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Software Developer calculations with worked answers

Updated 16 September 2026 · 2 min read

Developers do not spend their days on arithmetic, but a few calculations explain why some code is fast and some slow, and how computers store data. Knowing how many comparisons a binary search needs, or reading a hexadecimal value in a debugger, makes you a sharper programmer and helps with test questions on efficiency and data representation. Every example below was generated for this page and checked by separate code.

How this guide was made

Written for this site with AI assistance, then checked automatically before publishing. The topic table, sample questions, worked examples and study plan are generated from this site’s question bank: each worked example is recalculated by separate code, and each sample answer says where it came from.

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Worked examples

Try each one on paper first. Every figure is generated and then recalculated by separate code before it is published.

  1. Worst-case comparisons in a binary search

    A sorted array holds 10000 records. In the worst case, how many comparisons does a binary search need, and how many would a linear search need?

    Show the working
    1. Each binary-search comparison halves what is left, so the worst case is ⌊log₂ n⌋ + 1 comparisons.
    2. log₂(10000) ≈ 13.29, so ⌊13.29⌋ + 1 = 14
    3. A linear search may have to check every record: 10000 comparisons. That gap is what O(log n) versus O(n) means in practice.

    Answer: 14 comparisons (binary) versus 10000 (linear)

  2. Converting hexadecimal to decimal and binary

    Convert 0x2FB7 to decimal and to binary.

    Show the working
    1. Expand by place value: 2×16^3 + F×16^2 + B×16^1 + 7×16^0
    2. = 8192 + 3840 + 176 + 7 = 12215
    3. For binary, write each hex digit as four bits: 0010 1111 1011 0111

    Answer: 12215 decimal, 0010 1111 1011 0111 binary

  3. Converting an octet from decimal to binary

    Convert 244 to an 8-bit binary octet.

    Show the working
    1. Subtract the largest place value that fits, mark a 1, and repeat: 128 64 32 16 8 4 2 1.
    2. 244 = 128 + 64 + 32 + 16 + 4
    3. Mark those places with 1 and the rest with 0: 11110100

    Answer: 11110100

  4. Worst-case comparisons in a binary search

    A sorted array holds 100000 records. In the worst case, how many comparisons does a binary search need, and how many would a linear search need?

    Show the working
    1. Each binary-search comparison halves what is left, so the worst case is ⌊log₂ n⌋ + 1 comparisons.
    2. log₂(100000) ≈ 16.61, so ⌊16.61⌋ + 1 = 17
    3. A linear search may have to check every record: 100000 comparisons. That gap is what O(log n) versus O(n) means in practice.

    Answer: 17 comparisons (binary) versus 100000 (linear)

  5. Converting hexadecimal to decimal and binary

    Convert 0x5ED5 to decimal and to binary.

    Show the working
    1. Expand by place value: 5×16^3 + E×16^2 + D×16^1 + 5×16^0
    2. = 20480 + 3584 + 208 + 5 = 24277
    3. For binary, write each hex digit as four bits: 0101 1110 1101 0101

    Answer: 24277 decimal, 0101 1110 1101 0101 binary

  6. Converting an octet from binary to decimal

    Convert the binary octet 11011100 to decimal.

    Show the working
    1. Write the place values under each bit: 128 64 32 16 8 4 2 1.
    2. Add the place values where the bit is 1: 128 + 64 + 16 + 8 + 4
    3. = 220

    Answer: 220

  7. Worst-case comparisons in a binary search

    A sorted array holds 1000000 records. In the worst case, how many comparisons does a binary search need, and how many would a linear search need?

    Show the working
    1. Each binary-search comparison halves what is left, so the worst case is ⌊log₂ n⌋ + 1 comparisons.
    2. log₂(1000000) ≈ 19.93, so ⌊19.93⌋ + 1 = 20
    3. A linear search may have to check every record: 1000000 comparisons. That gap is what O(log n) versus O(n) means in practice.

    Answer: 20 comparisons (binary) versus 1000000 (linear)

  8. Converting hexadecimal to decimal and binary

    Convert 0xAB49 to decimal and to binary.

    Show the working
    1. Expand by place value: A×16^3 + B×16^2 + 4×16^1 + 9×16^0
    2. = 40960 + 2816 + 64 + 9 = 43849
    3. For binary, write each hex digit as four bits: 1010 1011 0100 1001

    Answer: 43849 decimal, 1010 1011 0100 1001 binary

Why these calculations matter to developers

Binary search steps show concretely what O(log n) means: a million sorted records need only about twenty comparisons, while a linear search may need a million. That difference decides whether an application stays fast as its data grows. Hexadecimal appears in colour codes, memory addresses, character encodings and hashes. Binary underlies bit flags, permissions and how integers are stored.

Using the results in design decisions

Efficiency estimates help you choose between approaches before writing code. If a feature searches a list on every request, estimate how large the list could become and how many comparisons that means. When the numbers grow quickly, a sorted structure, an index or a hash map is worth the extra effort.

Common questions

Why is binary search faster than linear search?

Each comparison halves the remaining search space, so the number of steps grows with the logarithm of the list size rather than with the size itself. It does require the data to be sorted first.

Why do programmers use hexadecimal?

Each hexadecimal digit stands for exactly four bits, so long binary values become short and readable. Colours, memory addresses and hash values are commonly written in hexadecimal for that reason.

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