Solution - Base conversion

$${\left(3B5\right)}_{16}$$

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

$$3\cdot {16}^2+11\cdot {16}^1+5\cdot {16}^0$$

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

$$949$$

Expand each digit by place value: $$3·{16}^2$$
+ $$11·{16}^1$$
+ $$5·{16}^0$$. This gives $$949$$ in base 10.

$$949\left(base10\right)$$

Use repeated division by the target base.

$$\frac{949}{10}$$

Record quotient and remainder for this division cycle.

$$949=10\cdot 94+9$$

Record quotient and remainder for this division cycle.

$$94=10\cdot 9+4$$

Record quotient and remainder for this division cycle.

$$9=10\cdot 0+9$$

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

$$949$$

Final converted representation in the target base.

$${\left(949\right)}_{10}$$

Repeatedly divide $$949$$ by $$10$$, collect remainders, and read them in reverse to get $$949$$.