Weird Binary Conversion Questions

I have been studying for the CCNA test. I have been using the
Trainsignal DVD which is very thorough. But when I tried some
parctice tests at the Cisco website I got a couple of binary
conversion questions that I did not understand. Can someone explain
the meaning of these two questions.

18. The decimal number 452 converted into a binary number, using
successive division by 2, is _____.

111000100
110000100
111001100
101000100




20. The binary number 11101000111 converted into a decimal number,
using powers of 2, is _____.

1183
1873
1638
1863

Thanks

 

Its asking you to do a couple binary conversions. Since IP addresses
and especially netmasks are numbers that can be expressed as binary
numbers for ease of use to illustrate what is really going on, its is
useful to know how to express decimal numbers as binary and back again.

Its asking you to convert 452 to binary. There's generally two
different ways you can do it by hand. Here's a website that outlines it.

http://www.allaboutcircuits.com/vol_4/chpt_1/6.html

Its asking you to convert 11101000111 to decimal. The same website
probably has some page about this explicitly, but this is simpler to
illustrate in a posting..

1 * 2^0
1 * 2^1
1 * 2^2
0 * 2^3
0 * 2^4
0 * 2^5
1 * 2^6
0 * 2^7
1 * 2^8
1 * 2^9
1 * 2^10 +
===========

Multiple it out, add it up, and it equals what?

Again, they are just asking you to do the binary to decimal and back
again by hand rather than by calculator (if yours supports it).

 

Thanks Doug. i understand it now. My problem earlier was I did not
understand the language e.g., powers of 2 and successive division by
2. I had learned the "trial-and-fit" method.

Thanks again for your help.

 

Isn't base conversion still taught in grade school? That's where I
learned it about 35 years ago.

But I'll bet most of you thought "I'll never need to use this in real
life."

 

This is real life???

Think yourself lucky, the window of opportunity
is closing

IPV6 will do away with it.

Of course you need a slightly different conversion
Hex-Binary but that of course is easy without
requiring formal methods. I suppose formal methods may
be needed in the initial learning stages.

 

I ran into this question too and could not figure it out but finaly I found a way to get to the right answer. I found it using this method.

We all know that if we have 8 bits we use :

128 64 32 16 8 4 2 1

In this question we do not have 8 bits but we have 11 bits

(11101000111)

When you continue counting from bit 128 you will get this.

1024 512 256 128 64 32 16 8 4 2 1

Then if we add the bits which are enabled you get :


1024 512 256 128 64 32 16 8 4 2 1
1 1 1 0 1 0 0 0 1 1 1

1024 + 512 + 256 + 64 + 4 + 2 + 1 = 1863










Op maandag 3 maart 2008 18:58:38 UTC+1 schreef tman het volgende:

 

1024 512 256 128 64 32 16 8 4 2 1
1 1 1 0 1 0 0 0 1 1 1






Op maandag 3 maart 2008 18:58:38 UTC+1 schreef tman het volgende: