1. The problem asks for the digit in the 150th position when writing natural numbers in sequence: 123456789101112... 2. We write numbers consecutively and count digits: single-digit numbers (1 to 9) contribute 9 digits. 3. Two-digit numbers (10 to 99) contribute $90 \times 2 = 180$ digits. 4. Since 150 is greater than 9 but less than 9 + 180 = 189, the 150th digit lies within the two-digit numbers. 5. Subtract the 9 digits from single-digit numbers: $150 - 9 = 141$ digits into the two-digit numbers. 6. Each two-digit number has 2 digits, so find which number contains the 141st digit: $\frac{141}{2} = 70.5$. 7. This means the 141st digit is the first digit of the 71st two-digit number. 8. The first two-digit number is 10, so the 71st two-digit number is $10 + 71 - 1 = 80$. 9. The 141st digit in the two-digit section is the first digit of 80, which is 8. Final answer: The digit in the 150th position is **8**.