A probability problem by Ayush Jain

we first count all zeroes at ones place Therefore the number will be X0 where X can take values from 1 to 333 =333

Then we count the zeroes at tens place Therefore the number will be X0Y where X can take values from 1 to 33 & Y can take values from 0 to 9=10 i.e 33 x 10 = 330

Now we count the zeroes at hundreds place Therefore the number will be X0YZ where X can take values from 1 to 33=33, Y can take values from 0 to 9=10 & Z can take values from 0 to9=10 i.e 3x10x10=300

Total= 333+330+300=963

First use Java to make this code-

       public class asdjh{
          public static void main(String[] args){
               int a = 0;
               for(int i = 0;i<3333; i++){
                   a++;
                   System.out.print(a);
               }
           }
       }

Then, copy the string that comes out and paste onto word.

Finally, press Ctrl+F and for Replace, put 0 for find and whatever for the other, and click REPLACE ALL.

It should say "Word has made $\boxed{963}$ replacements."

That's the shortcut way to do it.

I have just counted and wrote......
1-99=9 zeroes
100-199=20 zeroes
200-299=20 zeroes
.
.
.
900-999=20 zeroes
Therefore, 1-999= $\boxed{189}$ zeroes
1000-1099=120 zeroes
1100-1999=180 zeroes
Therefore, 1000-1999= $\boxed{300}$ zeroes
Therefore, 2000-2999= $\boxed{300}$ zeroes
3000-3099=120 zeroes
3100-3333=54 zeroes
Therefore, 3000-3333= $\boxed{174}$ zeroes
Therefore, 1-3333= $\boxed{963}$ zeroes