Convert the decimal number

• Dec 16th 2009, 10:17 PM
Draft
Convert the decimal number
82 to binary form

231 to octal form

• Dec 16th 2009, 10:32 PM
earboth
Quote:

Originally Posted by Draft
82 to binary form

231 to octal form

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 $\displaystyle 589_{dec} = 24D_{hex}$
• Dec 16th 2009, 10:34 PM
Draft
probably a stupid question, but what is the R??
• Dec 16th 2009, 10:48 PM
earboth
Quote:

Originally Posted by Draft
probably a stupid question, but what is the R??

R means remainder.

For instance $\displaystyle 589 = 36 \cdot 16 + 13$ . So if you divide 589 by 16 you get:

$\displaystyle 589 : 16 = \dfrac{36 \cdot 16}{16}+\dfrac{13}{16} = 36 \text{ with the remainder } R = 13$
• Dec 16th 2009, 11:09 PM
Draft
wrong topic mb
• Dec 17th 2009, 04:20 AM
HallsofIvy
Quote:

Originally Posted by Draft
82 to binary form

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.

Quote:

231 to octal form