Solución - Tiger Algebra Solver

$${\left(CDA\right)}_{16}$$

Pasos: $$12·{16}^2$$
+ $$13·{16}^1$$
+ $$10·{16}^0$$ = $$3.290$$.

$$12\cdot {16}^2+13\cdot {16}^1+10\cdot {16}^0$$

Pasos: $$12·{16}^2$$
+ $$13·{16}^1$$
+ $$10·{16}^0$$ = $$3.290$$.

$$3290$$

Pasos: $$12·{16}^2$$
+ $$13·{16}^1$$
+ $$10·{16}^0$$ = $$3,290$$.

$$3290\left(base10\right)$$

$${3.290}_{10}={3.290}_{10}$$.

$$\frac{3290}{10}$$

$${3.290}_{10}={3.290}_{10}$$.

$$3290=10\cdot 329+0$$

$${3.290}_{10}={3.290}_{10}$$.

$$329=10\cdot 32+9$$

$${3.290}_{10}={3.290}_{10}$$.

$$32=10\cdot 3+2$$

$${3.290}_{10}={3.290}_{10}$$.

$$3=10\cdot 0+3$$

$${3.290}_{10}={3.290}_{10}$$.

$$3290$$

$${3.290}_{10}={3.290}_{10}$$.

$${\left(3290\right)}_{10}$$

$$3,{290}_{10}=3,{290}_{10}$$.