Solution - Base conversion

$${\left(9D8\right)}_{16}$$

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

$$9\cdot {16}^2+13\cdot {16}^1+8\cdot {16}^0$$

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

$$2520$$

Expand each digit by place value: $$9·{16}^2$$
+ $$13·{16}^1$$
+ $$8·{16}^0$$. This gives $$2,520$$ in base 10.

$$2520\left(base10\right)$$

Use repeated division by the target base.

$$\frac{2520}{10}$$

Record quotient and remainder for this division cycle.

$$2520=10\cdot 252+0$$

Record quotient and remainder for this division cycle.

$$252=10\cdot 25+2$$

Record quotient and remainder for this division cycle.

$$25=10\cdot 2+5$$

Record quotient and remainder for this division cycle.

$$2=10\cdot 0+2$$

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

$$2520$$

Final converted representation in the target base.

$${\left(2520\right)}_{10}$$

Repeatedly divide $$2,520$$ by $$10$$, collect remainders, and read them in reverse to get $$2,520$$.