I'd like to know how to find the first 3 digits and the number of digits of the prime number 2^1257787-1. I think I'm supposed to use logarithms.

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- October 1st 2007, 04:53 PMfrankdent1Finding digits of a prime
I'd like to know how to find the first 3 digits and the number of digits of the prime number 2^1257787-1. I think I'm supposed to use logarithms.

- October 1st 2007, 05:11 PMgalactus
To find the number of digits, use

There are 378632 digits in the number, not subtracting the 1.

For the first few digits of the huge number, use

The first 3 digits are 412. - October 1st 2007, 06:04 PMfrankdent1How does the log of the base times the exponent equal the number of digits for b^n -1 ?

- October 2nd 2007, 06:47 PMfrankdent1bump
Ok, I understand there are 378632 digits, but why can we find the first few digits by using 10^.6151562...

- October 3rd 2007, 12:48 AMCaptainBlack
First the leading 3 digits of are the same as those of , so we need only consider the latter.

Now: .

Now the in the exponent above just tells you how far the decimal point in the answer is to the left of the most significant digit, and the fractional part tells you what the digits are.

RonL