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What is 55 using 8-bit unsigned binary?

Answer: 00110111

Thinking: Two possible ways of answering this question expected for A Level and even at GCSE, let's start with the most common.

- Method one - consider the column headers for binary numbers.

The question says we need 8 bits, so the column headers are: 128, 64, 32, 16, 6, 4, 2, 1. (Remember, you must start at one, right to left, and double at each increment, following the pattern of powers of two - post a question if you're struggling to do this)

Now, going right to left, fill in a one underneath each header if it can be used to make up the target number, 55. So, going left to right, 128 is too big, write a zero, 64 is too big, 32 is appropriate, so write a one, 16 is appropriate, and so on.

At the end, ensure the numbers add to the target. Here, 32+16+4+2+1 = 55.

- Method two - use remainder division. 

We will continually divide the number 55 by two, until we reach a result of zero. At each step, note the result of the division and the remainder, remember we're only using whole numbers here.

55/2 = 27 remainder 1

27/2 = 13 remainder 1

13/2 = 6 remainder 1

6/2 = 3 remainder 0

3/2 = 1 remainder 1

1/2 = 0 remainder 1, we have reached zero, so stop here.

Now, read back the remainders, bottom to top to give 110111. Remember the question asks for 8 bit signed binary, so add in two zeros on the left side, to make the number fit the requirement, giving our answer of 00110111.

- Note: the question asks for 8-bit unsigned binary. You may be wondering about the importance of the word unsigned. Remember, the use of signed/unsigned relates to how the number handles the sign (positive/negative) of the number. An unsigned number can only be positive, whereas a signed number can be either positive or negative, determined by the left-most bit (0 = positive, 1 = negative), if you are struggling with this, post a question.

Joseph C. GCSE Computing tutor, IB Computing tutor, A Level Computing...

10 months ago

Answered by Joseph, an A Level Computing tutor with MyTutor


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