At least 8 bits are needed to represent the number 231.
231 ÷ 2 = 115 r 1
115 ÷ 2 = 57 r 1
57 ÷ 2 = 28 r 1
28 ÷ 2 = 14 r 0
14 ÷ 2 = 7 r 0
7 ÷ 2 = 3 r 1
3 ÷ 2 = 1 r 1
1 ÷ 2 = 0 r 1
→ 231 is 1110 0111 in binary and has 8 binary digits. Thus 231 can be represented in 8 bits, but if more are provided, eg 16, it can still be represented (in 16 bits it would be 0000 0000 1110 0111, unless there is a binary point, with say 8 bits after it, then 231 would be 1110 0111 . 0000 0000).
The binary number 10000000 represents the decimal 128
8
The decimal equivalent of the binary number 1111111111111111 is 65535.
If 1101 is a decimal number, then its binary equivalent is 10001001101. If 1101 is a binary number, then its decimal equivalent is 13.
Binary ( 1 0 ) = decimal ( 2 )
5
The binary number 10000000 represents the decimal 128
231
The binary equivalent for the decimal number 23 is 10111
10111
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.
10 digits.
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For the simple reason that the 'places' in a binary number represent multiples of 2, rather than 10. Therefore, a binary '10' = a decimal '2.'
If that's binary, it's 57 base 10
It takes 7 digits.
The decimal number 101 is represented by the binary number 1100101.