Assume that all positive integers are written down consecutively from left to right as in 1234567891011...... The 6389th digit in this sequence is

The sequence is: 1 2 3 4 5 6 7 8 9 1 0 1 1 1 2 1 3 ...

Each number contributes a different number of digits depending on how many digits it has. So we group numbers by their digit-length and count how many digits each group contributes, until we reach the $6389^{th}$ digit.

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Range	How many numbers?	Digits per number	Total digits used
$1 - 9$	$9$	$1$	$9 \times 1 = 9$
$10 - 99$	$90$	$2$	$90 \times 2 = 180$
$100 - 999$	$900$	$3$	$900 \times 3 = 2700$
$1000 - 9999$	$9000$	$4$	$9000 \times 4 = 36000$

How do we get $90$ numbers from $10$ to $99$? → $99 - 10 + 1 = 90$. Same logic: $999 - 100 + 1 = 900$.

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Digits used up through all $3$-digit numbers:

$9 + 180 + 2700 = 2889$ digits

We've only used $2889$ digits so far, and we need the $6389^{th}$ digit. So we still need more digits from the $4$-digit numbers.

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Digits remaining:

$6389 - 2889 = 3500$ digits

These $3500$ digits will come from $4$-digit numbers (starting from $1000$). Each $4$-digit number uses $4$ digits, so:

$\dfrac{3500}{4} = 875$ (exactly, no remainder)

So we need exactly $875$ complete four-digit numbers.

Since the division is exact, we land perfectly on the last digit of the $875^{th}$ four-digit number.

If there was a remainder — say we got $875$ remainder $2$ — that would mean after writing $875$ complete $4$-digit numbers, we'd still need $2$ more digits, which would be the $1^{st}$ and $2^{nd}$ digits of the $876^{th}$ four-digit number.

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The first $4$-digit number is $1000$, so:

$875^{th}$ four-digit number $= 1000 + 875 - 1 = 1874$

Why subtract $1$? Because $1000$ itself is already the $1^{st}$ number, so the $875^{th}$ number is $1000 + 874 = 1874$, not $1000 + 875$.

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Since $3500 \div 4$ had no remainder, we land exactly on the last digit of $1874$.

The last digit of $1874$ is $4$.

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The $6389^{th}$ digit is $4$.