Forming largest two-digit number

Question

Form the largest two-digit number using two one-digit integers when repeating of digits is allowed.

I am not sure whether the answer is 99 or $9^9$. Is base considered a one-digit integer? How about the exponent?

Answer 1

I think answer is 99 (becouse it's largest two-digit number)

Answer 2

A number is just a number, not some digits with operators between them. If we would allow that, I could simply define a function $f$ that multiplies any $x$ with 3,000,000,000,000 (you get the idea), so $f(x)=3,000,000,000,000\cdot x$. Now $f(99)$ uses only two one-digit integers, but it is way bigger than your $9^9$.

Answer 3

Though $387420489$ is indeed a large number, with other properties using only $2$ integers, larger numbers are possible. Let us have the following problem:
n^^n
This is equal to $(n^n)^n$. Not only that, but we can extend this power tower as much as we like. And we still have $2$ integers.
Now, Hyper-E notation. We can use E9#9, which is ENORMOUS: 10^10^10^10^10^10^10^10^1000000000
But, debate is here: is it allowed? Since in the simplification, it has 26 integers. But, simplification doesn't count. The only thing that does is the actual problem.

So, that's your question answered. We can get numbers larger than googolplex, (10^10^100) googolduplex/googolplexian, (10^10^10^100) and even Graham's Number, with a large enough power tower! But, if boundaries exist such that we can only use 1 mathematical symbol/operator, where exponents don't count, $(9^9)!$ would probably be the right answer. But remember: getting to these numbers requires pushing the boundaries and thinking outside the box.