M45 and M46

Two new Mersenne Primes have been in the news recently. (Mersenne Primes are prime numbers of the form 2^n-1.) Finally, finally, after their primality (primeness?) was independently verified, they were revealed to the rest of the world:

M_{45}=2^{37,156,667}-1 and M_{46}=2^{43,112,609}-1.

There was a lot of speculation about the number of digits that these numbers would have. Not least for the fact that the first person to find a Mersenne prime with more than 10,000,000 digits would win $100,000. And indeed, the newly discovered primes have 11,185,272 and 12,978,189 digits respectively.

To put that in perspective, The Math Less Traveled shows that if you write out the number of atoms in the universe, that number would have a paltry 80 digits.

Of course, I think: how can I use this in my own classes? We don’t really talk about primes in Algebra II, Calculus, or MV Calculus. However, we do talk about logarithms in Algebra II.

Check it out.

How do you think we know how many digits are in M_{45} and M_{46}? It’s a simple application of logarithms.

We know that that a number is written in the form 10^N, it has N+1 digits (if N is an integer). Think about it: 10^1 has 2 digits; 10^3 has 4 digits; 10^5 has 6 digits.

If N isn’t an integer, it’s just a hop skip and jump away to saying that the number of digits is the next higher integer. So if we have 10^{3.1}, we have 4 digits.

Where do logarithms come into play?

Well, 2^{37,156,667}-1 has a certain number of digits. Since that 1 probably won’t affect anything since it’s such a huge number, we will ignore it. How many digits does 2^{37,156,667} have? Let’s use what we just discovered:

2^{37,156,667}=10^N

Then solving for N, we get N=37,156,667\log(2) \approx 11,185,271.306. Hence, we know there are 11,185,272 digits.

And a good question for the really ambitious student is to ask: can we be sure we can ignore that 1? (Answer: yes.)

(You can find the number of digits for M_{46} in the same process.)

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