How many times does the number 9 appear in total in all the even numbers from 2 to 9999? (This was the question that I came across in one of the Cracku questions set, but not able to find it in the book marks now,) Could you please help here with the solution?

I am getting 770..

Hi Aravind, If we consider all even numbers from 0 to 9999, we get (9999 + 1)/2 = 5000 such numbers. In these 5000 numbers, the unit's digit is always 0, 2, 4, 6 or 8. Hence, if we consider the number to be ABCD D can not be 9. A, B, and C can take any values from 0 to 9. So, the total number of digits comprising A, B and C in all 5000 appearances = 5000 x 3(for A, B and C) = 15000 Now, among these 15000 digits, digits 0 to 9 appear an equal number of times. Hence, 9 will appear 1/10 th of the times. Hence, total number of appearances of 9 = 15000/10 = 1500. Hope this helps!