If you mean 10 + 6 that's 16 which is 10000 in binary
5
5 bits are 5 binary digits. If they represent a decimal number, then that number can be anything from zero to 31, and can have either 1 or 2 digits.
The number of bits required to represent an integer number depends on the size of the integer, there is no absolute answer. Generally speaking, a binary number made up of some number of bits can be considered to directly represent an integer number according to the rules of binary arithmetic. If you specify the question a little differently you can come up with a question that does have an absolute answer. For example, if you ask what range of integers can be represented by a binary number with N bits, the answer is exactly 0 to 2^N (ignoring alternate ways to characterize binary numbers such as signed two's complement). More specifically, an 8 bit binary number can represent the integers from 0 to 255. There is also another way to restate your original question. That would be to assume that you are asking how many bits (at a minimum) does it take to represent a particular integer. In that case, again assuming unsigned binary arithmetic, the answer would be found by determining the next higher power of two than the number in question and finding log base 2 of that number. For example, the minimum number of bits required to represent the integer 60 would be 6, where the next higher power of two than 60 is 64, and the log base 2 (simply the power to which 2 must be raised to arrive at the number) is 6. In theory, an infinite number of bits could represent an infinity of integers, but that is probably not really what you are asking.
With 5 bits, you can represent (2^5) different numbers, which equals 32. This includes numbers ranging from 0 to 31 in unsigned binary representation. If using signed binary representation (like two's complement), the range would be from -16 to 15, still allowing for 32 distinct values.
My initial guess is as follows: binary code will render 2 possible digits, either 0 or 1. If there are 32 digits with two possible outcomes per digit then you will find the answer by calculating 2 to the 32nd power. The answer I got was 4294967296 seconds. Edit: The maximum number which can be represented by a binary string is always 2^n where n is the number of bits in the string.
Binary bits are necessary to represent 748 different numbers in the sense that binary bits are represented in digital wave form. Binary bits also have an exponent of one.
23 can be represented in binary as 10111 and would therefore require 5 bits to represent.
10 bits would be required. 10 bits long (10 digits long) can represent up to 1024.
5
45 in binary is 101101, so you need at least 6 bits to represent 45 characters.
Count them: 643(10)=1010000011(2)
32
18 in binary is 10010 Since 18 can't be written in term of 2 to the power x, the number of bits needed is 5. The answer is 5
When you convert this decimal number to the binary format, we have 111001001 that has 9 digits so 9bits is required to represent it in normal case. To convert decimals to binary visit http://acc6.its.brooklyn.cuny.edu/~gurwitz/core5/nav2tool.html
011000110110000101110100 is cat in Binary. That is 23 Bits, or just under 3 Bytes.
5 bits are 5 binary digits. If they represent a decimal number, then that number can be anything from zero to 31, and can have either 1 or 2 digits.
8