Math problem from book Contact: tens of billions binary = ? base 10

This came up last year, and I want to rephrase the question based on my understanding:

Going back over the text of the book, where they’re about to decrypt the message, the key is that it is being transmitted over and over as a number that is “tens of billions of bits long” that is the product of three prime numbers. I inferred from that that that is in binary, so. . .

With that, how long in base 10 would a number that is tens of billions of binary digits be?

(I figure with a number that big, there would be quite a few products of three primes that would provide a solution, so I’m not asking for that. :smiley: )

About ten billion decimal digits.

Assuming tens, means “a few” times ten". Where “a few” means 2 or three If “tens” means 33.2, the answer is exactly 10 billion.

Thanks. Not a particularly dramatic reduction in digits, so it would seem.

Should be about 3 billion-ish digits for 10 billion decimal digits. Wolfram alpha isn’t letting me get quite as high as 2^(10 000 000 000), but at 2^billion, it gives me a number of length around 300 milliion, and at 2^(100 000 000), it gives me a length around 30 million.

Basically, you could divide the number of bits by 3.3 and get a reasonable estimate of the length of your decimal number.

No, the three primes would be unique.

I got the same answer as others. Here’s how:

Say N is the number being transmitted.
log[sub]10[/sub] N would be the number of decimal digits of N, and log[sub]2[/sub] N would be the number of binary digits.

Then log[sub]2[/sub] N = log[sub]10[/sub] N / log[sub]10[/sub] 2, so the number of decimal digits = the number of binary digits times log[sub]10[/sub] 2, which is about 0.301.

Quick and dirty, 2[sup]10[/sup] ~= 10[sup]3[/sup].