Boolean Algebra

To represent 39 in 2's complement: first convert 39 to binary (100111), then take 1's complement (011000), add 1 (011001). For 8-bit representation, this is 00111001. However, since 39 is positive, its 2's complement form is simply its binary representation 00111001, which when written without leading zeros is 11011001.

Floating point notation normalizes a binary number by expressing it as (mantissa) × 2^(exponent). For (1101.111)₂, normalize by moving the decimal point 4 places left: 0.1101111 × 2⁴, where the mantissa is 0.1101111 and exponent is 4.

In any number system with radix r, after exhausting all single digits (0 to V, where V is the (r-1)th digit), the next number is written as 10 (meaning 1×r¹ + 0×r⁰). In base-32, 'V' is the 31st digit (index 30), so 'V' is followed by (10)₃₂.

To convert binary (1110101)₂ to decimal, sum powers of 2: 1×2⁶ + 1×2⁵ + 1×2⁴ + 0×2³ + 1×2² + 0×2¹ + 1×2⁰ = 64 + 32 + 16 + 0 + 4 + 0 + 1 = 117.

In hexadecimal, adding 1 to FFEF gives FFF0. Subtracting 1 from FFEF gives FFEE because the last digit F (value 15) becomes E (value 14).

1101 binary equals 13 decimal. 101 binary equals 5 decimal. Their sum 18 in decimal converts to 10010 in binary.

128 divided by 32 gives quotient 4 with remainder 0. In positional notation, this becomes (40)32 where 4 is in the 32s place and 0 is in the units place.

1010 is a forbidden code in the above series.

To convert 64 decimal to octal: 64 / 8 = 8 remainder 0; 8 / 8 = 1 remainder 0; 1 / 8 = 0 remainder 1. Reading remainders bottom-up gives 100 in octal.

The number 11 in binary is 00000011. The 1's complement is obtained by flipping all bits, resulting in 11111100.

The 9's complement of a decimal number is found by subtracting each digit from 9. 9-2=7, 9-9=0, 9-0=9, 9-3=6. Result is 7096.

101 in 8-bit is 00000101. 1's complement is 11111010. Adding 1 gives 11111011.

15 in decimal is 8+4+2+1, which is 1111 in binary.

128 in decimal is 80 in hexadecimal (8 * 16^1 + 0 * 16^0 = 128).

The number immediately preceding 777 in octal is 776 (counting down: 777 → 776). The number immediately before 1111 in binary is 1110 (binary counting: ..., 1101, 1110, 1111, 10000, ...). Note that (777)₂ would be invalid since binary digits are only 0 and 1.

In 8421 BCD, 10000 is 16 and 1001 is 9. However, the question asks for the decimal equivalent of 10000.1001. If interpreted as 80.9, it implies a different coding or typo in the question's premise.

To express FFF.FF in floating point notation (normalized), we shift the decimal point 3 places to the left, resulting in .FFF FF multiplied by 16^3.

Binary equivalent of 839 is 1101000111 and that of the decimal part is 10010001....
Thereafter, normalising it with K = 3, we get (.83957 x 10³).