把下面的二进制数改写成十进制数。
$(101001)_2$
$=(100000)_2+(1000)_2+(1)_2$
$=(\_\_\_\_\_\_×\_\_\_\_\_\_×\_\_\_\_\_\_×\_\_\_\_\_\_×\_\_\_\_\_\_)_{10}+(\_\_\_\_\_\_×\_\_\_\_\_\_×\_\_\_\_\_\_)_{10}+(\_\_\_\_\_\_)_{10}$
$=(\_\_\_\_\_\_)_{10}+(\_\_\_\_\_\_)_{10}+(\_\_\_\_\_\_)_{10}$
$=(\_\_\_\_\_\_)_{10}$
$(11101)_2$
$=(\_\_\_\_\_\_)_2+(\_\_\_\_\_\_)_2+(\_\_\_\_\_\_)_2+(\_\_\_\_\_\_)_2$
$=(\_\_\_\_\_\_)_{10}+(\_\_\_\_\_\_)_{10}+(\_\_\_\_\_\_)_{10}+(\_\_\_\_\_\_)_{10}$
$=(\_\_\_\_\_\_)_{10}$

$(101001)_2$
$=(100000)_2+(1000)_2+(1)_2$
$=(2×2×2×2×2)_{10}+(2×2×2)_{10}+(1)_{10}$
$=(32)_{10}+(8)_{10}+(1)_{10}$
$=(41)_{10}$
$(11101)_2$
$=(10000)_2+(1000)_2+(100)_2+(1)_2$
$=(16)_{10}+(8)_{10}+(4)_{10}+(1)_{10}$
$=(29)_{10}$

例2 把十进制数$(45)_{10}$改写成二进制数。

45 ÷ 2 = 22 余 1
22 ÷ 2 = 11 余 0
11 ÷ 2 = 5 余 1
5 ÷ 2 = 2 余 1
2 ÷ 2 = 1 余 0
1 ÷ 2 = 0 余 1
将所有余数从最后得到的余数开始倒序排列，得到二进制数101101。
答：$(45)_{10}$改写成二进制数是$(101101)_2$。