255. since number of 1's in "1111 1111" is 8 . we calculate decimal number as 2^8 = 256. starting from 0 to 255 .Hence the last(256th) number is 255.(sheetal)

General Purpose AnswerIn decimal numbers, the digits have values that are powers of ten. Starting at the implied decimal point and working to the left, the number 403 is equal to:3 x 100 + 0 x 101 + 4 x 102 = 3x1 + 0x10 + 4x100 = 3 + 0 + 400

In binary, the digits have values that are powers of two. Working to the left as before, the number 101 is equal to:

1 x 20 + 0 x 21 + 1 x 22 = 1x1 + 0x2+ 1x4 = (in decimal) 1 + 0 + 4 = 5

So, 1111 1111 = 1x1 + 1x2 + 1x4 + 1x8 + 1x16 + 1x32 + 1x64 + 1x128 = 255.

This also works to the right of the decimal point, but the powers of two are negative (i.e., fractions), so 101.011 in binary is:

1x1/8 + 1x1/4 + 0x1/2 + 1x1 + 0x2 + 1x4 = 5.375

Where the first digit to the right of the decimal point is 2-1 (one-half, or 1/2), the next is 2-2 (1/4), then 2-3 (1/8), etc.

Not sure I understand the logic behind the top answer. In this particular case, adding 1 to the original binary you get 1 0000 0000 which is 256, so take away the 1 again to get 255.

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01111 in binary is 15 in decimal.

All I know is that when a number is negative, you convert the decimal into binary and if it is negative you put 1111 before the binary digits.

111111 in binary is 255 in decimal which is FF in hexadecimal (i.e. 15 units and 15 16s)

111111112 = 128 + 64 + 32 + 16 + 8 + 4 + 2 + 1 = 255 101010102 = 128 + 32 + 8 + 2 = 170

In binary: 1111 1111 1111 1111 1111 1111 1111 1111 In octal: 37777777777 In hexadecimal: FFFFFFFF in decimal: 2³² - 1 = 4,294,967,295

The binary number 1111 is 15. The digits in a binary number are exponents of 2 rather than 10, so that for a four digit number in binary, the digit places represent 8, 4, 2, 1 instead of increasing values of 10. 1111 = 8+4+2+1 = 15

1111 can't be used for Binary Coded Decimal (BCD) because 1111=15 which is made of 2 digits 1 and 5. In BCD a 4-digit binary number is used for every decimal digit. ex. 1111 is incorrect 1 = 0001 5 = 0101 Answer: 0001 0101

1111 1111 1111 1111 = 2^16 = 65536

111110 = 10 001 010 1112

1111 converted from binary (base 2) to decimal (base 10) is 15 When you expand the steps... 1111 binary = (1 X 2^3) + (1 X 2^2) + (1 X 2^1) + (1 X 2^0) = 8 + 4 + 2 + 1 = 15

The answer is 1 0101 0111 1110 1011 1011 0011 1111 1010 0001 0111

In binary the largest number (using IEEE binary16) representable would be: 0111 1111 1111 1111 (grouping the bits in nybbles* for easier reading). This is split as |0|111 11|11 1111 1111| which represents: 0 = sign 111 11 = exponent 11 1111 1111 = mantissa. Using IEEE style, the exponent is offset by 011 11, making the maximum exponent 100 00 This is scientific notation using binary instead of decimal. As such there must be a non-zero digit before the binary point, but in binary this can only ever be a 1, so to save storage it is not stored and the mantissa effectively has an extra bit, which for the 10 bits specified makes it 11 bits long. Thus the mantissa represents: 1.11 1111 1111 This gives the largest number as: 1.1111 1111 11 × 10^10000 (all digits are binary, not decimal.) This expands to 1 1111 1111 1100 0000 (binary) = 0x1ffc0 = 131,008 Note that this is NOT accurate in storage - there are 6 bits which are forced to be zero, making the number only accurate to ±32 (decimal): the second largest possible real would be 1 1111 1111 1000 000 = 0x1ff80 = 130,944 - the numbers are only accurate to about 4 decimal digits; the largest decimal real number would be 1.310 × 10^5, and the next 1.309 × 10^5 and so on. However, with proper IEEE, an exponent with all bits set is used to identify special numbers, which makes the largest possible 0111 1101 1111 1111 which is 1.1111 1111 11 × 10^1111 = 1111 1111 1110 0000 = 0xffe0 = 65504 accurate to ±16, ie the largest is about 6.55 × 10^4. * a nybble is half a byte which is directly representable as a single hexadecimal digit.

The 1's complement is formed by inverting every binary digit (bit) of the number - if it is a 0 it becomes a 1, otherwise it is a 1 and becomes a 0. If 10 is in base 2, then its 1's compliment is 01 or just 1. If 10 is in base 10, then in binary it is 1010 and its 1's complement is 0101 = 5 in decimal. However, if more bits are being used to store it, there would be leading 0s that get inverted to 1s and so the resultant number is different; examples: 8 bits (a byte): decimal 10 = 0000 1010 → 1111 0101 = 245 in decimal 16 bits: decimal 10 = 0000 0000 0000 1010 → 1111 1111 1111 0101 = 65525 Next, if 2s complement is being used to represent negative numbers, the binary 1111 0101 represents decimal -11; similarly 1111 1111 1111 0101 represents decimal -11.

0xc = 1100 Hexadecimal digits use exactly 4 binary digits (bits). The 0x0 to 0xf of hexadecimal map to 0000 to 1111 of binary. Thinking of the hexadecimal digits as decimal numbers, ie 0x0 to 0x9 are 0 to 9 and 0xa to 0xf are 10 to 15, helps with the conversion to binary: 0xc is 12 decimal which is 8 + 4 → 1100 in [4 bit] binary.

9: 1001 10: 1010 11: 1011 12: 1100 13: 1101 14: 1110 15: 1111 16: 10000

0000 0000 1111 1000F ( or 15) = 1111 in binary, and 8 = 1000 in binary, so F is 1111 1000

15 written as a binary number is 1111.

16 is the 4th power of 2. So a hexadecimal number is converted to binary by replacing each hex digit by the 4-bit binary number having the same value. Conversely, in converting binary to hexadecimal, we group every 4 bits starting at the decimal (binary?) point and replace it with the equivalent hex digit. For example, the hexadecimal number 3F9 in binary is 1111111001, because 3 in binary is 11, F (decimal 15) is 1111, and 9 is 1001.

15 = 1111 14 = 1110 13 = 1101

BCD uses binary digits as if they were decimal digits. There are two BCD numbers per one 8-bit byte. For example, 23 decimal would be coded as 0010 0011, while in binary (not BDC), it would be 0001 0111. Similarly, 255 in BCD is 0000 0010 0101 0101, while in binary, it would b simply 1111 1111. As you can see, BCD is not as compact as binary, but it does have mathematical and software properties that makes it desirable (some games use BDC to store a "score", for example, because it takes fewer CPU cycles to display a BCD than a binary number).

The way I convert between decimal and hexadecimal is to first convert the decimal number to binary: 664062510 = 110010101010011111100012 Then split the binary number into 16-bit (4 digit) chunks: 0110 0101 0101 0011 1111 00012 Next, convert each chunk into a hexadecimal digit: 0110 0101 0101 0011 1111 00012 6 5 5 3 F 1 Finally, put all the digits together: 664062510 = 6553F116

Examples of binary numbers are...1010 = 101111 = 1510001 = 1710101 = 2111001 = 25

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