How many digits?

[kokakola]

If I'm working on an LL test of N = 2^48,000,000 - 1, for example, is there any way for me to find how many digits N has?

Thanks!

[Mini-Geek]

(with p being the exponent, the p in 2^p-1) The exact number is int(log_10(2^p)+1) = int(log_10(2)*p+1), which is roughly 0.3*p. Your example has exactly 14449440 (14.4 million) digits.
http://mersenne-aries.sili.net/digits.php calculates log_10(2^N), or the reverse (enter the digits to find the bits).
int(n) means the integer part of n. (e.g. int(4.8)=int(4.2)=int(4)=4)
log_10(2) is the base 10 logarithm of 2. (i.e. 10^(log_10(2))=2; it's about 0.3)

[kokakola]

Originally Posted by Mini-Geek

(with p being the exponent, the p in 2^p-1) The exact number is int(log_10(2^p)+1) = int(log_10(2)*p+1), which is roughly 0.3*p. Your example has exactly 14449440 (14.4 million) digits.
http://mersenne-aries.sili.net/digits.php calculates log_10(2^N), or the reverse (enter the digits to find the bits).
int(n) means the integer part of n. (e.g. int(4.8)=int(4.2)=int(4)=4)
log_10(2) is the base 10 logarithm of 2. (i.e. 10^(log_10(2))=2; it's about 0.3)

Thank you! Your formula is exactly what I was looking for.

[R.D. Silverman]

Originally Posted by kokakola

If I'm working on an LL test of N = 2^48,000,000 - 1, for example, is there any way for me to find how many digits N has?

Thanks!

How much math have you had? Normally, one sees logarithms in
the 2nd or (maybe) 3rd year of high school algebra.

[CRGreathouse]

Originally Posted by R.D. Silverman

How much math have you had? Normally, one sees logarithms in
the 2nd or (maybe) 3rd year of high school algebra.

High schools offer three years of algebra?

But I agree, logs are a high school topic.

[Mini-Geek]

Originally Posted by CRGreathouse

But I agree, logs are a high school topic.

True, but being able to connect the dots to realize that the exact formula for the number of digits in a number of the form 2^p-1 is int(log_10(2)*p+1) is not exactly what you learn when learning basic logarithms. Basic logarithmic theory would suggest that it's about log_10(2^p)=log_10(2)*p, but to recognize the problem as related to logarithms and rederive the exact formula would be a bit more difficult than you all seem to imply. And this is all assuming the OP has even learned logarithms.

[CRGreathouse]

Originally Posted by Mini-Geek

True, but being able to connect the dots to realize that the exact formula for the number of digits in a number of the form 2^p-1 is int(log_10(2)*p+1) is not exactly what you learn when learning basic logarithms. Basic logarithmic theory would suggest that it's about log_10(2^p)=log_10(2)*p, but to recognize the problem as related to logarithms and rederive the exact formula would be a bit more difficult than you all seem to imply. And this is all assuming the OP has even learned logarithms.

I have no argument with any of that, and would add the possibility that the OP learned it and forgot it.

[kokakola]

Originally Posted by R.D. Silverman

How much math have you had? Normally, one sees logarithms in
the 2nd or (maybe) 3rd year of high school algebra.

I have taken a number of college level math courses. As others have pointed out, I just didn't make the connection.

[joblack]

There is somewhere in the forum a Windows program which calculates the exact value ... search for it

[CRGreathouse]

Originally Posted by joblack

There is somewhere in the forum a Windows program which calculates the exact value ... search for it

That strikes me as overkill, since you can just as easily find it using Google as a calculator.

[R.D. Silverman]

Originally Posted by Mini-Geek

True, but being able to connect the dots to realize that the exact formula for the number of digits in a number of the form 2^p-1 is int(log_10(2)*p+1) is not exactly what you learn when learning basic logarithms.

Of course it is! Change of base is fundamental. I saw all of this,
IN CLASS in the 8th grade. i.e. the last year BEFORE high school.

Solving 2^x = 10^z is fundamental! It is totally trivial.