Number System Practice Questions

To find the 16-bit 2's complement representation of -28, we first find the 16-bit binary representation of +28, which is 0000 0000 0001 1100. Next, we find the 1's complement by inverting all the bits, which gives 1111 1111 1110 0011. Finally, we add 1 to the 1's complement to get the 2's complement: 1111 1111 1110 0100.

In an n-bit sign-magnitude representation, the largest possible magnitude is $2^{n-1}-1$. The operation $Z = X - Y$ can result in a magnitude that is the sum of the magnitudes of X and Y. The maximum possible magnitude of Z is $(2^{n-1}-1) + (2^{n-1}-1) = 2^n - 2$. To represent such a large magnitude, more than $n-1$ bits may be needed. For example, if $n=4$, let $X=+6$ and $Y=-5$. Then $Z = X-Y = 11$. Representing magnitude 11 ($1011_2$) requires 4 bits, so Z requires a total of $1+4=5$ bits, which is $n+1$ bits.

Given the equation $\sqrt{224_{r}} = 13_{r}$, we square both sides to get $224_{r} = (13_{r})^2$.
Converting to base 10, the left side is $2r^2 + 2r + 4$ and the right side is $(r + 3)^2 = r^2 + 6r + 9$.
Equating these expressions: $2r^2 + 2r + 4 = r^2 + 6r + 9$.
Rearranging the terms gives the quadratic equation $r^2 - 4r - 5 = 0$.
Factoring the quadratic yields $(r - 5)(r + 1) = 0$, resulting in $r = 5$ or $r = -1$.
Since the radix must be positive and greater than the digits (specifically $r > 4$), we have $r = 5$.

An $n$-bit binary string represents $2^n$ distinct values.
A ternary string with $x$ digits can represent $3^x$ distinct values.
To represent the same amount of information, we equate the capacities: $3^x = 2^n$.
Taking the logarithm base $3$ on both sides yields $\log_3(3^x) = \log_3(2^n)$.
Using the logarithmic property $\log_a(b^c) = c \log_a(b)$, we simplify to $x = n \log_3 2$.

The fixed-point representation has 3 bits for the fractional part. A decimal number can be represented exactly if its fractional part has a finite binary representation.
(i) 31.500: The fractional part $0.5$ is $1/2$, which is exactly $0.1_2$.
(ii) 0.875: The fractional part $0.875$ is $7/8$, which is exactly $0.111_2$.
(iii) 12.100: The fractional part $0.1$ has a non-terminating, repeating binary expansion ($0.000110011..._2$).
(iv) 3.001: The fractional part $0.001$ also has a non-terminating binary expansion.
Therefore, only numbers (iii) and (iv) cannot be represented exactly.

To represent integers from $0$ to $n$, we must accommodate $n+1$ distinct values. A bit-field of width $w$ can represent $2^w$ distinct values, so the condition $2^w \ge n+1$ must hold. Taking the logarithm base $2$ on both sides yields $w \ge \log_2(n+1)$. Since $w$ must be an integer, the minimum required width is $w = \lceil \log_2(n+1) \rceil$. Comparing this result with the provided options: Option A is $\lceil \log_2(n) \rceil + 1$, Option B is $\lceil \log_2(n-1) \rceil + 1$, and Option C is $\lceil \log_2(n+1) \rceil + 1$. Since none of these expressions match $\lceil \log_2(n+1) \rceil$, the

The representation uses i bits for the integer part and f bits for the fractional part. Since the number is unsigned, the minimum value is 0, which occurs when all bits are 0.

The maximum value occurs when all n bits are 1. The i integer bits being all 1s represent the value $2^i - 1$. The f fractional bits being all 1s represent the value $\sum_{k=1}^{f} 2^{-k} = 1 - 2^{-f}$. The total maximum value is the sum of these two parts: $(2^i - 1) + (1 - 2^{-f}) = 2^i - 2^{-f}$. Thus, the range is from 0 to $(2^i - 2^{-f})$.

To convert the hexadecimal number BCA9 to octal, we first convert it to binary and then group the binary digits into sets of three.

1. Convert hexadecimal to binary:
   Each hexadecimal digit corresponds to 4 binary digits:
   B = 11 (decimal) = 1011 (binary)
   C = 12 (decimal) = 1100 (binary)
   A = 10 (decimal) = 1010 (binary)
   9 = 9 (decimal) = 1001 (binary)
   So, BCA9 (hex) = 1011 1100 1010 1001 (binary).

2. Group binary digits into sets of three (from right to left) and convert to octal:
   Start from the rightmost bit and group every three bits:
   1 | 011 | 110 | 010 | 101 | 001
   Add leading zeros if necessary to complete the leftmost group:
   001 | 011 | 110 | 010 | 101 | 001

   Now convert each group of three binary digits to its octal equivalent:
   001 = 1 (octal)
   011 = 3 (octal)
   110 = 6 (octal)
   010 = 2 (octal)
   101 = 5 (octal)
   001 = 1 (octal)

   Combining these, we get 136251 (octal).

To convert $(1217)_{8}$ to hexadecimal, first convert it to binary:
$(1217)_{8} = (001\ 010\ 001\ 111)_{2}$
Next, group the binary digits in sets of four from the right and convert each group to its hexadecimal equivalent:
$(0010\ 1000\ 1111)_{2}$
$0010_2 = 2_{16}$
$1000_2 = 8_{16}$
$1111_2 = F_{16}$
Therefore, $(1217)_{8} = (28F)_{16}$. This is equivalent to $(028F)_{16}$.

The largest value representable by an $n$-bit binary number is $2^n - 1$. When adding two such numbers, the maximum possible sum is:
$(2^n - 1) + (2^n - 1) = 2(2^n) - 2 = 2^{n+1} - 2$
The maximum value representable by $m$ bits is $2^m - 1$. For $m = n+1$, the capacity is $2^{n+1} - 1$. Since $2^{n+1} - 2 \leq 2^{n+1} - 1$, the sum is guaranteed to fit within $n+1$ bits.

The 32-bit IEEE-754 number is parsed into three parts: sign (S), exponent (E), and mantissa (M).

• Sign bit (S): 0 (The number is positive).
• Exponent (E): 01111100 in binary, which is $124$ in decimal.
• Mantissa (M): 110110100...
  The decimal value is calculated using the formula $(-1)^S \times (1.M)_2 \times 2^{E-127}$.
  Substituting the values: $(-1)^0 \times (1.1101101)_2 \times 2^{124-127}$.
  This simplifies to $(1.1101101)_2 \times 2^{-3}$.
  The binary fraction $(1.1101101)_2$ is approximately $1.8515625$ in decimal.
  So, the value is $1.8515625 \times 2^{-3} = 1.8515625 / 8 \approx 0.2314$. This value is closest to $0.227$, which is $2.27 \times 10^{-1}$.

The Excess-3 code is a non-weighted BCD code obtained by adding $3$ to each decimal digit ($0-9$). A code is classified as self-complementing if the $9$'s complement of a decimal digit is equivalent to the $1$'s complement of its corresponding code word. For instance, the decimal digit $2$ is represented as $0101$ in Excess-3, and its $9$'s complement, which is $7$, is represented as $1010$. Since $1010$ is the bitwise inverse ($1$'s complement) of $0101$, the code satisfies the self-complementing property. Therefore, the Excess-3 code is specifically known as a self-complementing code.

To find the 2's complement of a binary number, invert all bits and add $1$ to the result.
For option C ($1000$):
First, invert the bits: $1000 \rightarrow 0111$.
Second, add $1$: $0111 + 1 = 1000$.
Since the result $1000$ is identical to the original value, it is its own 2's complement.

To convert the hexadecimal number $(12A7C)_{16}$ to octal, we first convert it to its decimal equivalent:
$(12A7C)_{16} = (1 \times 16^4) + (2 \times 16^3) + (10 \times 16^2) + (7 \times 16^1) + (12 \times 16^0)$
$= 65536 + 8192 + 2560 + 112 + 12 = 76412_{10}$
Next, we convert the decimal value $76412_{10}$ to octal by dividing by $8$ repeatedly:
$76412 \div 8 = 9551$ remainder $4$
$9551 \div 8 = 1193$ remainder $7$
$1193 \div 8 = 149$ remainder $1$
$149 \div 8 = 18$ remainder $5$
$18 \div 8 = 2$ remainder $2$
$2 \div 8 = 0$ remainder $2$
Reading the remainders from bottom to top gives $(225174)_8$.

For a 16-bit number, there are $2^{16}$ possible bit patterns.
In 2's complement representation (X), every bit pattern maps to a unique integer. Therefore, the number of distinct integers is X = $2^{16}$.
In sign-magnitude representation (Y), there are two representations for zero: +0 (a sign bit of 0 followed by all 0s) and -0 (a sign bit of 1 followed by all 0s). Since both represent the same value, the number of distinct integers is one less than the total number of patterns. Thus, Y = $2^{16} - 1$.
The difference is X - Y = $2^{16} - (2^{16} - 1) = 1$.

The given 16-bit number is 1111 1111 1111 0101. Since the most significant bit is 1, the number is negative in 2's complement representation. To find its magnitude, we take the 2's complement of the number.

1. Find the 1's complement by inverting all bits: 0000 0000 0000 1010.
2. Add 1 to the 1's complement: 0000 0000 0000 1011.

The resulting binary number, 1011, represents the magnitude. Converting to decimal: $8+2+1=11$. Since the original number was negative, the decimal value is -11.

In an $n$-bit 2's complement system, there are $2^n$ total distinct binary combinations available.
The most significant bit acts as the sign bit, determining whether the number is negative or non-negative.
The smallest (most negative) value is represented by the bit pattern $100\dots0$, which equals $-2^{n-1}$.
The largest positive value is represented by the bit pattern $011\dots1$, which equals $2^{n-1}-1$.
Therefore, the range of representable integers is $-2^{n-1}$ to $2^{n-1}-1$.

The IEEE-754 representation consists of sign bit $S=0$, exponent $E=10000000_2 = 128$, and mantissa $M=110..._2$.
The bias for 32-bit floating point is $127$, so the actual exponent is $E - 127 = 128 - 127 = 1$.
The mantissa includes an implicit leading $1$, so the significand is $1.110_2$.
The decimal value is calculated as $(-1)^S \times (1.M) \times 2^{\text{exponent}} = 1 \times (1.11_2) \times 2^1$.
Multiplying by $2^1$ shifts the binary point one place to the right: $11.1_2$.
This converts to decimal as $1 \times 2^1 + 1 \times 2^0 + 1 \times 2^{-1} = 2 + 1 + 0.5 = 3.5$.

ASCII (American Standard Code for Information Interchange) is a character encoding standard that uses $7$ bits to represent each character, allowing for $2^7 = 128$ unique symbols. In contrast, BCD (Binary Coded Decimal) typically uses $4$ bits for numeric representation, while EBCDIC (Extended Binary Coded Decimal Interchange Code) uses $8$ bits to encode a single character. Gray code is a binary numeral system where successive values differ by only one bit, rather than a character encoding standard. Therefore, ASCII is the specific system defined by a $7$-bit structure.

The given equation is $(43)_x = (y3)_8$. To solve for the number of possible integer solutions for x and y, we first convert both sides to base 10.
$4 \cdot x^1 + 3 \cdot x^0 = y \cdot 8^1 + 3 \cdot 8^0$
$4x + 3 = 8y + 3$
$4x = 8y \implies x = 2y$

Next, we consider the constraints on the bases and digits:

1. For $(43)_x$, the base $x$ must be greater than any digit, so $x > 4$.
2. For $(y3)_8$, the digit $y$ must be less than the base 8. Since it's a leading digit, $y \neq 0$. So, $1 \le y \le 7$.

Combining $x=2y$ with the constraints:

• Since $1 \le y \le 7$, the possible values for $y$ are $\{1, 2, 3, 4, 5, 6, 7\}$.
• The corresponding values for $x$ are $\{2, 4, 6, 8, 10, 12, 14\}$.
• Applying the constraint $x > 4$, the valid values for $x$ are $\{6, 8, 10, 12, 14\}$.
  This gives 5 possible solutions.