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You need to store 9 unique states, but don’t want to use up all of your bits.

If you’re like most people, you’re always looking for ways to save on storage space. Every bit counts, and you can’t afford to waste precious bytes on data that isn’t absolutely necessary.

The minimum number of bits required to represent 9 unique states are. This will allow you to store the information you need without using up all of your valuable space.

- How many distinct values may be expressed in 9 binary digits (bits) if n=9?My reasoning is that if I set each of those 9 bits to 1, I will get the largest number that those 9 digits can represent. As a result, the greatest value is 1 1111 1111, which is 511 in decimal. As a result, I infer that 9 binary digits may represent 511 distinct values. Is my way of thinking correct? If not, could you please clarify what I’m missing? How can I extend it to n bits?
- Because there are 512 possible combinations of zeroes and ones, 2 caps 9 = 512 values. However, what those numbers indicate depends on the system you’re using. If it’s an unsigned integer, the result will be:0 = 000000000 (min)

1 = 000000001… 510 = 111111110

511 = 111111111 (max)

You’ll get the following in two’s complement, which is widely used to represent numbers in binary:011111110 = 254 000000000 = 0 000000001 = 1…

255 = 011111111 (max)

hooray integer overflow! 100000000 = -256 (min)

-255 = 100000001…

111111110 x 2 = -2 111111111 x 1 = -1

In general, k bits may represent up to 2k values. Their range will vary depending on the system you use:Unsigned: from 0 to 2k

Signed -1: -2k-1 to 2k-1-1

- Starting small is a better method to fix things. Let’s begin with one bit. Which may be either 1 or 0. That is two values, or ten in binary. Now there are two bits that may be 00, 01, 10, or 11. That’s four values, or one hundred in binary… Do you see the pattern?
- Okay, because it’s already “leaked,” the true answer is 512 (511 is the biggest, but it’s 0 to 511, not 1 to 511).
- Without giving you the solution, here is the reasoning. Each digit might have two different values. You have nine of them. In base 10, you have 10 distinct values by digit, assume you have two of them (which makes from 0 to 99): 0 to 99 produces 100 numbers. If you do the math, you get an exponential function. base ^ number Of Digits:

10^2 = 100 ;

2^9 = 512 - There is a simpler way to look at this. Begin with one bit. This definitely represents two values (0 or 1). What happens when we add a little more? We can now represent twice as many values: the values that could previously be represented with a 0 attached and the values that could previously be represented with a 1 appended. As a result, the number of values we can express with n bits is just 2n (2 to the power n)
- What you don’t know is which encoding scheme is being utilized. Binary numbers may be encoded in a variety of ways. Investigate signed number representations. The ranges and quantity of integers that can be represented with 9 bits vary depending on the system utilized.

Is there a trend here? A single bit 0 may represent the number 0. A single bit 1 may represent the number 1. Similarly, 9 requires four bits: 1001

512

Binary number representation

Length of bit string (b) Number of possible values (N)

8 256

9 512

10 1024

1001

1.4. 2 Binary Numbers

4-Bit Binary Numbers Decimal Equivalents

1000 8

1001 9

1010 10

1011 11

two distinct values

A single bit can only hold two distinct values. That’s hardly much, but it’s sufficient to represent any two-valued state.

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