Check all true statements.

In duodecimal $($base $12),$ every digit is $0,1,2,3,4,5,6,7,8,9,A$ or B $.$

Expressed in base$-$n$,$ the integer n is $"10".$

The fast modular exponentiation algorithm takes advantage of the binary representation of the exponent.

The fast modular exponentiation algorithm computes bn mod $m$ in only about $lo{g}_2(n)$ steps. This makes it practical even when $n$ is large.

You can convert a number from binary to octal by grouping the digits $("$bits$")$ of the binary number into groups of $3,$ going from right to left. If the number of bits is not a multiple of $3,$ you may have to

add one or two leading $0$ bits on the left side. Then you convert each group of $3$ bits into one octal digit.

Among all base $b$ representations of a positive integer $n,$ the binary one is always at least as long as any other $($in terms of number of digits.$)$

You can convert a number from decimal to binary by replacing each decimal digit separately by its corresponding binary representation.

In ternary $($base $3),$ every digit is a $0,1$ or $2.$

If $k$ is an integer greater than $1$ and $n$ is a positive integer that is not a power of $k,$ then $n$ has $|$~$lo{g}_k(n)$~$|$ digits in base $k.$

In base $b,$ it is easy to see whether an integer is qumultiple of $b.$ Its last digit is zero in that case.

Given a positive integer $n$ and a base $b,$ we can find the last digit of the base bexpansion of $n$ by performing the division algorithm to find $n=bq+r.$ The remainder $r$ is the last digit. By repeating the process with $q$ instead of $n,$ we find the next digit, and so on$.$

Expressed in base$-$n$,$ the integer $n^2$ is $"100".$

You can convert a number from hexadecimal to binary by replacing each hexadecimal digit