Re: Binary number digits <- > Decimal number digits
- From: "mensanator@xxxxxxxxxxx" <mensanator@xxxxxxx>
- Date: Sun, 9 Dec 2007 09:28:13 -0800 (PST)
On Dec 9, 10:50�am, Jeremy Boden <jer...@xxxxxxxxxxxxxxx> wrote:
On Sun, 09 Dec 2007 16:58:11 +0100, Helmut Richter wrote:
On Sun, 9 Dec 2007, fc wrote:
For example, what calculus was made to arrive that a decimal number of
617 digits must have 2048 digits in binary? (Obviously, without make
the conversion of the decimal number 999999...99999 [617 9 numbers] to
binary) And: how to generalize for any decimal number of any length?
The number of digits of x in some base b is, not regarding rounding to
integers, the logarithm of x with respect to base b, which is �log x /
log b (log to arbitrary base).
So the number of decimal digits of x is �log 2 / log 10 = 0.30103 .
For a rough calculation, 2^10 = 1024,
So 10 bits is approximately 3 digits.
How come you didn't round up?
3.0102999566398119521373889472449 rounds to 4
giving you the correct answer: 10 bits makes
4 decimal digits.
--
Jeremy Boden
"64 bits good, 32 bits bad"
.
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