82 to binary form
231 to octal form
589 to hexadecimal form
Help please and ty!!
Use the method I've described here: http://www.mathhelpforum.com/math-he...ctal-form.html
I'm going to show you the last example so you can see that and how it works:
589 : 16 = 36 >>>>>> R = 13 that is D as hex number
36 : 16 = 2 >>>>>>> R = 4
2 : 16 = 0 >>>>>>> R = 2
Therefore $\displaystyle 589_{dec} = 24D_{hex}$
2 divides into 82 41 times with 0 remainder
82= 2(41)
2 divides into 41 20 times with remainder 1
41= 2(20)+ 1 so 82= 2(2(20)+ 1)
2 divids into 20 10 times with remainder 0
20= 2(10) so 82= 2(2(2(10))+ 1)
2 divides into 10 5 times with remainder 0
10= 2(5) so 82= 2(2(2(2(5)))+ 1
2 divides into 5 2 times with remainder 1
5= 2(2)+ 1 so 82= 2(2(2(2(2(2)+1))))+ 1
2 divides into 2 1 with remainder 0
2= 2(1) so 82= 2(2(2(2(2(2(1)+ 1))))+ 1= [tex]1(2^6)+ 0(2^5)+ 1(2^4)+ 0(2^3)+ 0(2^2)+ 1(2)+ 0
82 base 10 is 101001 base 2.
Try those two yourself. In hexadecimal, use "A" for "10", "B" for "11", "C" for "12", "D" for "13", "E" for "14", and "F" for "15".231 to octal form
589 to hexadecimal form
Help please and ty!!