Number Systems - 2

On subtracting 381 from 999, we get 618

(715)₈ = (461)₁₀
10's complement
461 - 999= 538 + 1 = 539.

In a sign magnitude representation, an additional bit of sign representation is attached with the number. Here 0 means +.

A 0
6 B

10 B

Excess-3 code is a self-complementing code because the 1's complement of any excess-3 encoded number yields the excess-3 representation of the 9's complement of the original decimal number. This property makes it useful for arithmetic operations.

Only option two gives 12A7C.

Excess-3 code is unweighted because there are no fixed weights assigned to bit positions. It is derived by adding 3 (0011 in binary) to each BCD digit, but this doesn't create positional weights like 8421 or 2421 codes do.

In sign magnitude representation, an additional left most bit of sign representation will be attached with the decimal number, here 9 means ‘_‘. Hence 9687 is the right answer.

In a sign magnitude representation an additional left most bit of sign representation will be attached with the number, here 1 mean '-'.

To find out the one’s complement of a binary number, we will simply replace the 1’s with 0’s and vice versa.
Hence, one’s complement of 1101101 will be 0010010.

Calculate: 15 * 256 = 3840, 5 * 16 = 80, plus 3 = 3923. Binary: 3923 = 11110101011 (11 bits). Counting 1's: 1+1+1+1+0+1+0+1+0+1+1 = 8 ones.

Calculate: 3 * 512 = 1536, 7 * 64 = 448, 5 * 8 = 40, plus 3 = 2027. Binary: 2027 = 11111101011. Counting 1's: 1+1+1+1+1+1+0+1+0+1+1 = 9 ones.

¹37
19

50

ASCII (American Standard Code for Information Interchange) uses 7 bits to represent characters, allowing for 128 unique characters. EBCDIC uses 8 bits (256 characters), BCD uses 4 bits per digit, and Gray code is a binary encoding system.

Left side: (12x)_3 = 1 * 3^2 + 2 * 3^1 + x * 3^0 = 9 + 6 + x = 15 + x. Right side: (123)_x = 1 * x^2 + 2 * x^1 + 3 * x^0 = x^2 + 2x + 3. Solving: 15 + x = x^2 + 2x + 3, so x^2 + x - 12 = 0, giving x = 3 or x = -4. Since base must be > 3 (digit 3 present), x = 3.

(123)_5 = 1×25 + 2×5 + 3 = 38 in decimal. For (x3)_y = 38, we need x×y + 3 = 38, so x×y = 35. Factor pairs of 35 with x < y (required for valid digit in base y) are (1,35) and (5,7), giving x = 1 or 5. The base y must exceed the highest digit (3), so y = 35 or 7 are both valid.

We will subtract in the same way as decimal, F = 15, D = 13
FD

• 88

75

BCD (Binary Coded Decimal) uses 4 bits to represent each decimal digit (0-9), making it an 8-4-2-1 weighted code. However, finding complements in BCD is NOT efficient compared to pure binary systems because BCD doesn't leverage binary arithmetic properties naturally.