Problems from Conversion from decimal base into another system of numeration

Development:

We have to do all the possible integer divisions of the numbers by $3$:

$\begin{array}{rclrcl}& (17{)}_{10}\Rightarrow & 17& |\underset{―}{3}& & \\ \\ & & \boxed{\text{2}}& 5& |\underset{―}{3}& \\ \\ & & & \boxed{\text{2}}& \boxed{\text{1}}\end{array}$

So that:

$(17{)}_{10}=(122{)}_3$

In the second case:

$\begin{array}{rclrclrc}& (89{)}_{10}\Rightarrow & 89& |\underset{―}{3}& & & & \\ \\ & & \boxed{\text{2}}& 29& |\underset{―}{3}& & & \\ \\ & & & \boxed{\text{2}}& 9& |\underset{―}{3}& & \\ \\ & & & & \boxed{\text{0}}& 3& |\underset{―}{3}\\ \\ & & & & & \boxed{\text{0}}& \boxed{\text{1}}\end{array}$

Then,

$(89{)}_{10}=(10022{)}_3$

In the third number of the exercise:

$\begin{array}{rclrclrc}& (121{)}_{10}\Rightarrow & 121& |\underset{―}{3}& & & & \\ \\ & & \boxed{\text{1}}& 40& |\underset{―}{3}& & & \\ \\ & & & \boxed{\text{1}}& 13& |\underset{―}{3}& & \\ \\ & & & & \boxed{\text{1}}& 4& |\underset{―}{3}\\ \\ & & & & & \boxed{\text{1}}& \boxed{\text{1}}\end{array}$

The equivalent is:

$(121{)}_{10}=(11111{)}_3$

Finally, in the last exercise it is necessary to combine the decomposition with the divisions, since the number is not in base $10$ but in base $18$.

First, into decimal:

$(3D{)}_{18}=(3(13){)}_{18}=3\cdot {18}^1+13\cdot {18}^0=54+13=67$

Now, to find the equivalent of $67$ in ternary, it is necessary to divide the above mentioned number successively by $3$:

$\begin{array}{rclrclr}& (67{)}_{10}\Rightarrow & 67& |\underset{―}{3}& & & \\ \\ & & \boxed{\text{1}}& 22& |\underset{―}{3}& & \\ \\ & & & \boxed{\text{1}}& 7& |\underset{―}{3}\\ \\ & & & & \boxed{\text{1}}& \boxed{\text{2}}\end{array}$

So

$(3D{)}_{18}=(2111{)}_3$

Solution:

$(122{)}_3$

$(10022{)}_3$

$(11111{)}_3$

$(2111{)}_3$

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