base 13 12 11 10 HELP

Question

Convert the last four digits 122917 number to base 13, where A, B, and C correspond to 10, 11 and 12

does anyone know how would i start this ?

Answer 1

We write $$2917=\sum_{i=0}^nx_i13^i=x_0+13(x_1+13(x_2+...))$$ All $x_i$ are between $0$ and $C=12$. From the above equation, $x_0$ is the reminder of the division of number $2917$ by $13$. Subtract the reminder from $2917$, divide by $13$ and you get $$\frac{2917-x_0}{13}=x_1+13(x_2+...)$$ This means that $x_1$ is the reminder of division of $\frac{2917-x_0}{13}$ by $13$. And continue like that, until the result of the division is less than $13$.

Answer 2

$122917 = 12*10^4 + 2917 = 12*10^4 + 13^3$

$10,000 = 4*2917 + 1212 = 4*13^3 + 12*10 + 12$

So $122917_{10} = C*(4*13^3 + C*A + C)+ 13^3=C*(4000_{13} + C*A + C)+1000_{13}$

Now $C$ is like $9$ but in base $13$. $C*k = 13*k-k$

So $C*A = A0 - A$ so

$122917 = C*(4000 + A0 -A+C) = C*(4000 + A0 + 2) = C*40A2+1000$

$= 40A20 - 40A2$.

$0 - 2= B $; borrow the $1$.

$(2-1)-A=1-A =14-A= 4$; borrow the $1$.

$(A-1)-0 = 9$

And $40 - 4 = 39$

So $= 40A20 - 40A2+ 1000=3994B+ 1000 = 3A94B$