Bases and value ranges

Question

How many bytes are requires to represent the decimal number 3,175,000 in binary? How many bytes would it take to store this number?

Answer 1

log2(3175000) = log(3175000)/log(2) = 21.5983251612 OR we will need 22 bits

But, computers store numbers in bytes (8 bits at a time),

so we will need 24 bits or 3 bytes at least to represent this number (part 1)

the extra two bits will be 0s (at the upper end of the number)

Most computers store NUMBERS as EVEN numbers of bytes (8 bits),

or short ints (16 bits or 2 bytes)

or ints (32 bits or 4 bytes)

or long ints (64 bits or 8 bytes)

or long long ints (128 bits or 16 bytes). (they don't use 6 bytes)

In your case,

3,175,000 must be stored as an int

or 4 bytes (part 2 ANSWER) (the upper 10 bits will be 0s)

Explanation of why this works is below...

A number in base 10 (like 123) means 1*10^2 + 2*10^1 + 3*10^0

Similarly, a number in base 2 or binary (like 101)

means 1 * 2^2 + 0*2^1 + 1*2^0

To find the number of places needed for 123 in base 10,

take log10(123) = 2.08990511144,

then take the next largest integer (3). So 3 places are needed

To find the number of places needed for 3,175,000 in base 2 (binary)

take log2(3175000) -- unfortunately, most calculators

don't have this function, BUT

logn(a) = log(a)/log(n), where log is log10 OR ln, and n is the desired base

log2(317500) = log(3175000)/log(2) == 21.5983251612

then take the next largest integer (22). So 22 bits are needed.

Answer 2

3,175,000 is 1100000111001001011000 in binary. Each 1 or 0 represents one bit of information. 8 bits make 1 byte. The binary number mentioned previously has 22 bits in it, which would take 3 bytes to store. However, in the real world, and depending on the programming language and data type used to store this number, it may actually be more than 3 bytes. Hope this helps!