Solution - Base conversion

$${\left(5D9\right)}_{16}$$

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

$$5\cdot {16}^2+13\cdot {16}^1+9\cdot {16}^0$$

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

$$1497$$

Expand each digit by place value: $$5·{16}^2$$
+ $$13·{16}^1$$
+ $$9·{16}^0$$. This gives $$1,497$$ in base 10.

$$1497\left(base10\right)$$

Use repeated division by the target base.

$$\frac{1497}{10}$$

Record quotient and remainder for this division cycle.

$$1497=10\cdot 149+7$$

Record quotient and remainder for this division cycle.

$$149=10\cdot 14+9$$

Record quotient and remainder for this division cycle.

$$14=10\cdot 1+4$$

Record quotient and remainder for this division cycle.

$$1=10\cdot 0+1$$

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

$$1497$$

Final converted representation in the target base.

$${\left(1497\right)}_{10}$$

Repeatedly divide $$1,497$$ by $$10$$, collect remainders, and read them in reverse to get $$1,497$$.