I have n = 7. How many different values can be represented using 7 digits in In Hexadecimal Systems?
I approached it this way:
I set each bit of the 7 to be equal to 1. Therefore the highest number I get is 111 1111 = 16^0 + 16^1 + 16^2 + 16^3 + 16^4 + 16^5 + 16^6 which is equal to 17895697. I also consider zero being part of the answer, so my range is 0 to 17895697 and therefore, I get 17895698 different values. however, I know that in binary you would do 2^7 = 128. does it apply to hexadecimals as well? If I do it this way I get 16^7 = 268435456 which does not equal to what I got before.
Is any of my answers correct? if not, could someone please explain what it the right way to do this question?
It may be more simple explanation that you would expect, but it does not mean it would not work.
One hexadecimal digit can represent one of 16 values (0x0 to 0xF, or 0 to 15 if you prefer), so 16^7 = 268,435,456 and that's how many different values you can achieve if you use all the bits.
0000 0000 0000 0000 0000 0000 0000 to 1111 1111 1111 1111 1111 1111 1111.
The general formula is the sum of digit * base^place
Consider decimal numbers. What's the largest number you can represent with 3 decimal digits? The largest possible digit is 9. Therefore the max number is:
9*10^2 + 9*10^1 + 9*10^0 = 9*100 + 9*10 + 9*1 = 999
The next largest number would with be 1*10^3 = 1000
Another way to calculate the max of decimal three digits, is to get the smallest number with 4 digits and subtract 1.
The smallest decimal with 4 digits is 1000 or
1*10^3 + 0*10^2 + 0*10^1 + 0*10^0 = 1000
Now subtract 1
1000 - 1 = 999
The possible digits for a hex num are 0..F (values 0..15)
For a 7 digits the places range from 6 down to 0.
The largest possible number is therefore:
15*16^6 + 15*16^5 .... 15*16^0
Or the smallest number in hex with 8 digits is #10000000
1*16^7 + 0*16^6 ... + 0*16^0 = 1*16^7
So the largest possible number with in hex with 7 digits is
1*16^7 -1
16^7 is the correct answer. Why? Because for every character you add there are 16 times the previous possibilities. So for 7 characters it's 16^7.
Consider binary - 2^1, you get 0 and 1, 2 possible values. 2^2, you get 0 through 3 (00,01,10,11), 4 possible values.
Using hex, it's 16^7 possible values, which is 268,435,456 possible values.
For example, if n=9, then how many different values can be represented in 9 binary digits (bits)?
My thinking is that if I set each of those 9 bits to 1, I will make the highest number possible that those 9 digits are able to represent. Therefore, the highest value is 1 1111 1111 which equals 511 in decimal. I conclude that, therefore, 9 digits of binary can represent 511 different values.
Is my thought process correct? If not, could someone kindly explain what I'm missing? How can I generalize it to n bits?
29 = 512 values, because that's how many combinations of zeroes and ones you can have.
What those values represent however will depend on the system you are using. If it's an unsigned integer, you will have:
000000000 = 0 (min)
000000001 = 1
...
111111110 = 510
111111111 = 511 (max)
In two's complement, which is commonly used to represent integers in binary, you'll have:
000000000 = 0
000000001 = 1
...
011111110 = 254
011111111 = 255 (max)
100000000 = -256 (min) <- yay integer overflow
100000001 = -255
...
111111110 = -2
111111111 = -1
In general, with k bits you can represent 2k values. Their range will depend on the system you are using:
Unsigned: 0 to 2k-1
Signed: -2k-1 to 2k-1-1
What you're missing: Zero is a value
A better way to solve it is to start small.
Let's start with 1 bit. Which can either be 1 or 0. That's 2 values, or 10 in binary.
Now 2 bits, which can either be 00, 01, 10 or 11 That's 4 values, or 100 in binary... See the pattern?
Okay, since it already "leaked": You're missing zero, so the correct answer is 512 (511 is the greatest one, but it's 0 to 511, not 1 to 511).
By the way, an good followup exercise would be to generalize this:
How many different values can be represented in n binary digits (bits)?
Without wanting to give you the answer here is the logic.
You have 2 possible values in each digit. you have 9 of them.
like in base 10 where you have 10 different values by digit say you have 2 of them (which makes from 0 to 99) : 0 to 99 makes 100 numbers. if you do the calcul you have an exponential function
base^numberOfDigits:
10^2 = 100 ;
2^9 = 512
There's an easier way to think about this. Start with 1 bit. This can obviously represent 2 values (0 or 1). What happens when we add a bit? We can now represent twice as many values: the values we could represent before with a 0 appended and the values we could represent before with a 1 appended.
So the the number of values we can represent with n bits is just 2^n (2 to the power n)
The thing you are missing is which encoding scheme is being used. There are different ways to encode binary numbers. Look into signed number representations. For 9 bits, the ranges and the amount of numbers that can be represented will differ depending on the system used.
I don't really understand how modulus division works.
I was calculating 27 % 16 and wound up with 11 and I don't understand why.
I can't seem to find an explanation in layman's terms online.
Can someone elaborate on a very high level as to what's going on here?
Most explanations miss one important step, let's fill the gap using another example.
Given the following:
Dividend: 16
Divisor: 6
The modulus function looks like this:
16 % 6 = 4
Let's determine why this is.
First, perform integer division, which is similar to normal division, except any fractional number (a.k.a. remainder) is discarded:
16 / 6 = 2
Then, multiply the result of the above division (2) with our divisor (6):
2 * 6 = 12
Finally, subtract the result of the above multiplication (12) from our dividend (16):
16 - 12 = 4
The result of this subtraction, 4, the remainder, is the same result of our modulus above!
The result of a modulo division is the remainder of an integer division of the given numbers.
That means:
27 / 16 = 1, remainder 11
=> 27 mod 16 = 11
Other examples:
30 / 3 = 10, remainder 0
=> 30 mod 3 = 0
35 / 3 = 11, remainder 2
=> 35 mod 3 = 2
The simple formula for calculating modulus is :-
[Dividend-{(Dividend/Divisor)*Divisor}]
So, 27 % 16 :-
27- {(27/16)*16}
27-{1*16}
Answer= 11
Note:
All calculations are with integers. In case of a decimal quotient, the part after the decimal is to be ignored/truncated.
eg: 27/16= 1.6875 is to be taken as just 1 in the above mentioned formula. 0.6875 is ignored.
Compilers of computer languages treat an integer with decimal part the same way (by truncating after the decimal) as well
Maybe the example with an clock could help you understand the modulo.
A familiar use of modular arithmetic is its use in the 12-hour clock, in which the day is divided into two 12 hour periods.
Lets say we have currently this time: 15:00
But you could also say it is 3 pm
This is exactly what modulo does:
15 / 12 = 1, remainder 3
You find this example better explained on wikipedia: Wikipedia Modulo Article
The modulus operator takes a division statement and returns whatever is left over from that calculation, the "remaining" data, so to speak, such as 13 / 5 = 2. Which means, there is 3 left over, or remaining from that calculation. Why? because 2 * 5 = 10. Thus, 13 - 10 = 3.
The modulus operator does all that calculation for you, 13 % 5 = 3.
modulus division is simply this : divide two numbers and return the remainder only
27 / 16 = 1 with 11 left over, therefore 27 % 16 = 11
ditto 43 / 16 = 2 with 11 left over so 43 % 16 = 11 too
Very simple: a % b is defined as the remainder of the division of a by b.
See the wikipedia article for more examples.
I would like to add one more thing:
it's easy to calculate modulo when dividend is greater/larger than divisor
dividend = 5
divisor = 3
5 % 3 = 2
3)5(1
3
-----
2
but what if divisor is smaller than dividend
dividend = 3
divisor = 5
3 % 5 = 3 ?? how
This is because, since 5 cannot divide 3 directly, modulo will be what dividend is
I hope these simple steps will help:
20 % 3 = 2
20 / 3 = 6; do not include the .6667 – just ignore it
3 * 6 = 18
20 - 18 = 2, which is the remainder of the modulo
Easier when your number after the decimal (0.xxx) is short. Then all you need to do is multiply that number with the number after the division.
Ex: 32 % 12 = 8
You do 32/12=2.666666667
Then you throw the 2 away, and focus on the 0.666666667
0.666666667*12=8 <-- That's your answer.
(again, only easy when the number after the decimal is short)
27 % 16 = 11
You can interpret it this way:
16 goes 1 time into 27 before passing it.
16 * 2 = 32.
So you could say that 16 goes one time in 27 with a remainder of 11.
In fact,
16 + 11 = 27
An other exemple:
20 % 3 = 2
Well 3 goes 6 times into 20 before passing it.
3 * 6 = 18
To add-up to 20 we need 2 so the remainder of the modulus expression is 2.
The only important thing to understand is that modulus (denoted here by % like in C) is defined through the Euclidean division.
For any two (d, q) integers the following is always true:
d = ( d / q ) * q + ( d % q )
As you can see the value of d%q depends on the value of d/q. Generally for positive integers d/q is truncated toward zero, for instance 5/2 gives 2, hence:
5 = (5/2)*2 + (5%2) => 5 = 2*2 + (5%2) => 5%2 = 1
However for negative integers the situation is less clear and depends on the language and/or the standard. For instance -5/2 can return -2 (truncated toward zero as before) but can also returns -3 (with another language).
In the first case:
-5 = (-5/2)*2 + (-5%2) => -5 = -2*2 + (-5%2) => -5%2 = -1
but in the second one:
-5 = (-5/2)*2 + (-5%2) => -5 = -3*2 + (-5%2) => -5%2 = +1
As said before, just remember the invariant, which is the Euclidean division.
Further details:
What is the behavior of integer division?
Division and Modulus for Computer Scientists
Modulus division gives you the remainder of a division, rather than the quotient.
It's simple, Modulus operator(%) returns remainder after integer division. Let's take the example of your question. How 27 % 16 = 11? When you simply divide 27 by 16 i.e (27/16) then you get remainder as 11, and that is why your answer is 11.
Lets say you have 17 mod 6.
what total of 6 will get you the closest to 17, it will be 12 because if you go over 12 you will have 18 which is more that the question of 17 mod 6. You will then take 12 and minus from 17 which will give you your answer, in this case 5.
17 mod 6=5
Modulus division is pretty simple. It uses the remainder instead of the quotient.
1.0833... <-- Quotient
__
12|13
12
1 <-- Remainder
1.00 <-- Remainder can be used to find decimal values
.96
.040
.036
.0040 <-- remainder of 4 starts repeating here, so the quotient is 1.083333...
13/12 = 1R1, ergo 13%12 = 1.
It helps to think of modulus as a "cycle".
In other words, for the expression n % 12, the result will always be < 12.
That means the sequence for the set 0..100 for n % 12 is:
{0,1,2,3,4,5,6,7,8,9,10,11,0,1,2,3,4,5,6,7,8,9,10,11,0,[...],4}
In that light, the modulus, as well as its uses, becomes much clearer.
Write out a table starting with 0.
{0,1,2,3,4}
Continue the table in rows.
{0,1,2,3,4}
{5,6,7,8,9}
{10,11,12,13,14}
Everything in column one is a multiple of 5. Everything in column 2 is a
multiple of 5 with 1 as a remainder. Now the abstract part: You can write
that (1) as 1/5 or as a decimal expansion. The modulus operator returns only
the column, or in another way of thinking, it returns the remainder on long
division. You are dealing in modulo(5). Different modulus, different table.
Think of a Hash Table.
When we divide two integers we will have an equation that looks like the following:
A/B​​ =Q remainder R
A is the dividend; B is the divisor; Q is the quotient and R is the remainder
Sometimes, we are only interested in what the remainder is when we divide A by B.
For these cases there is an operator called the modulo operator (abbreviated as mod).
Examples
16/5= 3 Remainder 1 i.e 16 Mod 5 is 1.
0/5= 0 Remainder 0 i.e 0 Mod 5 is 0.
-14/5= 3 Remainder 1 i.e. -14 Mod 5 is 1.
See Khan Academy Article for more information.
In Computer science, Hash table uses Mod operator to store the element where A will be the values after hashing, B will be the table size and R is the number of slots or key where element is inserted.
See How does a hash table works for more information
This was the best approach for me for understanding modulus operator. I will just explain to you through examples.
16 % 3
When you division these two number, remainder is the result. This is the way how i do it.
16 % 3 = 3 + 3 = 6; 6 + 3 = 9; 9 + 3 = 12; 12 + 3 = 15
So what is left to 16 is 1
16 % 3 = 1
Here is one more example: 16 % 7 = 7 + 7 = 14 what is left to 16? Is 2 16 % 7 = 2
One more: 24 % 6 = 6 + 6 = 12; 12 + 6 = 18; 18 + 6 = 24. So remainder is zero, 24 % 6 = 0
When multiplying (or doing any mathematics) to binary and decimal numbers, would you simply convert then multiply in decimals?
E.g., 3(base10) * 100(base2) would = 3 * 4 = 12?
You can multiply in any base as long as the base is the same for each operand.
In your example, you could have converted the 3(base10) to 11(base2) and multiplied:
11 * 100 = 1100
1100(Base2) = 12(base10)
Numbers are numbers. 3 * 0b100 will always equal 12, regardless of whether you use a lookup table or bit shifting to multiply them.
You would convert them to integers before multiplying, I would hope.
Thus they're all in binary.
convert Base 2 into base 10 number then multiply
For example:
1000 base 2 x 100 base 10
Converting 1000 base2
1000 base 2 = 2x2x2 = 8
so Multiplication result will be
8 base 10 x 100 base 10 = 800 base 10 = 800
Hope problem solved...
Or, what is the range of numbers that can be represented on a 4-bit machine using 2s-complement?
That would be -8 to +7
The range is -8 to 7, or 1000 to 0111. You can see the full range here.
4 bits (using 2's complement) will give you a range from -8 to 7.
This should be straightforward to work out yourself.
Range in twos complement will be:
-1 * 2 ^ (bits - 1)
to
2 ^ (bits - 1) - 1
So for 4 bits:
-1 * 2 ^ (4 - 1) = -1 * 2 ^ 3 = -8
to
2 ^ (4 - 1) - 1 = 2 ^ 3 - 1 = 7
Also, if you are interested and for others maybe browsing this question -
twos complement is used for easy binary arithmetic:
to add - you just add the two numbers without conversion and disregard the overflow:
-6 + 7 = 1
is
1010 = -6
0111 = 7
------
(1)0001 = 1 (ignoring the overflow)
...and more yet - to convert a negative binary number to its opposite positive number:
if the sign bit (highest order bit) is 1, indicating negative... reading from least significant to most significant bit (right to left), preserve every bit up through the first "1", then invert every bit after that.
So, with 8 bit
10011000 .. becomes
01101000 (* -1) = 104 * -1 = -104
and that is why 10000000 is your lowest negative number (or in X bit 1000.all zeroes..000), it translates to unsigned 10000000 * -1 = -128
Maybe a long winded answer but to those without the 1s and 0s background I figure it is useful
Well let's dissect this question.
Firstly, the question should be framed as - "The least*(because it is negative, so biggest is not the right usage)* possible negative number to be stored in 4-bit would be?"
Numbers are of two types -
Signed (holds both + and - numbers )
Unsigned (holds only + numbers)
We will be using binary representation to understand this.
For Unsigned -
4-bit minimum number = 0000 (0)
4-bit maximum number = 1111 (255)
So range would be Range : 0-15
For Signed 4-bits, first bit represent sign(+/-) and rest 3-bits represent number.
4-bit minimum will be a negative number.
So replace the first bit(MSB) with 1 in 0000(least unsigned number), making it 1000.
Calculate decimal equivalent of 1000 = 1*2^3 = 8
So, number would be -8 (- as we have 1 as the first bit in 1000)
4-bit maximum will be a positive number.
So replace the first bit(MSB) with 0 in 1111(largest unsigned number), making it 0111.
Calculate decimal equivalent of 0111 = 1*2^2 + 1*2^1 + 1*2^0 = 7
So, number would be +7 (+ as we have 0 as the first bit in 0111)
Range would be -8 to +7.