Solución - Tiger Algebra Solver

$${\left(7C3\right)}_{16}$$

Pasos: $$7·{16}^2$$
+ $$12·{16}^1$$
+ $$3·{16}^0$$ = $$1.987$$.

$$7\cdot {16}^2+12\cdot {16}^1+3\cdot {16}^0$$

Pasos: $$7·{16}^2$$
+ $$12·{16}^1$$
+ $$3·{16}^0$$ = $$1.987$$.

$$1987$$

Pasos: $$7·{16}^2$$
+ $$12·{16}^1$$
+ $$3·{16}^0$$ = $$1,987$$.

$$1987\left(base10\right)$$

$${1.987}_{10}={1.987}_{10}$$.

$$\frac{1987}{10}$$

$${1.987}_{10}={1.987}_{10}$$.

$$1987=10\cdot 198+7$$

$${1.987}_{10}={1.987}_{10}$$.

$$198=10\cdot 19+8$$

$${1.987}_{10}={1.987}_{10}$$.

$$19=10\cdot 1+9$$

$${1.987}_{10}={1.987}_{10}$$.

$$1=10\cdot 0+1$$

$${1.987}_{10}={1.987}_{10}$$.

$$1987$$

$${1.987}_{10}={1.987}_{10}$$.

$${\left(1987\right)}_{10}$$

$$1,{987}_{10}=1,{987}_{10}$$.