# Weird Binary Conversion Questions

• posted

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.

1. The decimal number 452 converted into a binary number, using successive division by 2, is _____.
111000100 110000100 111001100 101000100

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

1183 1873 1638 1863

Thanks

• posted

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.

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).

• posted

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

1. I had learned the "trial-and-fit" method.

• posted

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." :)

• posted

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.

• posted

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:

• posted
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:

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