Use bit width and binary range
Relate fixed-length binary patterns to the number of values a register can represent.
Exam scope and assessment notes
This sequence follows Cambridge IGCSE Computer Science 0984, 2026–2028 syllabus Version 4. It covers subject content sections 1.1–1.3 only. The separately published 2029–2031 syllabus is not mapped here. Practice is original, self-checked and does not award an official Cambridge grade. Full timed papers and human subject-teacher review remain outstanding. Cambridge states that 0984 has the same 2026–2028 subject content as 0478, with 9–1 grading. In June and November, Cambridge lists 0984 in Administrative zone 3 and for centres in Oman and the UAE. Confirm availability for your centre and exam series with Cambridge International. The separately published 2029–2031 syllabus changes to Python 3 only and removes pseudocode; this 2026–2028 route does not cover those changes.
Before you begin
- No programming language is required. Be comfortable with place value and multiplication; each lesson introduces the computer-science notation it uses.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
How many patterns can 3 bits represent?
What you will learn
- Explain why maximum conversion length is limited to 16 bits
- Find the largest unsigned value for a given bit width
- Preserve leading zeros when a width is specified
A fixed width sets a maximum
An n-bit unsigned number has n binary places and 2ⁿ different patterns, from all zeros to all ones. Its range is 0 to 2ⁿ−1.
Apply and check the rule
The syllabus limits integer conversions to a maximum of 16 bits. For 8 bits, the largest value is 255; for 16 bits, it is 65,535. Leading zeros fill the register width without changing the number.
Pause and explain
Which statement about an 8-bit value is correct?
Worked example
What is the largest unsigned value in a 12-bit register?
Show the worked solution
- There are 12 places, so the all-ones pattern is largest.
- Its value is 2¹²−1.
- 2¹²=4096, so the maximum is 4095.
Answer 4095
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Give the range of an unsigned 5-bit integer.
Give me a hint
Use 2⁵ patterns starting at zero.
Compare my reasoning
- There are 2⁵=32 patterns.
- The range is 0 to 31.
0 to 31
Look for these in your work
- 32 patterns found
- Inclusive range correct
Write denary 13 as an 8-bit unsigned number.
Give me a hint
Convert using 8, 4 and 1, then pad on the left.
Compare my reasoning
- 13 is 8+4+1, so its minimal pattern is 1101.
- Add four leading zeros to make eight bits.
00001101
Look for these in your work
- Value correct
- Eight digits shown
A device can store only a positive unsigned integer up to 255 in one register. How many bits does it need, and what happens if it stores 260?
Give me a hint
Find the smallest width whose maximum is at least 260.
Compare my reasoning
- An 8-bit register ends at 255, so 260 is outside its range.
- Nine bits can represent through 511.
- The device needs at least 9 bits; 8 bits would overflow.
At least 9 bits are needed. Storing 260 in an 8-bit register causes overflow because it exceeds 255.
Look for these in your work
- 8-bit limit identified
- 9-bit capacity identified
- Overflow explained
Common mistakes
- Using 2ⁿ as the largest value rather than the number of possible patterns
- Confusing the number of bit patterns with the largest pattern's value
What can you explain now?
What is the maximum unsigned value represented by 8 bits?
Compare with the explanation
255
Eight bits provide 2⁸=256 patterns, from 0 to 255.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Why is the largest value in an unsigned n-bit register 2ⁿ−1 instead of 2ⁿ?
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.