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Get access to the detailed solutions to the previous years questions asked in IIM IPMAT exam
There are nine single-digit numbers. Hence to write to them all, we’ll need 9 x 1 = 9 digits; similarly, there are 90 two-digit numbers. So, we’ll need 90 × 2 = 180 digits.
For 3 digit numbers, 900 x 3 = 2700 digits
It means that till the point we have written all the 3 digits numbers, we’ve used 9 + 180 + 2700 = 2889 digits.
Now, we have 6389 – 2889 = 3500 digits left.
That means we can write 3500/4 = 875 digits four-digit number, i.e., till 1874 (999 + 875)
Our sequence will look like 1234...........18731874
∴ The last digit will be 4