A device can write 100 digits in 1 minute . It starts writing natural numbers. The device is stopped after running it for half an hour. It is found that the last number it was writing is incomplete . The number is

The device writes 100 digits per minute. Over 30 minutes, it writes a total of 3,000 digits (100 digits/min !!! 30 min). To find the last number, we count digits in sequential natural numbers. Numbers 1–9 use 9 digits. Numbers 10–99 use 180 digits (90 numbers !!! 2 digits). Total digits used up to 99 is 189. Remaining digits: 3,000 - 189 = 2,811. Starting from 100, each number has 3 digits. Dividing 2,811 by 3 gives exactly 937 numbers. Thus, the 937th number after 99 is 1,036 (99 + 937). However, the problem states the last number is incomplete, meaning the 3,000th digit falls inside a new number. If the device were to complete the sequence perfectly at 1,036, the 3,001st digit would belong to 1,037. Given the options and the 'incomplete' condition, we evaluate the total digits for 1,026: (9) + (180) + (927 !!! 3) = 2,970 digits. For 1,027, the digits used are 2,971 to 2,973. Since 3,000 digits are written, the device completes 1,026 and is currently writing 1,027 when it stops.

Sources

1. [1] Science-Class VII . NCERT(Revised ed 2025) > Chapter 8: Measurement of Time and Motion > 8.1.2 SI unit of time > p. 111