Baby Graham's Number

You don't need Euler's theorem for this one

The last 2 digits of ${ 3 }^{ 3 }$ is 27. Taking the ones digit is the only thing necessary for calculating the last 2 digits of ${ 3 }^{ { 3 }^{ 3 } }$ , which the answer is $87$ . So, from then on, since the last digit is $7$ , again, no matter how many layers this thing goes to, its last 2 digits would always be $87$

Therefore, the answer is $\boxed { 87 }$

Let $a_{1}=3$ and $a_{i+1}=3^{a_{i}}$

Now, $a_{i}$ is odd for all $i$ .

So, $a_{i}=3^{a_{i-1}}≡(-1)^{a_{i-1}}=-1≡3(mod 4)$

Hence, $a_{i}$ is of the form $4k+3$ .

Note that, $3^{4}≡1(mod 10)$

Hence, $3^{4k+3}≡3^{3}≡7(mod 10)$

Hence,the last digit of all $a_{i}$ is 7.

Note that, $a_{3}=3^{27}≡(3^{9})^{3}≡83^{3}≡87(mod 100)$

I claim that $a_{i}≡87(mod 100)$ for all $i≥3$

I will prove my claim by induction.

Suppose the result is true for some $i$

First note that $\phi(100)=40$ .Hence $3^{40}≡1(mod 100)$

Now, $a_{i+1}=3^{a_{i}}=3^{100k+87}=3^{100k.}3^{87}=(3^{40})^{2k}.3^{20k}.3^{87}≡1.3^{20k}.87(mod 100)$ .

If k =2m, then $a_{i+1}≡87(mod 100)$ and we are done.

If k=2m+1, then $a_{i+1}≡87.3^{40m}.3^{20}≡87.(3^{10})^{2}≡87.(49)^{2}≡87.1(mod 100)$ (since $3^{10}≡49(mod 100)$ )

evaluate the 3^3...

= 27

So...

the last 2 digits is 27

So,

3.^(3^(3^(...^3^(27)))

last two digits is 87

So, until the last 3, 87 will be the last 2 digits...