1234567891011121314..........
What is the 2020th digit in the above sequence
Please answer my question
impatient much?
1-digit numbers: 9
2-digit numbers: 90
so starting at 100, that has used up 9+2*90 = 189 digits
2020-189 = 1831
That's 610 3-digit numbers, plus 1 digit.
so, what is the 611th 3-digit number?
To find the 2020th digit in the given sequence, we can identify that the sequence starts with the numbers 1, 2, 3,... and continues in the same pattern.
The first 9 digits (1 to 9) occupy positions 1 to 9 in the sequence. So, we can determine that the 2020th digit will lie beyond this range.
The next 90 digits (10 to 99) occupy positions 10 to 189 in the sequence. By subtracting 189 from 2020, we are left with 1831, which represents the number of remaining digits we need to account for.
The next 900 digits (100 to 999) occupy positions 190 to 2889 in the sequence. Subtracting 2889 from 1831 leaves us with -1058, indicating that the 2020th digit is not within this range.
The next 9000 digits (1000 to 9999) occupy positions 2890 to 11889 in the sequence. By subtracting 11889 from 1831, we are left with -10058, again indicating that the 2020th digit is not within this range.
So, we can conclude that the 2020th digit lies within the range of 10000 to 10999.
The number of digits in this range is (10999 - 10000 + 1) = 1000.
By subtracting 1831 from 1000, we are left with 169, which indicates that the 2020th digit is the 169th digit within the range of 10000 to 10999.
Since we know that each number within this range has 4 digits, we can divide 169 by 4 to find the position within a specific number.
169 divided by 4 equals 42 with a remainder of 1.
This means that the 2020th digit is the 1st digit in the number 10000 + 42 = 10042.
Therefore, the 2020th digit in the given sequence is 1.