treating it as three digit number, where each x ,y and z can be filled with digits 0 to 9.

when x = 0, y can be 3 in 10 times and z can be 3 in 10 times. There are 20 such 3's.

When x = 1, the same logic applies and there are 20 such 3's.

When x = 2, the same logic applilesm and will be 20 such 3's.

When x = 3, there will be 26 numbers, and so z can be 3 in 26 times. Y can not be 3. When z = 3, y has to be 0, 1 or 2. So z can be 3 in 3 times.

So totally there will be 89 3's printed in all the pages together.

I am not sure about this but this is how I did it.....

the pages start from 1

thus no. of digits in single digit nos. would be 9.

for 2 digit nos. it would be 90*2=180

for 3 digit nos. it would be 900*3=2700

but total no. of digits is 876

so we have

9+90*2+(x-99)*3= 867

where

x= total no. of pages

thus, we have

x-99=226

x=325

therefore no. of pages in the book is 325.

thus, no. of times digit 3 is printed

from 1-100 = 20

from 101-200= 20

from 201-300= 20+1=21

from 301-325= 25+3=28

there for total no. of times 3 is printed =89

I guess this should be the answer

The problem is impossible. If you start with 1, there is no way to end up with 867 digits since if you work it out, you would use 187 digits from pages 1 to 99 and then you are left with 680. 680 isn't divisable by 3 even though you need to divide by three in order to get the number of additional pages.