Question

\[\text{ Let } n \text{ be a positive integer with } n \ge 3. \text{ Show } 1989|(n^{n^{n^{n}}}-n^{n^{n}}) .\]

Answer 1

does
\[1989|(n^{n^{n^{n^n}}}-n^{n^{n^n}}) .\] too?

Answer 2

If this helps anyone this comes from http://www.math.muni.cz/~bulik/vyuka/pen-20070711.pdf . A10.

Answer 3

lol I don't know about that
I don't know how to do this one honestly

Answer 4

me neither

Answer 5

@dan815 @rational @kainui @ganeshie8 you guys have any thoughts totally blank right now

Answer 6

wow I'm actually disappointed http://www.wolframalpha.com/input/?i=%285%5E%285%5E%285%5E5%29%29-5%5E%285%5E5%29%29%2F%281989%29+is+integer

Answer 7

disappointed because I thought we were going to get to prove something

Answer 8

Are you sure it's not just such a huge number that it broke wolfram alpha? http://www.wolframalpha.com/input/?i=%285%5E%285%5E%285%5E5%29%29-5%5E%285%5E5%29%29

Answer 9

i think there are different interpretations of \(n^{n^n}\)

Answer 10

oh i guess it is possible to break wolfram :p

Answer 11

@satellite73 I don't think so, since the alternative is \(n^{n^n}= n^{2n}\) which seems kinda like a waste.

Answer 12

http://www.wolframalpha.com/input/?i=5%5E5%5E5%5E5+mod+1989
http://www.wolframalpha.com/input/?i=5%5E5%5E5+mod+1989

Answer 13

http://www.wolframalpha.com/input/?i=%283^3%29^3%2C3^%283^3%29

Answer 14

n^n^n w/o parenthesis is a tetration i think

Answer 15

5^5^5 = 5^(5^5)

Answer 16

@satellite73 here are some useful exponentiation rules you should memorize
\[(a^n)^m=a^{nm}\\
a^{n}*a^{m} = a^{n+m}\]

Answer 17

Pretty sure @satellite73 already memorized these rules lol.

Answer 18

how about starting here

n^n^n...n-n^n^n...n = 0 mod k
k times and k-1 times

Answer 19

ohh hhow about!