7 bits goes up to 2^7 - 1 = 127. Since 127 is itself prime, 127 is the largest prime that can be stored in 7 bits. I don't think there's a formula to generalize this for different numbers of bits, so you'd need to use a table of primes.
2006-10-22 21:53:19
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answer #1
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answered by DavidK93 7
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As often happens, the real answer is "it depends". 7 bits can be used to store non-negative integers, in which caseas others have said it is the largest prime number < 2^7, or 127.
If the 7 bits store positive or negative integers in one's complement, then it is the largest prime < 63, or 61.
While for 7 bits this would be a little odd, if you asked for a larger number it could be storing a real number, with mantissa and exponent.
Some older computers worked in decimal, where 4 bits gave a single decimal digit. So there are lots of possibilities, depending on how the bits are interpreted. (in a program I could use 7 bits to index the primes, in which case the largest prime number stored would be 719, I think , the 128th prime). But you probably want 127.
2006-10-23 00:33:03
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answer #2
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answered by sofarsogood 5
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I have no idea of calculating prime number in bit pattern. But the largest prime number that can be stored in a 7-bit binary pattern is 127. It will be written as: 1111111 in binary.
2006-10-22 22:03:41
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answer #3
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answered by mkm 2
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1111111, i.e. 127 base ten is prime.
A related question is to ask which numbers of the binary form 111111....1111 are prime. These numbers are called Mersenne primes.
2006-10-22 21:54:52
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answer #4
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answered by bag o' hot air 2
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1111111 i.e, 127
2006-10-22 22:25:22
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answer #5
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answered by . 3
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