In order to sort list of numbers using radix sort algorithm, we need to get…

To extract the i-th digit of a positive decimal integer n, counted from the right and starting at i = 1, first remove the rightmost (i − 1) digits with an integer (floor) division by 10 raised to the power (i − 1), then isolate the resulting units digit with modulo 10. Formally: digiti(n) = ⌊n / 10i − 1⌋ % 10. Using a ceiling function in place of the floor, or dividing by the wrong power of 10, generally returns a different digit — though not on every input, since ceiling and floor coincide when the quotient is already a whole number, and a shifted position can occasionally coincide with the target digit by chance.

1. Compute the divisor: 10 raised to the power (i − 1) = 10 raised to the power (3 − 1) = 102 = 100.

2. Divide n by the divisor: 250 / 100 = 2.5.

3. Apply the floor function: ⌊2.5⌋ = 2.

4. Apply modulo 10: 2 % 10 = 2 — this is the 3rd digit of 250 counted from the right (250 read right-to-left gives digits 0, 5, 2 at positions 1, 2, 3).

Cross-check with i = 1, the units digit (which is 0 for n = 250): ⌊250 / 100⌋ % 10 = ⌊250 / 1⌋ % 10 = 250 % 10 = 0 — correct, confirming the formula generalizes to every position.

• ⌈n / 10i − 1⌉ % 10 rounds the quotient up before the modulo instead of down, so at i = 3 it gives ⌈2.5⌉ % 10 = 3 % 10 = 3, not 2 — because the ceiling function rounds any non-integer quotient up to the next whole number, it returns a different digit than the floor-based formula whenever the true quotient is not already an integer.

• ⌊n / 10i⌋ % 10 divides by one extra power of 10, so at i = 3 it gives ⌊250 / 1000⌋ % 10 = 0 % 10 = 0 — the value that belongs to the (i + 1)-th position, not the i-th.

• ⌈n / 10i⌉ % 10 combines both deviations — the extra power of 10 and the upward rounding — giving ⌈250 / 1000⌉ % 10 = 1 % 10 = 1 at i = 3, which matches neither the i-th nor the (i + 1)-th digit.

So ⌊n / 10i − 1⌋ % 10 is the expression that returns the i-th digit from the right for every valid i.