2
$\begingroup$

I have the number $9243$ In Decimal (Base 10), I'm trying to convert it to octal and then Binary.

What I've done is (r IS THE REMAINDER)

$4096 | 9243| 2$

$ R = 1051$

$512|1051| 2$

$r = 27$

$8|27|3$

$R = 3$

$0|3|3$

SO I got a value of $2233$ In Octal

But the answer is 22033, where did I go Wrong? And how would I go about converting the correct answer to binary?

| cite | improve this question | | | | |
$\endgroup$
  • $\begingroup$ I'm not sure what the "p|q|r" notation means. This is how I would do this problem: 8 divides into 9423 1177 times with remainder 7: 9423= (1177)(8)+ 7. $\endgroup$ – user247327 Sep 26 '15 at 14:17
  • $\begingroup$ As in, 4096 divides into 9243 2 times, remainder 1051 $\endgroup$ – user20842454566 Sep 26 '15 at 14:20
  • $\begingroup$ i'm not sure how I would get 22033 with your method. Can you explain more? $\endgroup$ – user20842454566 Sep 26 '15 at 14:22
  • $\begingroup$ For what it's worth, I find it easier to convert to binary first and then to octal. Well, to be honest, I mostly let my calculator do these. $\endgroup$ – Robert Soupe Sep 26 '15 at 17:47
  • $\begingroup$ Unfortunately my professor does not permit calculators for this class. $\endgroup$ – user20842454566 Sep 26 '15 at 18:01
0
$\begingroup$

I really don't understand what you mean by the "p|q|" notation but this is how I would do the problem: 8 divides into 9243 1155 times with remainder 3 so 9243= (1155)8+ 3. 8 divides into 1155 144 times with remainder 3 so 1155= (144)8+ 3 and 9243= ((144)(8)+ 3)(8)+ 3= 144(8^2)+ 3(8)+ 3. 8 divides into 144 exactly 18 times with 0 remainder: 144= (18)(8) so 9243= (18)(8)(8^2)+ 3(8)+ 3= (18)(8^3)+ 3(8)+ 3. 8 divides into 18 twice with remainder 2: 18= 2(8)+ 2 so 9243= (2(8)+ 2)(8^3)+ 3(8)+ 3= 2(8^4)+ 2(8^3)+ 3(8)+ 3. Notice that there is no "8^2" term so we should write this as 2(8^4)+ 2(8^3)+ 0(8^2)+ 3(8)+ 3 so we are sure we haven't skipped a place. I suspect that was your error. In base 8, 9243 would be written as 22033. (You could have checked you answer by "expanding". 2233 in base 8 is 2(8^3)+ 2(8^2)+ 3(8)+ 3= 1179, not 9243.)

Now, 3= 2+ 1, and 8= 2^3 so 9423= 2(8^4)+ 2(8^3)+ 0(8^2)+3(8)+ 3= 2(2^12)+ 2(2^9)+ (2+ 1)(2^3)+ 2+ 1= 2^(13)+ 2^10+ 2^4+ 2^3+ 2+ 1. So in base 2, 9243 would be 10010000011011. Again, we have to put in 0s for the missing powers of 2.

(As you can see, I tried to use "comment" three times but couldn't get it to work!)

| cite | improve this answer | | | | |
$\endgroup$
  • $\begingroup$ Kind of hard to follow with that wall of text, but I understand your method now. $\endgroup$ – user20842454566 Sep 26 '15 at 15:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.