Conversion from decimal base into another system of numeration

Example

How would number $7$ be expressed in a numeration system of base $5$?

To find this, it will be necessary to divide $7$ by $5$ and keep the quotient and the remainder:

$(7{)}_{10}\Rightarrow \begin{array}{rcl}& 7& |\underset{―}{5}\\ & \boxed{\text{2}}& \boxed{\text{1}}\end{array}$

The first digit of the number we are looking for is the result of the division, and the second one is the remainder.

This way, the equivalent of $7$ in base $5$ will be:

$(12{)}_5$

We can prove that the operation is correct by decomposing the number:

$(12{)}_5=1\cdot 5^1+2\cdot 5^0=5+2=7$

Example

Following the same line, the number $13$ is equivalent in binary (system in base $2$) to:

$\begin{array}{rclrclr}& (13{)}_{10}\Rightarrow & 13& |\underset{―}{2}& & & \\ & & \boxed{\text{1}}& 6& |\underset{―}{2}& \\ & & & \boxed{\text{0}}& 3& |\underset{―}{2}\\ & & & & \boxed{\text{1}}& \boxed{\text{1}}\end{array}$

The first digit of the number we are looking for is the result of the last division, the second digit is the remainder of the same division, and the third and fourth digits are the remainders of the previous divisions, so the obtained number is:

$(13{)}_{10}=(1101{)}_2$

It is possible to prove again that the calculations are correct by decomposing the obtained number:

$(1101{)}_2=1\cdot 2^3+1\cdot 2^2+0\cdot 2^1+1\cdot 2^0=8+4+0+1=13$

Example

Convert $47$ into hexadecimal system:

Hexadecimal means base $16$, so it will be necessary to divide $47$ by $16$ as many times as possible to find the number:

$(47{)}_{10}\Rightarrow \begin{array}{rcl}& 47& |\underset{―}{16}\\ & \boxed{\text{15}}& \boxed{\text{2}}\end{array}$

So:

$(47{)}_{10}=(2(15){)}_{16}=(2F{)}_{16}$

Remember that the symbol to express $15$ in hexadecimal is $F$.

Decomposing the number in powers of $16$ is a way of proving that the result is correct:

$(2F{)}_{16}=2\cdot {16}^1+15\cdot {16}^0=32+15=47$

Example

To convert the following number into binary system:

$(115{)}_6$

First, it is necessary to convert it into decimal system and then into binary by doing all the possible integer divisions by $2$.

$(115{)}_6=1\cdot 6^2+1\cdot 6^1+5\cdot 6^0=36+6+5=47$

$\begin{array}{rclrclrcl}& (47{)}_{10}\Rightarrow & 47& |\underset{―}{2}& & & & & \\ & & \boxed{\text{1}}& 23& |\underset{―}{2}& & & & \\ & & & \boxed{\text{1}}& 11& |\underset{―}{2}& & & \\ & & & & \boxed{\text{1}}& 5& |\underset{―}{2}& & \\ & & & & & \boxed{\text{1}}& 2& |\underset{―}{2}\\ & & & & & & \boxed{\text{0}}& \boxed{\text{1}}\end{array}$

So that:

$(115{)}_6=(101111{)}_2$