Now I got all of yours attention, solve this problem.

A book has N pages, numbered the usual way, from 1 to N. The total number of digits in the page numbers is 1,095. How many pages does the book have?

Jhan bihana bihana ke dekhna paaiyela bhanera kholeko ta. Daka le jhukkai halyo ni

N(N+1)/2=1095.. Solve for N. that's the answer

From 1 to 9 ====> 9 digits
10 to 99 =====> 180 digits
.                           + ----------
Total of 1 digit and 2 digit no.s  = 189 digits
Now rest of the pages are the three numbered pages:
N - 189 = 1095 - 189 = 906
906/3 = 302

So the total pages: 9 + 90 + 302 = 401

1 to N = 1 to 129

Slackdemic got it right. Answer is 401.

Todays score:
Slackdemic +1
Sajhamitra -1
sara_solta11 -1

Here is my tricky question.
You have range of contiguous numbers in array from 1 to 1 million. Among of them one of the numbers is missing. Find out that missing number. You can use any data structure or use a simple math.

अहिलेको जमानामा यसो हल्का गुगल गर्नु पर्छ क्या, गुला खेलाउदै दिमाग चलाउन खोजेर मात्र हुदैन।

http://codepad.org/Hrj8BKZ7

int getMissingNo (int a[], int n)
{
    int i, total;
    total  = (n+1)*(n+2)/2;  
    for ( i = 0; i< n; i++)
       total -= a[i];
    return total;
}
 
/*program to test above function */
int main()
{
    int a[] = {1,2,4,5,6};
    int miss = getMissingNo(a,5);
    printf("%d", miss);
    /*getchar();*/
}

Do it with a datastructure.

if the numbers are in sequence, this function works fine. if its not in sequence its different story.

main method(){
int[] arrayofMillionNumbers = new int[100000];
for(int i = 1 ; i< arrayofMillionNumbers+1 ; i++){
arrayofMillionNumbers.add(i);
}
findMissing(arrayofMillionNumbers);
}
private int findMissing(int[] num) {
int missingNumber = 0;
for (int i = 0; i < (num.length - 1); i++) {
if ((num[i + 1] - num[i]) > 1) {
missingNumber = num[i] + 1;
break;
}
}
return missingNumber;
}

if the numbers are in sequence, this function works fine. if its not in sequence its different story.


m = missing number
x = 1;
for i = 1 to million
{
if x is not equal to i
{
m = x = missing number;
stop iteration;}
x = x + 1;
}
output m;

those numbers are continuous but distributed randomly. like 1,3,2,5,4.....etc

Add all the numbers in the array.
Then find n! ( that is 1,000,000!)
Subtract them, that would be number that is missing.

This is would be complete in O(n) which is better than sorting the number and then using one of above algorithm to find the missing one.

Yes that's true but there is another solution too. Hint: You can use an array.