Math Team Problem Involving Powers of Powers of 3

Question

So I am in my high school math team and I was given the following expression

$$3^{3^{3^{...}}}$$

Where there are multiple powers of 3 with a total of two thousand and fifteen 3's. The question asks for the last two digits of the expression. How exactly would I evaluate this expression?

Answer 1

You can use Euler's totient theorem. Since $\phi(100)=40$ you need to evaluate the expression with 2014 3's mod 40.

Since $\phi(40)=16$ you need to evaluate the expression with 2013 3's mod 16.

Then you need to evaluate the expression with 2012 3's mod 8, which requires that you evaluate the expression with 2011 3's mod 4.

But since the expression with 2010 3's is odd, and $3^{odd number} \equiv (-1)^{odd number} = 3$, so the expression with 2011 3's mod 4 is 3.

Then for the 2012 3's mod 8, you do $3^3 mod 8$ which is 3, then you do $3^3 mod 16$ which is 11 then you do $3^{11} mod 40$ which is a bit more difficult but turns out to be 27, and then finally you need to do $3^{27}$ mod 100 which I am too lazy to figure out right now.

Answer 2

Your number can be written as $3\uparrow\uparrow2015$, read as "3 tetrated by 3" meaning "3 to the power of itself 2015 times". It is an example of a hyperoperation (addition, multiplication, exponentiation, titration...) and is represented here with Knuth's up-arrow notation. Your question asks for only the last two digits of the number, so thankfully we don't have to calculate the entire thing.

Your number is a power tower, $3^{3^{3^{...}}}$ or $3\uparrow\uparrow n$ . Therefore its rightmost digits must satisfy certain properties common to all such towers. One of these properties is that all such towers of height greater than $d$ (say), have the same sequence of $d$ rightmost decimal digits. This property is used to calculate the digits of Graham's number, another power tower of the form $3\uparrow\uparrow n$, with a very, very large $n$.

You can see a chart of the rightmost digits of power towers of $3$ for certain heights on the page about Graham's number, but to answer your question, $3\uparrow\uparrow2015 = \ ...87$.