Question

What is the minimum number of bits needed to store the decimal value 200?

Answer 1

Answer:

8 bits

Explanation:

Given

Decimal Value = 200

Required

Determine minimum number of bits to store the value

Represent the decimal value with N and the number of bits with n

The relationship between n and N is:

$2^n = N$

Substitute 200 for N

$2^n = 200$

Take Log of both sides

$Log2^n = Log200$

Apply law of indices

$nLog2 = Log200$

Make n the subject

$n = \frac{Log200}{Log2}$

$n = 7.64385618977$

We need n to be an integer; So, we have to approximate

$n = 8$