Decimal Equivalent of Largest Unsigned Binary: 8bits=256, n bits

[ccky]

Homework Statement

What is the decimal equivalent of the largest unsigned binary that can be obtained with:
a.8bits. b:n bits

Homework Equations

The Attempt at a Solution

A:2^8=256. 2^8-1=255 how I can change the number to binary number?
B.I can't get the direction of n bits

 

[UltrafastPED]

Repeated division by two - they walk you through this both ways here: http://www.purplemath.com/modules/numbbase.htm

For n bits the maximum number is 2^n -1.

Actually for these two examples you don't need to do any arithmetic. Just ask yourself: what is the largest number that can be displayed on a 4 digit odometer?

 

[Magna Visus]

A:2^8=256. 2^8-1=255 how I can change the number to binary number?
B.I can't get the direction of n bits

Hope this can be some sort of a tool to help you:
Suppose your have a decimal number X and you want to convert it into the binairy number N made of n bits.

1st of all, in a general fashion:

X=$a_{0}$*$2^{0}$+$a_{1}$*$2^{1}$+$a_{2}$*$2^{2}$...$a_{n-1}$*$2^{n-1}$

If n=8, the highest number will be 255, and in binairy it is written as 11111111.
Keep in mind that each "number 1" is the value of a specific $a_{i}$

And the reading directions of these values are opposite i.e. $a_{0}$ is the first "1" from the right of 11111111, $a_{1}$ is the 2nd one.

For example: 11001001
$a_{0}$ =1
$a_{1}$ =0
$a_{2}$ =0
.
.
.
$a_{7}$ =1

2nd of all:
How did I transform 255 to the number 11111111?

Here's a general method (Bear in mind there are other methods):

Take your n bit number (let's consider 186 to add diversity, and in this case it's 8 bits):

Is 186 >= $2^{7}$=128? Yes.
thus: $a_{7}$=1.

Now take 186 and substract $2^{7}$ => 58.
And redo the operation:

Is 58 >= $2^{6}$=64? No.
thus: $a_{6}$=0.

Do not substract anything since 58<64

Now go to the 3rd digit.

Is 58 >= $2^{5}$=32? Yes.
thus $a_{5}$=1.

58-32=26.

Is 26 >= $2^{4}$=16? Yes.
thus $a_{4}$=1.

Is 26-16=10 > $2^{3}$=8? Yes.
thus $a_{3}$=1.

Is 10-8 >= $2^{2}$=4? No.
thus $a_{2}$=0.

Is 10-8>= $2^{1}$=2? Yes.
thus $a_{2}$=1.

Is 2-2 >= $2^{0}$=1? No.
thus the last digit on the right is 0.

Result: 10111010

Verification: 128+0+32+16+8+0+2+0=186. The operation is correct.

Hope this helps and clearify the idea of binairy to decimal transformation and vice-versa .