I mentioned in a throwway bit in the article on Goldbach’s Odd Conjecture being (apparently) proven that the number had been a bound in the conjecture. That is, it was proven in 1939 that numbers larger than that had to obey the conjecture, but that it was unproven for numbers smaller than that. I described it as a number that tekes something like seven million digits to write out in full, that is, in a decimal expansion rather than some powers-of-powers sort of thing.

So let me give it a little attention as a puzzle for people who want to pass a little time doing arithmetic. Am I right to say that would be a number with about seven million digits?

The obvious way to check is to see what Google comes up with if you put 3^(3^(15)), although that turns out to be Bible quotes. Its calculator gives back Infinity, which here just means “it’s a really, really big number”. My Mac’s calculator function *and* my copy of Octave agree on that. It’s possible to find a better calculator that gives a meaningful answer, but you can work out roughly how big the number is just by hand, and for that matter, without resorting to anything you have to look up. I promise.

### Like this:

Like Loading...

*Related*

## Author: Joseph Nebus

I was born 198 years to the day after Johnny Appleseed. The differences between us do not end there. He/him.
View all posts by Joseph Nebus

3^(ln10/ln3) = 10, so 3^(x * ln10/ln3) will have floor(x) + 1 digits to the left of the decimal point, for x >= 0.

Solving x * ln10/ln3 = 3^15 gives x = 3^15 * ln3/ln10 = 6,846,168.5117.

So 3^(3^15) will have 6,846,169 digits.

LikeLike

Exactly so.

For bonus points, can you work out the estimate without knowing the log of 3?

LikeLike