Binary digits 10101000 to decimal number?

Answer 1

101110102 = BA16

Its the same as any other base conversion. Iteratively divide the number by the desired base, using the rules of arithmetic for the original base, until the result is zero, and then record the remainders in reverse order. In the case of conversion from base 2 to base 16, you are dividing by 100002, which is the same as shifting to the right by four until you have nothing left.

101110102 divided by 100002 = 10112 remainder 10102

10112 divided by 100002 = 02 remainder 10112

Convert 10102 to A16 and convert 10112 to B16, and reverse write as BA16.

You can also do this by inspection, by simply grouping the original binary number, 10111010, into groups of four bits, 1011, 1010, and writing down the conversion, B, A.

Answer 2

N = 1001 0010 0100 0000 1101 1000 1101 1110 1110 1100 1001 0100 1000 0001 1110 0101

If N is already in binary and you want to convert it to our normal base-10 system, then the value would be:

10,538,661,580,010,455,525

If N is already in base-10 and you want to convert it to binary, then the value would be:

0010 0110 1110 1110 1100 1010 1000 1110 0110 1101 1111 1100 0011 0100 1111 1110 1101 1101 1111 0101 1110 1000 1000 1101 1110 0100 0011 1001 1011 1001 0101 0001 0010 1010 0100 0111 0000 1100 0010 0101 0110 1010 1010 1011 1011 1101 1000 0101 1001 1100 0000 1100 0101

Answer 3

Converting from binary to hexadecimal is fairly simple, because the base from which we're converting (2) is a factor of the base to which we're converting (16). The easiest way to do it then, is to group the digits into sets of four (because 24 = 16), and convert them individually:

10111010 => {1011}{1010} => {B}{A}

So 101110102 = BA16

Answer 4

To convert binary to hex, starting from the low-order bit (working right-to-left), split the binary digits into groups of four-bit nybbles (half-a-byte) and then translate each nybble to its hexadecimal equivalent. This works because there are only 16 possible nybbles and we can represent each with a single hex digit:

0000 = 0

0001 = 1

0010 = 2

0011 = 3

0100 = 4

0101 = 5

0110 = 6

0111 = 7

1000 = 8

1001 = 9

1010 = a

1011 = b

1100 = c

1101 = d

1110 = e

1111 = f

Thus, given the binary value 01100011110, we find there are 3 four-bit nybbles:

011 0001 1110

in this case, the high-order nybble has only 3 bits, but we always assume leading zeroes for any missing bits.

Translating these three nybbles to hexadecimal we get:

0011 = 3 0001 = 1

1110 = e

Thus 01100011110 binary is 31e hexadecimal, which is often written 0x31e.

Converting to octal is similar except we use groups of 3 bits with the following conversions:

000 = 0

001 = 1

010 = 2

011 = 3

100 = 4

101 = 5

110 = 6

111 = 7

Thus 01100011110 splits into 01 100 011 110 with the following translation:

001 = 1

100 = 4

011 = 3

110 = 6

Thus 01100011110 binary is 1436 octal which is often written 01436.

Answer 5

311 The number 100,110,111 in binary code would be: 101111101111000111100011111

Answer 6

The binary number 10101000 = 168

Answer 7

BA5DBA11

Answer 8

Ans

Answer 9

UK