Question

Let N is the number of times digit 5 is written when listing all natural numbers from 1 to $10^5$. Then the sum of digits of N is

Answer 1

5

Answer 2

Consider numbers from 00000 to 99999 (100,000 numbers). In each of the 5 digit positions, each digit (0–9) appears equally often, i.e., 100000/10 = 10,000 times.

• Thus, digit 5 appears 5 × 10,000 = 50,000 times.
• The number 100000 (which is outside the 00000 to 99999 range) does not contain the digit 5.
• So, N = 50,000 and the sum of its digits is 5 + 0 + 0 + 0 + 0 = 5.