Throwing away!

In this case the title is only a clue.

In a number base, $b$ , apply the following throwingAway procedure on a number $n$ : while the current version of $n$ is greater or equal to $b$ , replace $n$ with the sum of its digits in base $b$ arithmetic. If the single digit result is equal to $b-1$ , then replace that result with $0$ .

Two examples:

$n=57$ and $b=10$ , since $57\geq 10$ , add $5+7\Longrightarrow 12$ , since $12\geq 10$ , add $1+2\Longrightarrow 3$ , since $3\ngeq 10$ , the answer is $3$ .

$n=\text{DEADC0DE}_{16}$ and $b={10}_{16}$ , since $\text{DEADC0DE}_{16}\geq {10}_{16}$ , add $\text{D}_{16}+\text{E}_{16}+\text{A}_{16}+\text{D}_{16}+\text{C}_{16}+\text{0}_{16}+\text{D}_{16}+\text{E}_{16}\Longrightarrow 59_{16}$ , since $59_{16}\geq {10}_{16}$ , add $5_{16}+9_{16}\Longrightarrow \text{E}_{16}$ , since $\text{E}_{16}\ngeq 10$ , the answer is $\text{E}_{16}$ .

Next, form a histogram of applying the throwingAway procedure in base 60 to all positive integers from 1 to $({10}_{60})^6-1$ .

Select the most frequently occurring answer, if more than digit is the most frequent, then any will do.

The answer is the frequency selected. What is that frequency?