UIL CS: Number Systems

This is part of the UIL CS study series. See also:

• Java Fundamentals
• Bitwise & Boolean
• Recursion & Loops
• OOP & Generics
• Collections & Algorithms
• Practice Resources

Main topic list: https://www.uiltexas.org/files/academics/UILCS-JavaTopicList2526.pdf

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1. Number Systems

A. Converting from decimal to any base

Method: repeated division (for integers).

To convert decimal $N$ to base $b$:

1. Divide $N$ by $b$.
2. Record the remainder.
3. Replace $N$ with the quotient.
4. Repeat until the quotient is $0$.
5. Read remainders bottom to top.

Example: convert $156_{10}$ to base $4$

• $156 \div 4 = 39$ remainder $0$
• $39 \div 4 = 9$ remainder $3$
• $9 \div 4 = 2$ remainder $1$
• $2 \div 4 = 0$ remainder $2$

$156_{10} = 2130_{4}$

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B. Shortcuts between binary, hex, and decimal

Binary to Decimal

Binary is base $2$, so each bit represents a power of $2$.

Example: $101101_2$

$$1\cdot2^{5} + 0\cdot2^{4} + 1\cdot2^{3} + 1\cdot2^{2} + 0\cdot2^{1} + 1\cdot2^{0} = 32 + 8 + 4 + 1 = 45$$

So $101101_2 = 45_{10}$

Hex to Decimal

Hex is base $16$.

Digits: $0$-$9$, $A=10$, $B=11$, $C=12$, $D=13$, $E=14$, $F=15$.

Example: $2AF_{16}$

$$2\cdot16^{2} + 10\cdot16^{1} + 15\cdot16^{0} = 512 + 160 + 15 = 687$$

So $2AF_{16} = 687_{10}$

Binary to Hex shortcut (VERY common on UIL)

Group bits in chunks of $4$ from the right.

Example: $110101111001_2$

Group: $1101\;0111\;1001$

Convert each nibble:

• $1101 = D$
• $0111 = 7$
• $1001 = 9$

So: $110101111001_2 = D79_{16}$

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C. Arithmetic in a base other than decimal

Example: addition in base $4$

$$\begin{array}{r} 312_{4} \\ + 233_{4} \\ \hline \end{array}$$

Work right to left:

• $2 + 3 = 5$. In base $4$, $5 = 11_{4}$. Write $1$ carry $1$.
• $1 + 3 + 3 = 7$. In base $4$, $7 = 13_{4}$. Write $3$ carry $1$.
• $3 + 2 + 1 = 6$. In base $4$, $6 = 12_{4}$. Write $2$ carry $1$.
• carry $1$

Answer: $312_{4} + 233_{4} = 12131_{4}$

Note: Understand positional values (e.g., $2^{n}$, $16^{n}$).