Solution - Base conversion

$${\left(FAD\right)}_{16}$$

Expand digits with place-value powers in the source base.

$$15\cdot {16}^2+10\cdot {16}^1+13\cdot {16}^0$$

Sum the expanded terms to obtain the base-10 value.

$$4013$$

Expand each digit by place value: $$15·{16}^2$$
+ $$10·{16}^1$$
+ $$13·{16}^0$$. This gives $$4,013$$ in base 10.

$$4013\left(base10\right)$$

Use repeated division by the target base.

$$\frac{4013}{10}$$

Record quotient and remainder for this division cycle.

$$4013=10\cdot 401+3$$

Record quotient and remainder for this division cycle.

$$401=10\cdot 40+1$$

Record quotient and remainder for this division cycle.

$$40=10\cdot 4+0$$

Record quotient and remainder for this division cycle.

$$4=10\cdot 0+4$$

Read remainders from last to first to form target-base digits.

$$4013$$

Final converted representation in the target base.

$${\left(4013\right)}_{10}$$

Repeatedly divide $$4,013$$ by $$10$$, collect remainders, and read them in reverse to get $$4,013$$.