Number Systems - 1

The decimal number 16 in binary is 10000, which is 5 binary digits. This is calculated by repeatedly dividing by 2: 16 = 8 remainder 0, 8 = 4 remainder 0, 4 = 2 remainder 0, 2 = 1 remainder 0, 1 = 0 remainder 1, reading remainders backwards gives 10000.

With 8 bits, you can represent 2^8 = 256 different numbers (from 0 to 255). Each bit can be 0 or 1, so with 8 bits you have 2^8 possible combinations.

The decimal number 9 in binary is 1001, which has 4 binary digits. Converting: 9 = 4 remainder 1, 4 = 2 remainder 0, 2 = 1 remainder 0, 1 = 0 remainder 1, reading remainders backwards gives 1001.

2^-48 equals 1 divided by 2^48, which requires 48 binary fraction digits to represent. In binary, this is 0.000...001 with 47 zeros followed by a 1 at the 48th position after the decimal point.

The number of numbers which can be represented with a 6- bit binary word will be 2⁶ = 64 and the numbers will range from 0 - 63. The maximum count which a 6-bit binary word can represent is 63.

In binary representation, the least significant bit determines parity. Odd numbers always have LSB=1 because they can be written as 2k+1, while even numbers have LSB=0 because they equal 2k.

Converting hex to binary: each hex digit becomes 4 binary digits. B=11 decimal=1011 binary, 3=3 decimal=0011 binary. Combining: 10110011. Options B and D are invalid binary (contain digits >1).

 (1110000)² = 2⁴ +2⁵ + 2⁶ = 112.
Hence, option 2 is correct.

Divide 511 by 16 repeatedly: 511 = 3116 + 15 (F), 31 = 116 + 15 (F), 1 = 0*16 + 1. Reading remainders bottom to top gives 1FF. Also note 511 = 2^9 - 1 = 512 - 1 = 0x1FF.

In BCD, each decimal digit becomes 4 binary digits. For 221: 2 becomes 0010, 2 becomes 0010, 1 becomes 0001. Concatenating gives 001000100001.

The number of numbers which can be represented with a 6-digit binary number will be 2⁶ = 64.

Group binary digits into sets of 4 from the right: 1010 (10 decimal = A hex) and 1111 (15 decimal = F hex). This gives AF in hexadecimal.

The number of numbers which can be represented by a 10-digit binary number will be 2¹⁰ = 1024 and the numbers range from 0 - 1023. The maximum count which a 10-bit binary word can represent is 1023.

In a self-complementing weighted code, the 9's complement is obtained by inverting all bits. For this property to hold in a 4-bit BCD code, the sum of the weights must be 9. This ensures that the weighted sum of inverted bits equals 9 minus the original digit.

The ASCII value for the character '5' is 53 in decimal, which converts to 35 in hexadecimal (3*16 + 5 = 53). Option A (5) is the character itself, B (25) is 37 decimal, and D (65) is the ASCII code for 'A'.

Roman numerals are not positional - they use additive notation where symbol values are fixed regardless of position (VII always means 5+1+1=7). Binary, decimal, and hexadecimal are all positional systems where digit value depends on place value (units, sixteens, etc.).

ASCII is a 7-bit character encoding with exactly 128 characters (2^7 = 128), including 32 control characters (0-31) for functions like tab and newline, plus 96 printable characters. EBCDIC is an incompatible 8-bit encoding with 256 characters.

A 12-bit binary number has 2^12 = 4096 possible values, so its resolution or accuracy is 1/4096. This means it can represent decimal fractions with a precision of approximately 0.000244. Option A (1/1024) would be for 10-bit, B (1/2048) for 11-bit.

Taking the (r-1)'s complement twice returns the original number - it's an involution operation. For example, in binary (r=2, so r-1=1), the 1's complement of the 1's complement of 1010 (0101 → 1010) gives back 1010. This is a fundamental property of complement operations.

Converting decimal 324.987 to octal: 324/8 = 40 remainder 4, 40/8 = 5 remainder 0 → 504 for integer part. For fractional 0.987: 0.987×8 = 7.896, 0.896×8 = 7.168, 0.168×8 = 1.344 → .771. So 324.987 decimal = 504.771 octal.